The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

I. E. Irodov Solutions for Physics Class 11 & Class 12

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The document The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
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Q. 91. A car moves uniformly along a horizontal sine curve y = a sin (x/α), where a and α are certain constants. The coefficient of friction between the wheels and the road is equal to k. At what velocity will the car ride without sliding?

Ans. Since the car follows a curve, so the maximum velocity at which it can ride without sliding at the point of minimum radius of curvature is the sought velocity and obviously in this case the static friction between the car and the road is limiting.

Hence from the equation Fn = mw 

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

so    The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev    (1)

We know that, radius of curvature for a curve at any point (x, y) is given as,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev    (2)

For the given curve,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Substituting this value in (2) we get,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

For the minimum  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

and therefore, corresponding radius of curvature

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev     (3)

Hence from (1) and (2)

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 92. A chain of mass m forming a circle of radius R is slipped on a smooth round cone with half-angle θ. Find the tension of the chain if it rotates with a constant angular velocity co about a vertical axis coinciding with the symmetry axis of the cone. 

Ans. The sought tensile stress acts on each element of the chain. Hence divide the chain into small, similar elements so that each element may be assumed as a particle. We consider one such element of mass dm, which subtends angle d α at the centre. The chain moves along a circle of known radius R with a known angular speed ω and certain forces act on it We have to find one of these forces.

From Newton’s second law in projection form, Fx = mwx we get

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Then putting   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRevThe Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 93. A fixed pulley carries a weightless thread with masses m1 and m2  at its ends. There is friction between the thread and the pulley. It is such that the thread starts slipping when the ratio m2/m1 = η0. Find:
 (a) the friction coefficient;
 (b) the acceleration of the masses when m2/m1  = η > η0.

Ans. Let, us consider a small element of the thread and draw free body diagram for this element, (a) Applying Newton’s second law of motion in projection form, Fn = mwn for this element,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

or,    The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev        (1)

Also,    The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev    (2)

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

(b) When The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev Svhich is greater than    η the blocks will move with same value of acceleration, (say w) and clearly m2 moves downward. From Newton’s second law in projection form (downward for m2 and upward for m1) we get : 

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev     (4)
and   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev     (5)

Also   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev     (6)

Simultaneous solution of Eqs. (4), (5) and (6) yields : 

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 94. A particle of mass m moves along the internal smooth surface of a vertical cylinder of radius R. Find the force with which the particle acts on the cylinder wall if at the initial moment of time its velocity equals v0  and forms an angle α with the horizontal. 

Ans. The force with which the cylinder wall acts on the particle will provide centripetal force necessary for the motion of the particle> and since there is no acceleration acting in the horizontal direction, horizontal component of the velocity will remain constant througout the motion.

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

So vx = vcos α

Using, Fn = m wn, for the particle of mass m,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

which is the required normal force.


Q. 95. Find the magnitude and direction of the force acting on the particle of mass m during its motion in the plane xy according to the law x = a sin ωt, y = b cos ωt, where a, b, and ω are constants. 

Ans. Obviously the radius vector describing the position of the particle relative to the origin of coordinate is 

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Differentiating twice with respect the time :

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev   (1)

Thus  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 96. A body of mass m is thrown at an angle to the horizontal with the initial velocity v0. Assuming the air drag to be negligible, find:
 (a) the momentum increment Δp that the body acquires over the first t seconds of motion;
 (b) the modulus of the momentum increment Δp during the total time of motion. 

Ans.     The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev     (1)

(b) Using the solution of problem 1.28 (b), the total time of motion The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Hence using The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 97. At the moment t = 0 a stationary particle of mass in experiences a time-dependent force F = at (τ — t), where a is a constant vector, τ is the time during which the given force acts. Find:
 (a) the momentum of the particle when the action of the force discontinued;
 (b) the distance covered by the particle while the force acted. 

Ans. From the equation o f the g iven time dependence force  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev  the force vanishes,

(a) Thus  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

or,   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
but   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

(b) Again from the equation   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
or,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
Integrating w ithin the limits for   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
or,   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Thus   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Hence distance covered during the tim e interval t = τ, 

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 98. At the moment t = 0 a particle of mass m starts moving due to a force F = F0 sin ωt, where F0 and ω are constants. Find the distance covered by the particle as a function of t. Draw the approximate plot of this function. 

Ans. We have The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

or  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

On integrating,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

When 

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Thus    The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 99. At the moment t = 0 a particle of mass m starts moving due to a force F = F0 cos ωt, where F0 and ω are constants. How long will it be moving until it stops for the first time? What distance will it traverse during that time? What is the maximum velocity of the particle over this distance?

Ans. According to the problem, the force acting on the particle of mass m The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev  cos ω

So,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Integrating, within the limits. 

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

It is clear from equation (1), that after starting at t = 0, die particle comes to rest fro the first time at The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

From Eqn.  (1),   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev    (2)

Thus during the time interval t = π/ω, the sought distance 

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

From Eq. (1)

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 100. A motorboat of mass m moves along a lake with velocity v0. At the moment t = 0 the engine of the boat is shut down. Assuming the resistance of water to be proportional to the velocity of the boat F = —rv, find:
 (a) how long the motorboat moved with the shutdown engine;
 (b) the velocity of the motorboat as a function of the distance covered with the shutdown engine, as well as the total distance covered till the complete stop;
 (c) the mean velocity of the motorboat over the time interval (beginning with the moment t = 0), during which its velocity decreases η  times. 

Ans. (a) From the problem  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Thus   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

or,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

On integrating   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

But at  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

or,   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Thus for   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

(b) We have  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Integrating within the given limits to obtain v (s):

or,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Thus for  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Now, average velocity over this time interval,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 101.  Having gone through a plank of thickness h, a bullet changed its velocity from v0 to v. Find the time of motion of the bullet in the plank, assuming the resistance force to be proportional to the square of the velocity.

Ans. According to the problem

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Integrating, withing the limits,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

To find the valufc of k, rewrite

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

On integrating

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

So,   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev   (2)

Putting the value of k from (2) in (1), we get

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Q. 102. A small bar starts sliding down an inclined plane forming an angle α with the horizontal. The friction coefficient depends on the distance x covered as k = ax, where a is a constant. Find the distance covered by the bar till it stops, and its maximum velocity over this distance.

Ans. From Newton’s second law for the bar in projection from, Fx = m wx along x direction 

we get  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

or,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


or,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

or,   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
As the motion of the bar is unidirectional it stops after going through a distance of   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
Hence, the maximum velocity will be at the distance, x = tan α/a Putting this value of x in (1) the maximum velocity, 

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 103. A body of mass m rests on a horizontal plane with the friction coefficient lc. At the moment t = 0 a horizontal force is applied to it, which varies with time as F = at, where a is a constant vector. Find the distance traversed by the body during the first t seconds after the force action began. 

Ans. Since, the applied force is proportional to the time and the frictional force also exists, the motion does not start just after applying the force. The body starts its motion when F equals the limiting friction.

Let the motion start after time t0 , then

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev the body remains at rest and for t > t0 obviously 

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Q. 104. A body of mass m is thrown straight up with velocity vo. Find the velocity v' with which the body comes down if the air drag equals kv2, where k is a constant and v is the velocity of the body. 

Ans. While going upward, from Newton’s second law in vertical direction :

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

At the maximum height h, the speed v = 0, so

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Integrating and solving, we get, 

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev    (1)

When the body falls downward, the net force acting on the body in downward direction equals The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Hence net acceleration, in downward direction, according to second law of motion

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Thus,   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Integrating and putting the value of h from (1), we get, 

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 105. A particle of mass m moves in a certain plane P due to a force F whose magnitude is constant and whose vector rotates in that plane with a constant angular velocity ω. Assuming the particle to be stationary at the moment t = 0, find:
 (a) its velocity as a function of time;
 (b) the distance covered by the particle between two successive stops, and the mean velocity over this time. 

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Ans. Let us fix x - y co-ordinate system to the given plane, taking x-axis in the direction along which the force vector was oriented at the moment  t = 0 , then the fundamental equation of dynamics expressed via the projection on x and y-axes gives,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev      (1)
and  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev    (2)

(a) Using the condition  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev   (3)

and  

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev    (4)

Hence,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

(b) It is seen from this that the velocity v turns into zero after the time interval Δt, which can be found from the relation ,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev  Consequently, the sought distance, is 

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 106. A small disc A is placed on an inclined plane forming an angle a with the horizontal (Fig. 1.27) and is imparted an initial velocity v0. Find how the velocity of the disc depends on the angle y if the friction coefficient k = tan α and at the initial moment yo  = = π/2. 

Ans. The acceleration of the disc along the plane is determined by the projection of the force of gravity on this plane Fx = mg sin α and the friction force fr = kmg cos α . In our case k = tan α and therefore

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Let us find the projection of the acceleration on the derection of the tangent to the trajectory and on the x-axis :

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

It is seen fromthis that wt = - wx, which means that the velocity v and its projection vx differ only by a constant value C which does not change with time, i.e.

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

where  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev . The constant C is found from the initial condition v = v0, whence The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev initially. Finally we obtain

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

In the cource of time The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev (Motion then is unaccelerated.)


Q. 107. A chain of length l is placed on a smooth spherical surface of radius R with one of its ends fixed at the top of the sphere. What will be the acceleration w of each element of the chain when its upper end is released? It is assumed that the length of the chain The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Ans. Let us consider an element of length ds at an angle qp from the vertical diameter. As the speed of this element is zero at initial instant of time, it's centripetal acceleration is zero, and hence,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev  where λ is the linear mass density o f the chain Let  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRevbe the tension at the upper and the lower ends of ds. we have from, The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
or,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

If we sum the above equation for all elements, the term   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev because there is no tension at the free ends, so

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 108. A small body is placed on the top of a smooth sphere of radius R. Then the sphere is imparted a constant acceleration w0 in the horizontal direction and the body begins sliding down. Find:
 (a) the velocity of the body relative to the sphere at the moment of break-off;
 (b) the angle θ0 between the vertical and the radius vector drawn from the centre of the sphere to the break-off point; calculate θ0 for w0 = g.

Ans. In the problem, we require the velocity of the body, realtive to the sphere, which itself moves with an acceleration w0 in horizontal direction (say towards left). Hence it is advisible to solve the problem in the frame of sphere (non-inertial frame).
At an arbitary moment, when the body is at an angle θ with the vertical, we sketch the force diagram for the body and write the second law of motion in projection form 

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev   (3)

Note that the Eq. (3) can also be obtained by the work-energy theorem A = ΔT (in the frame of sphere)

therefore, The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
or,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
Solving Eqs. (2) and (3) we get,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Hence  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 109. A particle moves in a plane under the action of a force which is always perpendicular to the particle's velocity and depends on a distance to a certain point on the plane as 1/rn, where n is a constant. At what value of n will the motion of the particle along the circle be steady? 

Ans. This is not central force problem unless the path is a circle about the said point. Rather here Ft (tangential force) vanishes. Thus equation of motion becomes,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

and,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

We can consider the latter equation as the equilibrium under two forces. When the motion is perturbed, we write r = r0 + x and the net force acting on the particle is,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

This is opposite to the displacement  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev is an outward directed centrifiigul force while  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRevis thd inward directed external force).


Q. 110. A sleeve A can slide freely along a smooth rod bent in the shape of a half-circle of radius R (Fig. 1.28). The system is set in rotation with a constant angular velocity ω about a vertical axis OO'. Find the angle θ corresponding to the steady position of the sleeve. 

Ans. There are two forces on the sleeve, the weight F1 and the centrifugal force F2. We resolve both forces into tangential and normal component then the net downward tangential force on the sleeve is,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

is always positive for small values of 0 and hence the net tangential force near θ = 0  opposes any displacement away from it. θ = 0 is then stable.

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

However The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev is stable because the force tends to bring the sleeve near the equilibrium position The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

If ω2R = g, the two positions coincide and becomes a stable equilibrium point.


Q. 111. A rifle was aimed at the vertical line on the target located precisely in the northern direction, and then fired. Assuming the air drag to be negligible, find how much off the line, and in what direction, will the bullet hit the target. The shot was fired in the horizontal direction at the latitude φ = 60°, the bullet velocity v = 900 m/s, and the distance from the target equals s = 1.0 km. 

Ans. Define the axes as shown with z along the local vertical, x due east and y due north. (We assume we are in the northern hemisphere). Then the Coriolis force has the components.

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRevThe Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 112. A horizontal disc rotates with a constant angular velocity ω = 6.0 rad/s about a vertical axis passing through its centre. A small body of mass m = 0.50 kg moves along a diameter of the disc with a velocity v' = 50 cm/s which is constant relative to the disc. Find the force that the disc exerts on the body at the moment when it is located at the distance r = 30 cm from the rotation axis. 

Ans. The disc exerts three forces which are mutually perpendicular. They are the reaction of the weight, mgy vertically upward, the Coriolis force 2mv' ω perpendicular to the plane of the vertical and along the diameter, and mω2r outward along the diameter. The resultant force is,

  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 113. A horizontal smooth rod AB rotates with a constant angular velocity ω = 2.00 rad/s about a vertical axis passing through its end A. A freely sliding sleeve of mass m = 0.50 kg moves along the rod from the point A with the initial velocity v0  = 1.00 m/s. Find the Coriolis force acting on the sleeve (in the reference frame fixed to the rotating rod) at the moment when the sleeve is located at the distance r = 50 cm from the rotation axis. 

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Ans. The sleeve is free to slide along the rod AB. Thus only the centrifugal force acts on it.

The equation is,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

so,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
or,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

v0 being the initial velocity when r = 0. The Coriolis force is then,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 114. A horizontal disc of radius R rotates with a constant angular velocity ω about a stationary vertical axis passing through its edge. Along the circumference of the disc a particle of mass m moves with a velocity that is constant relative to the disc. At the moment when the particle is at the maximum distance from the rotation axis, the resultant of the inertial forces Fin acting on the particle in the reference frame fixed to the disc turns into zero. Find:
 (a) the acceleration ω' of the particle relative to the disc;
 (b) the dependence of Fin  on the distance from the rotation axis. 

Ans. The disc OBAC is rotating with angular velocity ω about the axis OO' passing through the edge point O. The equation of motion in rotating frame is,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

where  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev is the resultant inertial force. (pseudo force) which is the vector sum of centrifugal and Coriolis forces.

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

where  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev the inward drawn unit vector to the centre from the point in question, here A.

Thus,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

so,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

(b) At B  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

its magnitude is  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 115. A small body of mass m = 0.30 kg starts sliding down from the top of a smooth sphere of radius R = 1.00 m. The sphere rotates with a constant angular velocity ω = 6.0 rad/s about a vertical axis passing through its centre. Find the centrifugal force of inertia and the Coriolis force at the moment when the body breaks off the surface of the sphere in the reference frame fixed to the sphere.

Ans. The equation of motion in the rotating coordinate system is,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Thus the equation of motion are,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRevThe Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRevThe Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

A result that is easy to understant by considering the motion in non-rotating frame. The eliminating φ we get,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Integrating the last equation

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Hence

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

So the body must fly off for  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev exactly as if the sphere were nonrotating.

Now, at this point  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q.116. A train of mass m = 2000 tons moves in the latitude φ = 60° North. Find:
(a) the magnitude and direction of the lateral force that the train exerts on the rails if it moves along a meridian with a velocity v = 54 km per hour;
(b) in what direction and with what velocity the train should move for the resultant of the inertial forces acting on the train in the reference frame fixed to the Earth to be equal to zero.

Ans. (a) When the train is moving along a meridian only the Coriolis force has a lateral component and its magnitude (see the previous problem) is,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

So,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

= 3.77 kN, (we write λ for the latitude)

(b) The resultant of the inertial forces acting on the train is,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Thus  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

(We write λ for the latitude here)

Thus the train must move from the east to west along the 60th parallel with a speed,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 117. At the equator a stationary (relative to the Earth) body falls down from the height h = 500 m. Assuming the air drag to be negligible, find how much off the vertical, and in what direction, the body will deviate when it hits the ground. 

Ans. We go to the equation given in 1.111. Here vy = 0 so we can take y = 0, thus we get for the motion in the The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev .

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

and    The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Integrating,   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

So The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

    The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

There is thus a displacement to the east of

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 106. A small disc A is placed on an inclined plane forming an angle a with the horizontal (Fig. 1.27) and is imparted an initial velocity v0. Find how the velocity of the disc depends on the angle y if the friction coefficient k = tan α and at the initial moment yo  = = π/2. 

Ans. The acceleration of the disc along the plane is determined by the projection of the force of gravity on this plane Fx = mg sin α and the friction force fr = kmg cos α . In our case k = tan α and therefore

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Let us find the projection of the acceleration on the derection of the tangent to the trajectory and on the x-axis :

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

It is seen fromthis that wt = - wx, which means that the velocity v and its projection vx differ only by a constant value C which does not change with time, i.e.

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

where  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev . The constant C is found from the initial condition v = v0, whence The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev initially. Finally we obtain

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

In the cource of time The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev (Motion then is unaccelerated.)


Q. 107. A chain of length l is placed on a smooth spherical surface of radius R with one of its ends fixed at the top of the sphere. What will be the acceleration w of each element of the chain when its upper end is released? It is assumed that the length of the chain The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Ans. Let us consider an element of length ds at an angle qp from the vertical diameter. As the speed of this element is zero at initial instant of time, it's centripetal acceleration is zero, and hence,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev  where λ is the linear mass density o f the chain Let  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRevbe the tension at the upper and the lower ends of ds. we have from, The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
or,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

If we sum the above equation for all elements, the term   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev because there is no tension at the free ends, so

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 108. A small body is placed on the top of a smooth sphere of radius R. Then the sphere is imparted a constant acceleration w0 in the horizontal direction and the body begins sliding down. Find:
 (a) the velocity of the body relative to the sphere at the moment of break-off;
 (b) the angle θ0 between the vertical and the radius vector drawn from the centre of the sphere to the break-off point; calculate θ0 for w0 = g.

Ans. In the problem, we require the velocity of the body, realtive to the sphere, which itself moves with an acceleration w0 in horizontal direction (say towards left). Hence it is advisible to solve the problem in the frame of sphere (non-inertial frame).
At an arbitary moment, when the body is at an angle θ with the vertical, we sketch the force diagram for the body and write the second law of motion in projection form 

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev   (3)

Note that the Eq. (3) can also be obtained by the work-energy theorem A = ΔT (in the frame of sphere)

therefore, The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
or,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
Solving Eqs. (2) and (3) we get,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Hence  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 109. A particle moves in a plane under the action of a force which is always perpendicular to the particle's velocity and depends on a distance to a certain point on the plane as 1/rn, where n is a constant. At what value of n will the motion of the particle along the circle be steady? 

Ans. This is not central force problem unless the path is a circle about the said point. Rather here Ft (tangential force) vanishes. Thus equation of motion becomes,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

and,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

We can consider the latter equation as the equilibrium under two forces. When the motion is perturbed, we write r = r0 + x and the net force acting on the particle is,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

This is opposite to the displacement  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev is an outward directed centrifiigul force while  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRevis thd inward directed external force).


Q. 110. A sleeve A can slide freely along a smooth rod bent in the shape of a half-circle of radius R (Fig. 1.28). The system is set in rotation with a constant angular velocity ω about a vertical axis OO'. Find the angle θ corresponding to the steady position of the sleeve. 

Ans. There are two forces on the sleeve, the weight F1 and the centrifugal force F2. We resolve both forces into tangential and normal component then the net downward tangential force on the sleeve is,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

is always positive for small values of 0 and hence the net tangential force near θ = 0  opposes any displacement away from it. θ = 0 is then stable.

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

However The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev is stable because the force tends to bring the sleeve near the equilibrium position The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

If ω2R = g, the two positions coincide and becomes a stable equilibrium point.


Q. 111. A rifle was aimed at the vertical line on the target located precisely in the northern direction, and then fired. Assuming the air drag to be negligible, find how much off the line, and in what direction, will the bullet hit the target. The shot was fired in the horizontal direction at the latitude φ = 60°, the bullet velocity v = 900 m/s, and the distance from the target equals s = 1.0 km. 

Ans. Define the axes as shown with z along the local vertical, x due east and y due north. (We assume we are in the northern hemisphere). Then the Coriolis force has the components.

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRevThe Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 112. A horizontal disc rotates with a constant angular velocity ω = 6.0 rad/s about a vertical axis passing through its centre. A small body of mass m = 0.50 kg moves along a diameter of the disc with a velocity v' = 50 cm/s which is constant relative to the disc. Find the force that the disc exerts on the body at the moment when it is located at the distance r = 30 cm from the rotation axis. 

Ans. The disc exerts three forces which are mutually perpendicular. They are the reaction of the weight, mgy vertically upward, the Coriolis force 2mv' ω perpendicular to the plane of the vertical and along the diameter, and mω2r outward along the diameter. The resultant force is,

  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 113. A horizontal smooth rod AB rotates with a constant angular velocity ω = 2.00 rad/s about a vertical axis passing through its end A. A freely sliding sleeve of mass m = 0.50 kg moves along the rod from the point A with the initial velocity v0  = 1.00 m/s. Find the Coriolis force acting on the sleeve (in the reference frame fixed to the rotating rod) at the moment when the sleeve is located at the distance r = 50 cm from the rotation axis. 

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Ans. The sleeve is free to slide along the rod AB. Thus only the centrifugal force acts on it.

The equation is,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

so,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
or,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

v0 being the initial velocity when r = 0. The Coriolis force is then,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 114. A horizontal disc of radius R rotates with a constant angular velocity ω about a stationary vertical axis passing through its edge. Along the circumference of the disc a particle of mass m moves with a velocity that is constant relative to the disc. At the moment when the particle is at the maximum distance from the rotation axis, the resultant of the inertial forces Fin acting on the particle in the reference frame fixed to the disc turns into zero. Find:
 (a) the acceleration ω' of the particle relative to the disc;
 (b) the dependence of Fin  on the distance from the rotation axis. 

Ans. The disc OBAC is rotating with angular velocity ω about the axis OO' passing through the edge point O. The equation of motion in rotating frame is,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

where  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev is the resultant inertial force. (pseudo force) which is the vector sum of centrifugal and Coriolis forces.

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

where  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev the inward drawn unit vector to the centre from the point in question, here A.

Thus,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

so,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

(b) At B  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

its magnitude is  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 115. A small body of mass m = 0.30 kg starts sliding down from the top of a smooth sphere of radius R = 1.00 m. The sphere rotates with a constant angular velocity ω = 6.0 rad/s about a vertical axis passing through its centre. Find the centrifugal force of inertia and the Coriolis force at the moment when the body breaks off the surface of the sphere in the reference frame fixed to the sphere.

Ans. The equation of motion in the rotating coordinate system is,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Thus the equation of motion are,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRevThe Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRevThe Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

A result that is easy to understant by considering the motion in non-rotating frame. The eliminating φ we get,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Integrating the last equation

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Hence

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

So the body must fly off for  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev exactly as if the sphere were nonrotating.

Now, at this point  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 116. A train of mass m = 2000 tons moves in the latitude φ = 60° North. Find:
 (a) the magnitude and direction of the lateral force that the train exerts on the rails if it moves along a meridian with a velocity v = 54 km per hour;
 (b) in what direction and with what velocity the train should move for the resultant of the inertial forces acting on the train in the reference frame fixed to the Earth to be equal to zero.

Ans. (a) When the train is moving along a meridian only the Coriolis force has a lateral component and its magnitude (see the previous problem) is,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

So,  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

= 3.77 kN, (we write λ for the latitude)

(b) The resultant of the inertial forces acting on the train is,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Thus  The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

(We write λ for the latitude here)

Thus the train must move from the east to west along the 60th parallel with a speed,

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 117. At the equator a stationary (relative to the Earth) body falls down from the height h = 500 m. Assuming the air drag to be negligible, find how much off the vertical, and in what direction, the body will deviate when it hits the ground. 

Ans. We go to the equation given in 1.111. Here vy = 0 so we can take y = 0, thus we get for the motion in the The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev .

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

and    The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Integrating,   The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

So The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

    The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

There is thus a displacement to the east of

The Fundamental Equation Of Dynamics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

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