Irodov Solutions: The Fundamental Equation of Dynamics | Physics Class 11 - NEET PDF Download

Ans. Let R be the constant upward thurst on the aerostat of mass m, coming down with a constant acceleration w. Applying Newton’s second law of motion for the aerostat in projection form

F_{y =} mw_y

mg - R - mw      (1)

Now, if Am be the mass, to be dumped, then using the Eq. F_{y =} mw_y

  (2)

From Eqs. (1) and (2), w e get,   

Ans. Let us write the fundamental equation of dynamics for all the three blocks in terms of projections, having taken the positive direction of x and y axes as shown in Fig; and using the fact that kinematical relation between the accelerations is such that the blocks move with same value of acceleration (say w)

(1)
(2)
and          (3)

The simultaneous solution of Eqs. (1), (2) and (3) yields,

As the block m₀ moves down with acceleration w, so in vector form

Ans. Let us indicate the positive direction of Jt-axis along the incline (Fig.). Figures show the force diagram for the blocks.
Let, R be the force of interaction between the bars and they are obviously sliding down with the same constant acceleration w.

Newton’s second law of motion in projection form along x-axis for the blocks gives :

    (1)

     (2)

Solving Eqs. (1) and (2) simultaneously, we get

(b) when the blocks just slide down the plane, w = 0 , so from Eqn. (3)

Hence  

Ans.  Case 1. When the body is launched up:

Let k be the coefficient of friction, u the velocity of projection and l the distance traversed along the incline. Retarding force on the block = mg sin α + k mg cos α and hence the retardation = g sin α + k g cos α .
Using the equation of particle kinematics along the incline,

0 - u²-2 (g sin α + k g cos α) l

or,  (1)
and

or,         (2)

   (3)

Case (2). When the block comes downward, the net force on the body - mg sin a - km g cos a and hence its acceleration - g sin a - k g cos a Let, t be the time required then

       (4)

From Eqs. (3) and (4) 

   (according to the question),

Hence on solving we get

Ans. At the initial moment, obviously the tension in the thread connecting m₁ and m₂ equals the weight of m₂.

(a) For the block m₂ to come down or the block m₁ to go up, the conditions is

where T is tension and f_r is friction which in the limiting case equals km₁g cosα. Then 

or  

or  

(b) Similarly in the case

or,  

(c) For this case, neither kind o f motion is possible, and fr need not be limiting.

Hence,  

Ans. From the conditions, obtained in the previous problem, first we will check whether the mass m₂ goes up or down.

Here, m₂/ m₁ = η > sin α + k cos α , (substituting the values). Hence the mass m₂ will come down with an acceleration (say w). From the free body diagram of previous problem,

      (1)

and        (2)

Adding (1) and (2), we get, 

Substituting all the values,  

As m₂ moves down with acceleration of magnitude w = .05 g > 0, thus in vector form acceleration of m₂:

Ans. Let us write the Newton’s second law in projection form along positive x-axis for the plank and the bar

fr - m₁ w₁, fr = m₂ w₂            (1)

At the initial moment, fr represents the static friction, and as the force F grows so does the friction force fr, but up to it’s limiting value 

Unless this value is reached, both bodies moves as a single body with equal acceleration. But as soon as the force fr reaches the limit, the bar starts sliding over the plank i.e.  

Substituting here the values of w₁ and w₂ taken from Eq. (1) and taking into account that 

   corresponds to the moment  

Hence,  

If  

Ans.  Apply F_x = m w_x for body A

or,  

Now, from kinematical equation :

or,  

(using Eq. (1)).

i.e.  

or,  

and putting the values of α , k and l in Eq . (2) we get

Ans. Let us fix the x - y co-ordinate system to the wedge, taking the x - axis up, along the incline and the y - axis perpendicular to it (Fig.).

Let us apply Newton’s second law in projection form along x and y axis for the bar :

   (3)

For T_min the value of (cos β + k sin β ) should be maximum 

So,

Putting this value of β in Eq. (3) we get,

Ans. First of all let us draw the free body diagram for the small body of mass m and indicate x - axis along the horizontal plane and y - axis, perpendicular to it, as shown in the figure.
Let the block breaks off the plane at

So,  

or,        (1)

From  for the body under investigation :

 Integrating within the limits for v (t)

So,     (2)

Integrating, Eqn. (2) for s (t)

       (3)

Using the value of t = t₀ from Eq. (1), into Eqs. (2) and (3)

Ans. Newton’s second law of motion in projection form, along horizontal or x - axis i.e. 

Integrating, over the limits for v (s)

which is the sought relationship.

Ans. From the Newton’s second law in projection from :

For the bar,

T - 2 kwg - (2 m) w      (1)

For the motor,

    (2)

Now, from the equation of kinematics inthe frame of bar or motor

    (3)

From (1), (2) and (3) wc get on eliminating T and W

Ans. Let us write Newton’s second law in vector from  for both the blocks (in the frame ofjround).

These two equations contain three unknown quantities  The third equation is provided by the kinematic relationship between the accelerations :

where   is th acceleration of the mass m₁ with respect to the pulley or elevator car.
Summing up termwise the lelt hand and the right-hand sides of these kinematical equations, we get

   (4)

The simultaneous solution of Eqs* (1), (2) and (4) yield s

Using this result in Eq. (3) , we get,

Using the results in Eq. (3) we get  

(b) obviously the force exerted by the pulley on the celing of the car

Note : one could also solve this problem in the frame of elevator car.

Ans. Let us write Newton’s second law for both, bar 1 and body 2 in terms of projection having taken the positive direction of x₁ and x₂ as shown in the figure and assuming that body 2 starts sliding, say, upward along the incline 

For the pulley, moving in vertical direction from the equation  

(as mass of the pulley m_p = 0 ) 

or   (3)

As the length of the threads are constant, the kinematical relation ship of accelerations becomes 

      (4)

Simultaneous solutions of all these equations yields :

As η > 1, w is directed vertically downward, and hence in vector form 

Ans. Let us write Newton’s second law for masses m₁ and m₂ and moving pully in vertical direction along positive x - axis (Fig.) : 

or       (3)

Again using Newton’s second law in projection form for mass m₀ along positive x₁ direction

(Fig.), we get 

   (4)

The kinematical relationship between the accelerations of masses gives in terms of projection on the x - axis

Simultaneous solution of the obtained five equations yields 

In vector form

Ans. As the thread is not tied with m, so if there were no friction between the thread and the ball m, the tension in the thread would be zero and as a result both bodies will have free fall motion. Obviously in the given problem it is the friction force exerted by the ball on the thread, which becomes the tension in the thread. From the condition or language of the problem w_M > w_m and as both are directed downward so , relative acceleration of M = w_M > w_m and is directed downward. Kinematical equation for the ball in the frame of rod in projection form along upward direction gives :

Newton’s second law in projection form along vertically down direction for both, rod and ball gives,

Multiplying Eq. (2) by m and Eq. (3) by M and then subtracting Eq. (3) from (2) and after using Eq. (1) we get

Ans. Suppose, the ball goes up with accleration w₁ and the rod comes down with the acceleration w₂. As the length of the thread is constant,

2 = w₂    (1)

From Newton's second law in projection form along vertically upward for the ball and vertically downward for the rod respectively gives,

    (2)

and       (3)

but T = 2T   (because pulley is massless)   (4)

From Eqs. (1), (2), (3) and (4)

From kinematical equation in projection form, we get 

as, w₁ and w₂ are in the opposite direction.
Putting the values of w₁ and w₂, the sought time becomes

Ans. Using Newton's second law in projection form along x - axis for the body 1 and along negative x - axis for the body 2 respectively, we get 

For the pulley lowering in downward direction from Newton's law along x axis,

As the length of the thread is constant so,

w₂ =  2w₁          (4)

The simultaneous solution of above equations yields,

    (5)

Obviously during the time interval in which the body 1 comes to the horizontal floor covering the distance h, the body 2 moves upward the distance 2h. At the moment when the body 2 is at the height 2h from the floor its velocity is given by the expression :

After the body m₁ touches the floor the thread becomes slack or the tension in the thread zero, thus as a result body 2 is only under gravity for it's subsequent motion.

Owing to the velocity v₂ at that moment or at the height 2h from the floor, the body 2 further goes up under gravity by the distance,

Thus the sought maximum height attained by the body 2 :

Ans. Let us draw free body diagram of each body, i.e. of rod A and of wedge B and also draw the kinemetical diagram for accelerations, after analysing the directions of motion of A and B. Kinematical relationship of accelarations is :

    (1)

Let us write Newton’s second law for both bodies in terms of projections having taken positive directions of y and x axes as shown in the figure.

    (2)

and     (3)

Simultaneous solution of (1), (2) and (3) yields : 

Note : We may also solve this problem using conservation of mechanical energy instead of Newton's second law.

Ans. Let us draw free body diagram of each body and fix the coordinate system, as shown in the figure. After analysing the motion of M and m on the basis of force diagrams, let us draw the kinematical diagram for accelerations (Fig.).
As the length of threads are constant so,

 do not change their directions that why

so, from the triangle law of vector addition

      (1)

From the Eq. Fx = mw_x , for the wedge and block :

T - N = Mw,        (2)

and    N = mw    (3)

Now, from the Eq. F_y = mw_y , for the block 

mg - T - kN= mw     (4)

Simultaneous solution of Eqs. (2), (3) and (4) yields :     

Hence using Eq. (1)

Ans. Bodies 1 and 2 will remain at rest with repect to bar  the sought minimum acceleration of the bar. Beyond these limits there will be a relative motion betw een b ar and the bodies.  the tendency o f body 1 in relation to the bar A is to move towards right and is in the opposite sense for  On the basis of above argument the static friction on 2 by A is directed upward and on 1 by A is directed towards le ft for the purpose o f calculating .

Let us write Newton’s second law for bodies 1 and 2 in terms of projection along positive x - axis (Fig.).

As body 2 has no acceleration in vertical direction, so

From (4) and (5)

Hence

Ans. On the basis of the initial aigument of the solution of Q.79, the tendency of bar 2 with respect to 1 will be to move up along the plane.
Let us fix (x - y) coordinate system in the frame o f ground as shown in the figure.
From second law of motion in projection form along y and x axes :

m g cos α - N = m w sin a

So, the sought maximum acceleration of the wedge :

Ans. Let us draw the force diagram of each body, and on this basis we observe that the prism moves towards right say with an acceleration  nd the bar 2 of mass moves down the plane with respect to 1, say with acceleration   (Fig.)

Let us write Newton’s second law for both bodies in projection form along positive y₂ and x₁ axes as shown in the Fig.

or,    (1)
and      (2)

Solving (1) and (2), we get

Ans. To analyse the kinematic relations between the bodies, sketch the force diagram of each body as shown in the figure.

On the basis of force diagram, it is obvious that the wedge M will move towards right and the block will move down along the wedge. As the length of the thread is constant, the distance travelled by the block on the wedge must be equal to the distance travelled by the wedge on the flo&r. Hence   do not change their directions and acceleration that’s why  (say) and accordingly the diagram of kinematical dependence is shown in figure.

 so from triangle law o f vector addition.

  (1)

From F_x = m w_x , (for the wedge), 

   (2)

For the bar m let us fix (x - y) coordinate system in the frame of ground Newton ’s law in projection form along x and y axes (Fig.) gives

(3)
        (4)

Solving the above Eqs. simultaneously, we get

Note : We can study the motion of the block m in the frame of wedge also, alternately we may solve this problem using conservation of mechanical energy. 

Ans. Let us sketch the diagram for the motion of the particle of mass m along the circle of radius R and indicate x and y axis, as shown Jn the figure.

(a) For the particle, change in momentum  

so ,  

and time taken in describing quarter of the circle, 

Ans. While moving in a loop, normal reaction exerted by the flyer on the loop at different points and uncom pensated weight if any contribute to the weight of flyer at those points,

(a) When the aircraft is at the lowermost point, Newton’s second law of motion in projection form F_n = mw_n gives

(b) When it is at the upper most point, again from F_n - mw_n we get

(c) When the aircraft is at the middle point of the loop, again from  Fn = m w_n 

The uncompensated weight is mg. Thus effective weight    obliquely.

Ans. Let us depict the forces acting on the small sphere m, (at an arbitrary position when the thread makes an angle θ from the vertical) and write equation  via projection on A 
the unit vectors

(as vertical is reference line of angular position)

or  

Integrating both the sides : 

or,    

(Eq. (1) can be easily obtained by the conservation of mechanical energy).
From

(b) Vertical component of velocity, v_y = v sin θ

So,  

For maximum  

which yields  

Therefore from  

(c) We have  

But in accordance with the problem w_y = 0 

So,    

or,    

or,    

Ans. The ball has only normal acceleration at the lowest position and only tangential acceleration at any of the extreme position. Let v be the speed of the ball at its lowest position and l be the length of the thread, then according to the problem

where α is the maximum deflection angle

From Newton’s law in projection form : F_t = mw_t

or,  

On integrating both the sides within their limits.

   or,

Note : Eq. (2) can easily be obtained by the conservation of mechanical energy of the ball in the uniform field of gravity.
From Eqs. (1) and (2) with θ = α

or,  

Ans. Let us depict the forces acting on the body A(which are the force of gravity  and the normal reaction  and write equation   via projection on the unit vectors  

or,  

Integrating both side for obtaining v (θ)

At the mom ent the body loses contact with the surface, N = 0 and therefore the Eq. (2) becomes

v² = gR cos θ

where v and θ correspond to the moment when the body loses contact with the surface. Solving Eqs. (1) and (3) we obtain

Ans. At first draw the free body diagram of the device as, shown. The forces, acting on the sleeve are it's weight, acting vertically downward, spring force, along the length of the spring and normal reaction by the rod, perpendicular to its length.
Let F be the spring force, and Δl be the elongation.

(1)

    (2)

and it is independent of the direction of rotation.

Ans.  According to the question, the cyclist moves along the circular path and the centripetal force is provided by the frictional force. Thus from die equation  F_n = m w_n

or,     (1)

Ans.  As initial velocity is zero thus

v² = 2 w_ts

As w_t > 0 the speed of the car increases with time or distance. Till the moment, sliding starts, the static friction provides the required centripetal acceleration to the car.

Thus  

So,  

or,  

Hence  

so, from Eqn. (1), the sought distance  

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 F_n = mw 

so        (1)

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

    (2)

For the given curve,

Substituting this value in (2) we get,

For the minimum  

and therefore, corresponding radius of curvature

     (3)

Hence from (1) and (2)

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, F_x = mw_x we get

Then putting  

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, F_n = mw_n for this element,

or,            (1)

Also,        (2)

(b) When which is greater than η₀  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 : 

     (4)and        (5)

Also        (6)

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

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.

So v_x = v₀ cos α

Using, F_n = m w_n, for the particle of mass m,

which is the required normal force.

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

Differentiating twice with respect the time :

   (1)

Thus  

Ans. 

     (1)

(b) Using the solution of problem 1.28 (b), the total time of motion

Hence using

Ans. From the equation o f the g iven time dependence force   the force vanishes,

(a) Thus  

or,   but  

(b) Again from the equation   or,  Integrating w ithin the limits for  

or,  

Thus  

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

Ans. We have 

or  

On integrating,

When 

Thus    

Ans. According to the problem, the force acting on the particle of mass m cos ωt 

So,  

Integrating, within the limits. 

It is clear from equation (1), that after starting at t = 0, die particle comes to rest fro the first time at

From Eqn.  (1),       (2)

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

From Eq. (1)

Ans. (a) From the problem

Thus  

or,  

On integrating  

But at  

or,  

Thus for  

(b) We have  

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

or,  

Thus for  

Now, average velocity over this time interval,

Ans. According to the problem

Integrating, withing the limits,

To find the valufc of k, rewrite

On integrating

So,      (2)

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

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

we get  

or,  

or,

or,  

As the motion of the bar is unidirectional it stops after going through a distance of  

Hence, the maximum velocity will be at the distance, x = tan α/a Putting this value of x in (1) the maximum velocity,

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 body remains at rest and for t > t₀ obviously 

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

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

Integrating and solving, we get, 

    (1)

When the body falls downward, the net force acting on the body in downward direction equals

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

Thus,  

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

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,

      (1)and      (2)

(a) Using the condition     (3)

and  

    (4)

Hence,  

(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 ,

  Consequently, the sought distance, is

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

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

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

where   . The constant C is found from the initial condition v = v₀, whence  initially. Finally we obtain

In the cource of time  (Motion then is unaccelerated.)

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,   where λ is the linear mass density o f the chain Let  be the tension at the upper and the lower ends of ds. we have from,

or,  

If we sum the above equation for all elements, the term    because there is no tension at the free ends, so

Ans. In the problem, we require the velocity of the body, relative 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 

   (3)

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

therefore, or,  Solving Eqs. (2) and (3) we get,

Hence  

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

and,  

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,

This is opposite to the displacement   is an outward directed centrifugal force while  is the inward directed external force).

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,

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.

However  is stable because the force tends to bring the sleeve near the equilibrium position

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

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.

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,

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

The equation is,

so,  or,  

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

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,

where   is the resultant inertial force. (pseudo force) which is the vector sum of centrifugal and Coriolis forces.

where   the inward drawn unit vector to the centre from the point in question, here A.

Thus,  

so,  

(b) At B  

its magnitude is  

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

Thus the equation of motion are,

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

Integrating the last equation

Hence

So the body must fly off for   exactly as if the sphere were nonrotating.

Now, at this point  

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,

So,  

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

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

Thus  

(We write λ for the latitude here)

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

Ans. We go to the equation given in 1.111. Here v_y = 0 so we can take y = 0, thus we get for the motion in the  .

and    

Integrating,  

So

There is thus a displacement to the east of