Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

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

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Q. 31. A ball starts falling with zero initial velocity on a smooth inclined plane forming an angle a with the horizontal. Having fallen the distance h, the ball rebounds elastically off the inclined plane. At what distance from the impact point will the -ball rebound for the second time? 

Ans. The ball strikes the inclined plane (Ox) at point O (origin) with velocity  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev As the ball elastically rebounds, it recalls with same velocity v0, at the same angle a from the normal or y axis (Fig.). Let the ball strikes the incline second time at P, which is at a distance l (say) from the point O, along the incline. From the equation Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

where τ is the time of motion of ball in air while moving from O to P.

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev  (2) 

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Now from me equation.

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

so, Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Hence the sought distance,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev


Q. 32. A cannon and a target are 5.10 km apart and located at the same level. How soon will the shell launched with the initial velocity 240 m/s reach the target in the absence of air drag? 

Ans. Total time of motion

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev   (1)

and horizontal range 

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev    (2)

From Eqs. (1) and (2) 

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

On simplifying Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Solving for τ2 wc get :

Thus

τ =  42.39 s = 0.71 min and
τ = 24-55 s = 0.41 min depending on the angle α.


Q. 33. A cannon fires successively two shells with velocity v0  = 250 m/s; the first at the angle θ1 =  60° and the second at the angle θ2  = 45° to the horizontal, the azimuth being the same. Neglecting the air drag, find the time interval between firings leading to the collision of the shells. 

Ans. Let the shells collide at the point P (x, y). If the first shell takes t s to collide with second and Δt be the time interval between the firings, then 

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev


Q. 34. A balloon starts rising from the surface of the Earth. The ascension rate is constant and equal to v0. Due to the wind the balloon gathers the horizontal velocity component vx  = ay, where a is a constant and y is the height of ascent. Find how the following quantities depend on the height of ascent:
 (a) the horizontal drift of the balloon x (y);
 (b) the total, tangential, and normal accelerations of the balloon. 

Ans. According to the problem

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Integrating  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev    (1)

And also we have  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

So,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

(b) According to the problem

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev     (2)

So, Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Therefore Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Diff. Eq. (2) with respect to time.

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

So,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Hence Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev


Q. 35. A particle moves in the plane xy with velocity v = ai + bxj, where i and j are the unit vectors of the x and y axes, and a and b are constants. At the initial moment of time the particle was located at the point x = y = 0. Find:
 (a) the equation of the particle's trajectory y (x);
 (b) the curvature radius of trajectory as a function of x. 

Ans. (a) The velocity vector of the particle

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

So,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev   (1)

From (1) Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev    (2)

And Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Integrating  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev     (3)

From Eqs. (2) and (3) , we get,   Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev (4)

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Let us differentiate the path Eq.  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev w ith resp ect to x,

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

From Eqs. (5) and (6), the sought curvature radius :

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev


Q. 36. A particle A moves in one direction along a given trajectory with a tangential acceleration wτ = aτ, where a is a constant vector coinciding in direction with the x axis (Fig. 1.4), and τ is a unit vector coinciding in direction with the velocity vector at a given point. Find how the velocity of the particle depends on x provided that its velocity is negligible at the point x = 0. 

Ans. In accordance with the problem  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

But  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

So,   Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

or,   Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev (because  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev  directed towards the jc-axis)

So,   Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Hence  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev


Q. 37. A point moves along a circle with a velocity v = at, where a = 0.50 m/s2. Find the total acceleration of the point at the mo- merit when it covered the n-th (n = 0.10) fraction of the circle after the beginning of motion. 

Ans. The velocity of the particle v = at

So,   Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev      (1)

And  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev    (2)

From  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

So,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev    (3)

From Eqs. (2) and (3) wn = 4πaη

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev


Q. 38. A point moves with deceleration along the circle of radius R so that at any moment of time its tangential and normal accelerations are equal in moduli. At the initial moment t = 0 the velocity of the point equals v0. Find:
 (a) the velocity of the point as a function of time and as a function of the distance covered s;
 (b) the total acceleration of the point as a function of velocity and the distance covered. 

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Ans.  (a) According to the problem

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

For v (t),  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Integrating this equation from  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev Integrating this equation from Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

So,   Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Hence  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

(b) The normal acceleration of the point 

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

And as accordance with the problem  

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

so, Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev


Q. 39. A point moves along an arc of a circle of radius R. Its velocity depends on the distance covered Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev where a is a constant. Find the angle α between the vector of the total acceleration and the vector of velocity as a function of s.

Ans. From the equation  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

As w, is a positive constant, the speed of the particle increases with time, and the tangential acceleration vector and velocity vector coincides in direction.

Hence the angle between  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev is equal to between Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev can be found by means of the formula  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev


Q. 40. A particle moves along an arc of a circle of radius R according to the law l = a sin ωt, where l is the displacement from the initial position measured along the arc, and a and ω are constants. Assuming R = 1.00 m, a = 0.80 m, and co = 2.00 rad/s, find:
 (a) the magnitude of the total acceleration of the particle at the points l = 0 and l = ±a;
 (b) the minimum value of the total acceleration wmin  and the corresponding displacement lm.

Ans. From the equation    Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev     (1)

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev   (2)

(a) At the point Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Hence  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Similarly  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Hence Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev


Q. 41. A point moves in the plane so that its tangential acceleration wτ = a, and its normal acceleration wn  = bt4, where a and b are positive constants, and t is time. At the moment t = 0 the point was at rest. Find how the curvature radius R of the point's trajectory and the total acceleration w depend on the distance covered s. 

Ans. As wt = a and at t = 0, the point is at rest

So,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev     (1)

Let R be the curvature radius, then 

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

But according to the problem

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Therefore  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Hence  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev


Q. 42. A particle moves along the plane trajectory y (x) with velocity v whose modulus is constant. Find the acceleration of the particle at the point x = 0 and the curvature radius of the trajectory at that point if the trajectory has the form 

(a) of a parabola y = ax2;
 (b) of an ellipse (xla)2  (y/b)2  = 1; a and b are constants here.

Ans. (a) Let us differentiate twice the path equation y (x) with respect to time.

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Since the particle moves uniformly, its acceleration at all points of the path is normal and at the point  x = 0 it coincides with the direction of derivative Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev Keeping in mind that at the point  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

We get  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

So, Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Note that we can also calculate it from the formula of problem (1.35 b)

(b) Differentiating the equation of the trajectory with respect to time we see that

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev  (1)

which implies that the vector Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev is normal to the velocity vector  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev which, of course, is along the tangent. Thus the former vactor is along the normal and the normal component of acceleration is clearly

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

on using  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Differentiating (1)

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Also from (1) Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

So,   Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev (since tangential velocity is constant = v)

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

and  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

This gives R = a2/ b .


Q. 43. A particle A moves along a circle of radius R = 50 cm so that its radius vector r relative to the point O (Fig. 1.5) rotates with the constant angular velocity ω = 0.40 rad/s. Find the modulus of the velocity of the particle, and the modulus and direction of its total acceleration. 

Ans. Let us fix the co-ordinate system at the point O as shown in the figure, such that the radius vector  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev makes an angle θ with x axis at the moment shown.
Note that the radius vector of the particle A rotates clockwise and we here take line ox as reference line, so in this case obviously the angular velocity  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev taking anticlockwise sense of angular displacement as positive.

Also from the geometry of the triangle OAC  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Let us write, 

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Differentiating with respect to time. 

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

As ω is constant, v is also constant and Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

So,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Alternate : From the Fig. the angular velocity of the point A, with respect to centre of the circle C becomes

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Thus we have the problem of finding the velocity and acceleration of a particle moving along a circle of radius R with constant angular velocity 2 ω

Hence

. Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev


Q. 44. A wheel rotates around a stationary axis so that the rotation angle φ varies with time as φ = at2, where a = 0.20 rad/s2. Find the total acceleration w of the point A at the rim at the moment t = 2.5 s if the linear velocity of the point A at this moment v = 0.65 m/s. 

Ans. Differentiating φ (t) with respect to time

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev   (1)

For fixed axis rotation, the speed of the point A:  

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev   (2)

Differentiating with respect to time 

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

But    Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

So,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev    


Q. 45. A shell acquires the initial velocity v = 320 m/s, having made n = 2.0 turns inside the barrel whose length is equal to l = 2.0 m. Assuming that the shell moves inside the barrel with a uniform acceleration, find the angular velocity of its axial rotation at the moment when the shell escapes the barrel. 

Ans. The shell acquires a constant angular acceleration at the same time as it accelerates linearly. The two are related by (assuming both are constant)

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Where w = linear acceleration and β = angular acceleration 

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev



Q. 46. A solid body rotates about a stationary axis according to the law φ = at - bt3, where a = 6.0 rad/s and b = 2.0 rad/s3. Find:
 (a) the mean values of the angular velocity and angular acceleration averaged over the time interval between t = 0 and the complete stop;
 (b) the angular acceleration at the moment when the body stops. 

Ans. Let us take the rotation axis as z-axis whose positive direction is associated with the positive direction o f the cordinate <p, the rotation angle, in accordance w ith the right-hand screw rule (Fig.)

(a) Defferentiating φ (t) with respect to time.

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

From (1) the solid comes to stop at  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev
The angular velocity  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRevKinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Sim ilarly  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev for all values of t.

So, Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev
So,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev
Hence  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev


Q. 47. A solid body starts rotating about a stationary axis with an angular acceleration β = at, where a = 2.0.10-2  rad/s3. How soon after the beginning of rotation will the total acceleration vector of an arbitrary point of the body form an angle α = 60° with its velocity vector?

Ans. Angle a is related with |wt| and wn by means of the fomula :

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev    (1)

where R is die radius of die circle which an arbitrary point of the body circumscribes. From die given equation Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev is positive for all values of t)

Integrating within the limit  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

So,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev
and  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Putting the values of |wt| and wn in Eq. (1), we get,

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev


Q. 48. A solid body rotates with deceleration about a stationary axis with an angular deceleration Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev where ω is its angular velocity. Find the mean angular velocity of the body averaged over the whole time of rotation if at the initial moment of time its angular velocity was equal to ω0

Ans. In accordance with the problem, βz < 0

Thus Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev where A: is proportionality constant

or,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev    (1)

When ω = 0 , total time o f rotation  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Average angular velocity   Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev


Q.49. A solid body rotates about a stationary axis so that its angular velocity depends on the rotation angle φ as ω = ω0  — aφ, where ω0 and a are positive constants. At the moment t = 0 the angle = 0. Find the time dependence of
 (a) the rotation angle;
 (b) the angular velocity. 

Ans. We have  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Integratin this Eq. within its limit for (φ) f

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Hence  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev   (1)

(b) From the Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev or by differentiating Eq. (1)

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev 


Q.50. A solid body starts rotating about a stationary axis with an angular acceleration β = β0  cos φ, where β0  is a constant vector and φ is an angle of rotation from the initial position. Find the angular velocity of the body as a function of the angle φ. Draw the plot of this dependence. 

Ans. Let us choose the positive direction of z-axis (stationary rotation axis) along the vector  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

In accordance with the equation  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Integrating this Eq. within its limit for

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

The plot  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev is shown in the Fig. It can be seen that as the angle qp grows, the vector  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev increases, coinciding with the direction of the vector Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev reaches the maximum at  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev then starts decreasing and finally turns into zero at φ = π. After that the body starts rotating in the opposite direction in a similar fashion (ωz < 0). As a result, the body will oscillate about the position Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev with an amplitude equal to π/2.


Q. 51. A rotating disc (Fig. 1.6) moves in the positive direction of the x axis. Find the equation y (x) describing the position of the instantaneous axis of rotation, if at the initial moment the axis C of the disc was located at the point O after which it moved
 (a) with a constant velocity v, while the disc started rotating counterclockwise with a constant angular acceleration β (the initial angular velocity is equal to zero);
 (b) with a constant acceleration w (and the zero initial velocity), while the disc rotates counterclockwise with a constant angular velo- city ω.

Ans.  Rotating disc moves along the x-axis, in plane motion in x - y plane. Plane motion of a solid can be imagined to be in pure rotation about a point (say 7) at a certain instant known as instantaneous centre of rotation. The instantaneous axis whose positive sense is directed along Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev the solid and which passes through the point/, is known as instantaneous axis of rotation.

Therefore the velocity vector of an aibitrary point (P) of the solid can be represented as :

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev    (1)

On the basis of Eq. (1) for the C. M. (C) of the disc

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev   (2)

According to the problem Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev and Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev plane, so to satisy the  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev Hence point  l is at a distance  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev  above the centre of the disc along y - axis. Using all these facts in Eq. (2), we get

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev  (3)

(a) From the angular kinematical equation 

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev    (4)

On the other hand x = v t, (where x is the x coordinate of the C.M.) 

or, t = x/y    (5)

From Eqs. (4) and (5), Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Using this value o f co in Eq. (3) w e get Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

(b) As centre C moves with constant acceleration w, with zero initial velocity 

So,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Therefore,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Hence  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev


Q. 52.  A point A is located on the rim of a wheel of radius R = 0.50 m which rolls without slipping along a horizontal surface with velocity v = 1.00 m/s. Find:
 (a) the modulus and the direction of the acceleration vector of the point A;
 (b) the total distance s traversed by the point A between the two successive moments at which it touches the surface. 

Ans. The plane motion of a solid can be imagined as the combination of translation of the C.M . and rotation about C.M.

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev is the position o f vector o f A with respect to C.

In the problem  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev constant, and the rolling is without slipping Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev Using these conditions in Eq. (2)

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Here,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev is the unit vector directed along Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Hence  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev or directed toward the centre of the wheel.

(b) Let the centre o f the wheel m ove toward right (positive jc-axis) then for pure tolling on the rigid horizontal surface, wheel will have to rotate in clockwise sense. If co be the angular velocity o f the wheel then Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Let the point A touches the horizontal surface at t = 0, further let us locate the point A at t = ty

When it makes θ = ω t at the centre of the wheel.

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev
Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev
H ence distance covered b y the point A during Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev


Q. 53. A ball of radius R = 10.0 cm rolls without slipping down an inclined plane so that its centre moves with constant acceleration w = 2.50 cm/s2; t = 2.00 s after the beginning of motion its position corresponds to that shown in Fig. 1.7. Find:
 (a) the velocities of the points A, B, and 0;
 (b) the accelerations of these points. 

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRevKinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Ans. Let us fix the co-ordinate axis xyz as shown in the fig. As the ball rolls without slipping along the rigid surface so, on the basis o f the solution o f problem Q.52 :

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

At the position corresponding to that of Fig., in accordance with the problem, 

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

and Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

(a) Let us fix the co-ordinate system with the frame attached with the rigid surface as shown in the Fig.

As point O is the instantaneous centre of rotation of the ball at the moment shown in Fig.

so,    Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev
Now,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

So,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

So,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev
Similarly  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

So,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev


Q. 54. A cylinder rolls without slipping over a horizontal plane. The radius of the cylinder is equal to r. Find the curvature radii of trajectories traced out by the points A and B (see Fig. 1.7). 

Ans. Let us draw the kinematical diagram of the rolling cylinder on the basis of the solutioi of problem Q.53.

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRevKinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

As, an arbitrary point of the cylinder follows a curve, its normal acceleration and radius of curvature are related by the well known equation

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

so, for point A,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

or,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Similarly for point B, 

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

or,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev


Q. 55. Two solid bodies rotate about stationary mutually perpendicular intersecting axes with constant angular velocities ω1 = 3.0 rad/s and ω2  = 4.0 rad/s. Find the angular velocity and angular acceleration of one body relative to the other. 

Ans. The angular velocity is a vector as infinitesimal rotation commute. Then file relative angular velocity of the body 1 with respect to the body 2 is dearly.

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

as fo r relative lin ear velocity. The relative acceleration of 1 w.r.t  2 is

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

where S ' is a fram e corotating with the second body and S is a space fixed frame with origin coinciding with the point of intersection of the two axes,

but Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Since S ' rotates with angular velocity  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev as the first b ody rotates with constant angular velocity in space, thus

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Note that for any vector  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev relation in space forced frame (k) and a frame (Jd) rotating with angular velocity Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev


Q. 56. A solid body rotates with angular velocity ω = ati + bt2j, where a = 0.50 rad/s2, b = 0.060 rad/s3, and i and j are the unit vectors of the x and y axes. Find:
 (a) the moduli of the angular velocity and the angular acceleration at the moment t = 10.0 s;
 (b) the angle between the vectors of the angular velocity and the angular acceleration at that moment. 

Ans. We have  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev    (1)

So,   Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

D ifferen tia tin g E q. (1) with respect to time

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev   (2)

So,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

and  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

(b)  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Putting the values of (a) and (b) and ' taking t =10s, we get

α = 17°


Q. 57.  A round cone with half-angle α = 30° and the radius of the base R = 5.0 cm rolls uniformly and without slipping over a horizontal plane as shown in Fig. 1.8. The cone apex is hinged at the point O which is on the same level with the point C, the cone base centre. The velocity of point C is v = 10.0 cm/s. Find the moduli of  
(a) the vector of the angular velocity of the cone and the angle it forms with the vertical;
 (b) the vector of the angular acceleration of the cone.

Ans.  Let the axis of the cone (OC) rotates in an ticlockw ise sense w ith constant angular velocity Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev and the cone itself about it’s own axis (OC) in clockwise sense with angular velocity  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev Then the resultant angular velocity of the cone.

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev     (1)

As the rolling is pure the magnitudes of the vectors 

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev  can be easily found from Fig.

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev (2)

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

(b) Vector of angular acceleration

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

The vector  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev which rotates about the OO' axis with the angular velocity Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev retains i magnitude. This increment in the time interval dt is equal to

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Thus Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

The magnitude of the vector  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev
So,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev


Q. 58. A solid body rotates with a constant angular velocity ω0  = 0.50 rad/s about a horizontal axis AB. At the moment t = 0 the axis AB starts turning about the vertical with a constant angular acceleration β0 = 0.10 rad/s2. Find the angular velocity and angular acceleration of the body after t = 3.5 s. 

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Ans. The axis AB acquired the angular velocity

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev    (1)

Using the facts of the solution of 1.57, the angular velocity of the body

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

And the angular acceleration.

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

But  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

So,  Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 3) - Mechanics, Irodov JEE Notes | EduRev

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