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Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE PDF Download

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.

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

From (1) the solid comes to stop at  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
The angular velocity  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEEIrodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Sim ilarly  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE for all values of t.

So, Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
So,  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
Hence  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


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 :

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE    (1)

where R is die radius of die circle which an arbitrary point of the body circumscribes. From die given equation Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE is positive for all values of t)

Integrating within the limit  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

So,  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
and  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

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

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


Q. 48. A solid body rotates with deceleration about a stationary axis with an angular deceleration Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE 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 Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE where A: is proportionality constant

or,  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE    (1)

When ω = 0 , total time o f rotation  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Average angular velocity   Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


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  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

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

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Hence  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE   (1)

(b) From the Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE or by differentiating Eq. (1)

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE 


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  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

In accordance with the equation  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Integrating this Eq. within its limit for

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

The plot  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE is shown in the Fig. It can be seen that as the angle qp grows, the vector  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE increases, coinciding with the direction of the vector Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE reaches the maximum at  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE 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 Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE 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 Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE 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 :

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE    (1)

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

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE   (2)

According to the problem Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE and Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE plane, so to satisy the  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE Hence point  l is at a distance  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE  above the centre of the disc along y - axis. Using all these facts in Eq. (2), we get

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE  (3)

(a) From the angular kinematical equation 

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE    (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), Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Using this value o f co in Eq. (3) w e get Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

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

So,  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Therefore,  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Hence  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


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.

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE is the position o f vector o f A with respect to C.

In the problem  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE constant, and the rolling is without slipping Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE Using these conditions in Eq. (2)

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Here,  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE is the unit vector directed along Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Hence  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE 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 Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

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.

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
H ence distance covered b y the point A during Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


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. 

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEEIrodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

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 :

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

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

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

and Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

(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,    Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
Now,  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

So,  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

So,  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
Similarly  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

So,  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


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.

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEEIrodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

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

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

so, for point A,  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

or,  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Similarly for point B, 

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

or,  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


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.

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

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

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

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 Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Since S ' rotates with angular velocity  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE as the first b ody rotates with constant angular velocity in space, thus

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Note that for any vector  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE relation in space forced frame (k) and a frame (Jd) rotating with angular velocity Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


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  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE    (1)

So,   Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

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

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE   (2)

So,  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

and  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

(b)  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

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 Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE and the cone itself about it’s own axis (OC) in clockwise sense with angular velocity  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE Then the resultant angular velocity of the cone.

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE     (1)

As the rolling is pure the magnitudes of the vectors 

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE  can be easily found from Fig.

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE (2)

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

(b) Vector of angular acceleration

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

The vector  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE which rotates about the OO' axis with the angular velocity Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE retains i magnitude. This increment in the time interval dt is equal to

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Thus Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

The magnitude of the vector  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE
So,  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE


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. 

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Ans. The axis AB acquired the angular velocity

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE    (1)

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

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

And the angular acceleration.

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

But  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

So,  Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

The document Irodov Solutions: Kinematics- 4 | I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
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FAQs on Irodov Solutions: Kinematics- 4 - I. E. Irodov Solutions for Physics Class 11 & Class 12 - JEE

1. What is kinematics?
Ans. Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. It involves studying the position, velocity, and acceleration of objects, as well as the relationships between them.
2. How is kinematics different from dynamics?
Ans. Kinematics and dynamics are both branches of physics that study motion, but they approach it from different perspectives. Kinematics focuses on describing the motion of objects without considering the forces involved, while dynamics deals with the causes of motion and the effect of forces on objects.
3. What are the basic quantities in kinematics?
Ans. The basic quantities in kinematics are displacement, velocity, and acceleration. Displacement measures the change in position of an object, velocity measures the rate at which an object's position changes, and acceleration measures the rate at which an object's velocity changes.
4. How is average velocity calculated in kinematics?
Ans. Average velocity can be calculated by dividing the total displacement of an object by the total time taken. It is a vector quantity that includes both magnitude (speed) and direction. The formula for average velocity is: average velocity = total displacement / total time.
5. What is the difference between speed and velocity in kinematics?
Ans. Speed and velocity are related but distinct concepts in kinematics. Speed is a scalar quantity that measures how fast an object is moving, while velocity is a vector quantity that measures both the speed and direction of an object's motion.
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