Q: A sphere of radius R is cut from a larger solid sphere of radius 2 R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y

aligned & I _1= 2 5 M (2 R )^2 \\ & I _2= 2 5 mr ^2+ md ^2 \\ & = 2 5 ( M 8 ) R ^2+ M 8 ( R )^2 \\ & = 7 MR ^2 40 \\ & I = I _1- I _2= 8 MR ^2 5 - 7 MR ^2 40 = 57 40 MR ^2 aligned I _2 I _1 = 7 57

Question 1

A solid sphere A of radius R and mass M is attached at a point to a smaller solid sphere B of radius r

Accepted Answer

The distance between the centres of the two spheres A and B is d = R + r. Let I_ A, cm and I_ B, cm be the moments of inertia of spheres A and B about their respective vertical centroidal axes. For solid spheres, I_ A, cm = 2 5 MR^2 and I_ B, cm = 2 5 mr^2. The moment of inertia of the system about a vertical axis passing through the centre of A is given by the sum of the moment of inertia of A about its own centre and the moment of inertia of B about the centre of A. Using the parallel axis the

Question 2

The angular speed of a flywheel is increased from 600 rpm to 1200 rpm in 10 s. The number of revolutions completed by the flywheel during this time is :

Accepted Answer

Initial angular speed, _0 = 600 2 60 = 20 rad/s Final angular speed, = 1200 2 60 = 40 rad/s Angular acceleration, = - _0 t = 40 - 20 10 = 2 rad/s^2 Angular displacement, = _0 t + 1 2 t^2 = 20 10 + 1 2 2 (10)^2 = 200 + 100 = 300 rad Number of revolutions, N = 2 = 300 2 = 150 Answer: 150

Question 3

A thin wire of length 'L' and linear mass density 'm' is bent into a circular ring (in x-y plane) with centre 'C' as shown in figure. The moment of inertia of the ring about an axis yy' will be :

Accepted Answer

Total mass of the ring, M = m L. Since the wire of length L is bent into a circular ring of radius R, its circumference is L. 2 R = L R = L 2 The axis yy' is a tangent to the ring in its plane. The moment of inertia of a ring about its diametric axis is I_ d = 1 2 MR^2. Using the parallel axis theorem, the moment of inertia of the ring about the tangent yy' is: I_ yy' = I_ d + MR^2 I_ yy' = 1 2 MR^2 + MR^2 = 3 2 MR^2 Substituting the values of M and R: I_ yy' = 3 2 (mL) ( L 2 )^2 I_ yy' = 3 2 mL

Question 4

A thin horizontal disc is rotating about a vertical axis passing through its fixed centre O. Its angular momentum is L_A and L_B computed about points A and B, respectively, with OB=2 OA. The value of

Accepted Answer

The angular momentum of a rigid body about any arbitrary point P is given by the relation: L _P = L _ cm + r _ cm/P M v _ cm where L _ cm is the angular momentum of the body about its center of mass, r _ cm/P is the position vector of the center of mass with respect to point P, M is the total mass of the body, and v _ cm is the velocity of the center of mass. In the given problem, the disc is rotating about a vertical axis passing through its fixed center O. Since the center of mass of the disc

Question 5

The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sp

Accepted Answer

Angular Momentum conservation array l|l I _1 _1= I _2 _2 & M = same \ _2= _1 4 & R _2=2 r _1 \ T= 2 & I r ^2 \ & I _2=4 I _1 \ & ~T _1=27 day array aligned & T _2=4 ~T _1=4 27 \ & ~T _2=108 days aligned

Question 6

A sphere of radius R is cut from a larger solid sphere of radius 2 R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y

Accepted Answer

aligned & I _1= 2 5 M (2 R )^2 \ & I _2= 2 5 mr ^2+ md ^2 \ & = 2 5 ( M 8 ) R ^2+ M 8 ( R )^2 \ & = 7 MR ^2 40 \ & I = I _1- I _2= 8 MR ^2 5 - 7 MR ^2 40 = 57 40 MR ^2 aligned I _2 I _1 = 7 57

Question 7

The radius of gyration of a solid sphere of mass 5 kg about X Y is 5 m as shown in figure. The radius of the sphere is 5 x 7 ~m , then the value of x is:

Accepted Answer

I_ X Y =I_ C M +M R^2= 2 5 M R^2+M R^2= 7 5 M R^2= 7 5 5 R^2=7 R^2...(1) I_ X Y =M K^2=5 5^2...(ii) 5 5^2=7 R^2 [ From (1) and (2)] R= 5 7 5= 5 x 7 (Given) x= 5

Question 8

A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is v in the direction shown, which one of the following options is correct ( P and Q are any high

Accepted Answer

In the case of pure rolling, The topmost point will have velocity 2 v while point Q i.e. lowest point will have zero velocity. Hence point P moves faster than point Q.

Question 9

The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is 2400 ~g ~cm ^2. The length of the 400 g rod is nearly:

Accepted Answer

Moment of inertia of rod =I= m ^2 12 aligned & 2400=400 ^2 12 \ & 72= ^2 \ & = 72 =8.48 ~cm 8.5 ~cm aligned

Rotational Motion — Previous Year Questions & Solutions