Title Exercise-1(Rotational Motion).pdf ✅

ROTATIONAL MOTION 21 EXERCISE – 1: Basic Objective Questions Introduction to Rotational Motion and Moment of Inertia 1. A fan is moving around its axis. What will be its motion regarded as? (a) pure rolling (b) rolling with slipping (c) skidding (d) pure rotation Ans. (d) Sol. As axis of rotation of fan is fixed hence this motion is pure rotation. 2. A body is rotating with angular velocity ( ) ˆ ˆ ˆ  = − + 3i 4j k . The linear velocity of a point having position vector ( ) ˆ ˆ ˆ r 5i 6j 6k = − + is: (a) ˆ ˆ ˆ 6 2 3 i j k + − (b) ˆ ˆ ˆ 18 13 2 i j k + − (c) ˆ ˆ ˆ − − + 18 13 2 i j k (d) ˆ ˆ ˆ 6 2 8 i j k − + Ans. (c) Sol. Body is rotating with angular velocity,  = − + 3i 4j k Position vector, r 5i 6j 6k = − + Linear velocity, v r v 18i 13j 2k  =   = − − + 3. The moment of inertia of a body does not depend on: (a) the mass of the body (b) the angular velocity of the body (c) the axis of rotation of the body (d) the distribution of the mass in the body Ans. (b) Sol. n 2 i i i 1 I m r = =   Moment of inertia is independent of angular velocity of the body. 4. Three point masses m1, m2 and m3 are located at the vertices of an equilateral triangle of side ‘a’. What is the moment of inertia of the system about an axis along the altitude of the triangle passing through m1? (a) ( ) 2 1 2 a m m 4 + (b) ( ) 2 2 3 a m m 4 + (c) ( ) 2 1 3 a m m 4 + (d) ( ) 2 1 2 3 a m m m 4 + + Ans. (b) Sol. ( ) ( ) 2 2 2 1 2 3 2 2 3 a a I m 0 m m 2 2 a m m 4     = + +         = + 5. The ratio of the squares of radii of gyration of a circular disc and a circular ring of the same radius about a tangential axis in the plane of bodies is: (a) 1 : 2 (b) 5 : 6 (c) 2 : 3 (d) 2 : 1 Ans. (b) Sol. Moment of inertia of a ring about axis in the plane of ring and passing through the com, ( ) 2 cm R MR I 2 = Moment of inertia of a disk about axis in the plane of disk and passing through the com, ( ) 2 cm D MR I 4 = Moment of inertia about tangential axis ( ) ( ) 2 t cm 2 2 2 2 R 1 t 2 2 2 2 D 2 t I I MR MR 3 I MR MR MK 2 2 M R 5 I M R M R M K 4 4 = + = + = =  = + = =   
22 ROTATIONAL MOTION 2 2 1 K 5 2 5 K 4 3 6    =  =     6. For the same total mass which of the following will have the largest moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of the body? (a) A disc of radius a (b) A ring of radius a (c) A square lamina of side 2a (d) Four roads forming square of side 2a Ans. (d) Sol. 2 Disc 1 I Ma 2 = ( ) ( ) 2 Ring 2 2 Plate 2 2 2 4rods I Ma M 2a 2 I Ma 6 3 M 2a 4 M 4 I 4 a Ma 12 4 3 = = =     = + =       Moment of inertia of 4 rods of side 2a is largest. 7. The radius of gyration of a body depends upon : (a) translation motion (b) axis of rotation (c) area of the body (d) all of these Ans. (b) Sol. Moment of inertia of a rigid body is defined as; 2 I MK I K M  =  = Where, K is the radius of gyration. We know that I depend on mass and axis of rotation So, radius of gyration of a body depends upon the axis of rotation. 8. A circular disc X of radius R is made from an iron plate of thickness t, and another disc Y of radius 4R is made from an iron plate of thickness t/4. Then the relation between the moment of inertia IX and IY about a axis passing through their center of mass and perpendicular to the plane of disk is : (a) I Y = 32 IX (b) I Y = 16 IX (c) I Y = IX (d) I Y = 64 IX Ans. (d) Sol. Moment of inertia of disc, 1 2 I MR 2 = ( ) ( ) 2 2 x 2 2 Y Y X 1 I R tR 2 1 t I 4R 4R 2 4 I 64I =    =       = 9. Moment of inertia of a uniform annular disc of internal radius r and external radius R and mass M about an axis through its centre and perpendicular to its plane is: (a) 1 2 2 M(R r ) 2 − (b) 1 2 2 M(R r ) 2 + (c) 4 4 2 2 M(R r ) 2(R r ) + + (d) 4 4 2 2 1 M(R r ) 2 (R r ) + − Ans. (b) Sol. I of given disc = I of disc of radius R - I of disc of radius r ( ) ( ) 1 1 2 2 2 2 I = R R r r 2 2   −   Also,
ROTATIONAL MOTION 23 ( )   ( ) ( ) ( ) 2 2 4 4 4 4 2 2 2 2 M R r 1 1 M I R r R r 2 2 R r 1 M R r 2 =   −   =  − = − − = + 10. If the radius of a solid sphere is 35 cm, calculate the radius of gyration when the axis is along a tangent: (a) 7 10 cm (b) 7 35 cm (c) 7 cm 5 (d) 2 cm 5 Ans. (b) Sol. 2 2 2 2 t cm 2 7 I I MR MR MR MR 5 5 = + = + = ( ) 2 2 2 7 MR MK 5 7 K 35 7 35 cm 5 = = = 11. The moment of inertia of a straight thin rod of mass M, length L about an axis perpendicular to its length and passing through its one end is: (a) 1 2 ML 12 (b) 1 2 ML 3 (c) 1 2 ML 2 (d) 2 ML Ans. (b) Sol. 2 2 2 end cm ML L I I Md M 12 2   = + = +     2 ML 3 = 12. What is the moment of inertia I of a uniform hollow sphere of mass M and radius R, pivoted about an axis that is tangent to the surface of the sphere? (a) 2 2 MR 3 (b) 3 2 MR 5 (c) 6 2 MR 5 (d) 5 2 MR 3 Ans. (d) Sol. 2 2 t cm cm I I md I MR = + = + 2 5 2 2 2 MR MR MR 3 3 = + = 13. The moment of inertia of a metre stick of mass 300 gm, about an axis at right angles to the stick and located at 30 cm mark, is: (a) 5 2 8.3 10 g cm  − (b) 2 5.8g cm − (c) 5 2 3.7 10 g cm  − (d) none of these Ans. (c) Sol. Since, it’s a meter stick, the mark is 20cm from the center. ( ) ( ) 2 Mark cm 2 2 4 4 4 2 5 2 I I Md 300 100 300 20 12 25 10 12 10 37 10 g - cm Or 3.7 10 g - cm = + = + =  +  =   14. moment of inertia of a solid sphere about an axis passing through centre of gravity is 2 2 MR 5 ; then its radius of gyration about a parallel axis at a distance 2R from first axis is: (a) 5R (b) 22 / 5 R (c) 5 R 2 (d) 12 / 5 R Ans. (b) Sol. By parallel axis theorem
24 ROTATIONAL MOTION ( ) 2 A cm 2 2 2 2 I I Md 2 MR M 2R 5 22 MR MK 5 22 K R 5 = + = + = =  = 15. Four spheres of diameter 2a and mass M are placed with their centres on the four corners of a square of side b. Then the moment of inertia of the system about an axis along one of the sides of the square is: (a) 4 2 2 Ma 2Mb 5 + (b) 8 2 2 Ma 2Mb 5 + (c) 8 2 Ma 5 (d) 4 2 2 Ma 4Mb 5 + Ans. (b) Sol. 1 2 3 4 2 2 2 2 2 I I I I I 2 2 Ma 2 Ma Mb 2 5 5 8 Ma 2Mb 5 = + + +     = + +         = + 16. For the given uniform square lamina ABCD, whose centre is O (a) AC EF 2 I I = (b) I AD = 3IEF (c) I AC = IEF (d) AC EF I 2 I = Ans. (c) Sol. 2 Ef ML I 12 = 2 2 2 O EF GH ML ML ML I I I 12 12 6 = + = + = (By perp. axis theorem due to symmetry EF GH I I = ) AC BD O I I I + = (By perp. axis theorem and symmetry AC BD I I = ) 2 AC ML I 12  = AC EF  = I I 2 AD EF L I I M 2   = +     {parallel axis theorem} 2 2 2 AD ML ML ML I 12 4 3  = + = AD EF  = I 4I 17. Moment of inertia of a circular wire of mass M and radius R about its diameter is : (a) MR2 /2 (b) MR2 (c) 2 MR2 (d) MR2 /4 Ans. (a) Sol. D F C A B E O