Q#9
One end of a uniform rod of mass m and length l is clamped. The rod lies on a smooth horizontal surface and rotates on it about the clamped end at a uniform angular velocity ω. The force exerted by the clamp on the rod has a horizontal component(a) mω²l (b) zero (c) mg (d) ½mω²l.
Answer: (d)
Since the rod rotates on the horizontal surface, the horizontal component of the force applied by the clamp is the centripetal force
= mω²r = mω²(l/2) = ½mω²l.
Q#10
A uniform rod is kept vertically on a horizontal smooth surface at a point O. If it is rotated slightly and released, it falls down on the horizontal surface. The lower end will remain
(a) at O
(b) at a distance less than l/2 from O
(c) at a distance l/2 from O
(d) at a distance larger than l/2 from O.
Answer: (c)
Since the rod is uniform its Center of mass (CoM) will be in the middle i.e. at l/2 distance from the ends. When the rod is vertical the CoM will bel/2 distance vertically above the point O. Since there is no force in the horizontal direction the CoM will be on the point O after it falls down. Hence the lower end will remain at a distance l/2 from O.
Q#11
A circular disc A of radius r is made from an iron plate of thickness t and another circular disc B of radius 4r is made from an iron plate of thickness t/4. The relation between the moments of inertia IA and IB is
(a) IA > IB (b) IA = IB (c) IA < IB (d) depends on the actual value of t and r.
Answer: (c)
Let the density of iron plate be ρ.
Mass of first disc m = πr²tρ
M.I. = IA = ½mr² = ½πr²tρr² = ½πr4tρ
Mass of second disc = M = π(4r)²(t/4)ρ = 4πr²tρ
M.I. = IB = ½M(4r)² = ½4πr²tρ16r² = 32πr4tρ
Clearly, IA < IB.
Q#12
Equal torques act on the discs A and B of the previous problem, initially both being at rest. At a later instant, the linear speeds of a point on the rim of A and another point on the rim of B are vA and vB respectively. We have
(a) vA > vB (b) vA = vB (c) vA < vB
(d) the relation depends on the actual magnitude of the torques.
ANSWER: (a)
Explanation: Let T be the torque and A and A' be the angular accelerations. A = T/IA and A' = T/IB. Clearly A > A'. In fact, the ratio of the moment of Inertias is 64. So at a later instant, the angular velocity of the first disc will be much greater than the second. Hence option (a).
Q#13
A closed cylindrical tube containing some water (not filling the entire tube) lies in a horizontal plane. If the tube is rotated about a perpendicular bisector, the moment of inertia of water about the axis
(a) increases (b) decreases (c) remains constant (d) increases if the rotation is clockwise and decreases if it is anticlockwise.
Answer: (a)
During the above rotation water will move away from the axis of rotation. And the moment of inertia is directly proportional to the square of the distance of the mass from the axis of rotation. Hence the option (a).
Q#14
The moment of inertia of a uniform semicircular wire of mass M and radius r about a line perpendicular to the plane of the wire through the center is
(a) Mr² (b) ½Mr² (c) ¼Mr² (d) (2/5)Mr².
Answer: (a)
Since the mass of the wire is at a distance r away from the axis. Hence the option (a).
Q#15
Let I1 and I2 be the moments of inertia of two bodies of identical geometrical shape, the first made of aluminum and the second of iron.
(a) I1 < I2 (b) I1 = I2 (c) I1 > I2 (d) relation betwen I1 and I2 depends on the actual shape of the bodies.
Answer: (a)
Aluminum has a lower density than iron hence the first body has lesser mass than the second. And if the shapes of the bodies are same the M.I. is directly proportional to the mass. So, the first body will have lower M.I. than the second body.
Q#16
A body having its center of mass at the origin has three of its particle at (a, 0, 0), (0, a, 0), (0, 0, a). The moments of inertia of the body about the X and Y axes are 0.20 kgm² each. The moment of inertia about the Z-axis
(a) is 0.20 kg-m² (b) 0.40 kg-m² (c) is 0.20/2 kg-m² (d) Cannot be deduced with the information.
Answer: (d)
To know the moment of inertia we need to know the mass and shape of the body. None is available in the problem, hence the option (d).
Q#17
A cubical block of mass M and edge a slides down a rough inclined plane of inclination θ with a uniform velocity. The torque of the normal force on the block about its center has a magnitude
(a) zero (b) Mga (c) Mga.sinθ (d) ½Mga.sinθ.
Answer: (d)
The weight Mg of the cubical block can be resolved along the plane and perpendicular to it. Since the block is sliding down with a uniform velocity, the resultant force along the plane is zero. There are only two forces along the plane - Weight component (Mgsinθ) and the force of friction F in the opposite direction.
f + Mg.sinθ =0
f = –Mgsinθ
So these two forces are equal and opposite in direction. But they are not in the same line. While the weight component acts through CoM, the Friction force acts along the contact surface. Since the edge of the block is 'a' it is clear that the distance between these two forces = a/2. The torque produced by these two forces = ½Mga.sinθ. In the figure, it is in the anti-clockwise direction.
Now consider the weight component perpendicular to the plane acting through the CoM which is balanced by the equal and opposite Normal Force N. But N is not in line with the perpendicular weight component i.e. it does not pass through the Com. These two equal and opposite forces create a torque in the clockwise direction and its magnitude is same as the first torque because the block does not have angular motion. So, the torque produced by the normal force is equal to ½Mgasinθ.
f + Mg.sinθ =0
f = –Mgsinθ
So these two forces are equal and opposite in direction. But they are not in the same line. While the weight component acts through CoM, the Friction force acts along the contact surface. Since the edge of the block is 'a' it is clear that the distance between these two forces = a/2. The torque produced by these two forces = ½Mga.sinθ. In the figure, it is in the anti-clockwise direction.
Now consider the weight component perpendicular to the plane acting through the CoM which is balanced by the equal and opposite Normal Force N. But N is not in line with the perpendicular weight component i.e. it does not pass through the Com. These two equal and opposite forces create a torque in the clockwise direction and its magnitude is same as the first torque because the block does not have angular motion. So, the torque produced by the normal force is equal to ½Mgasinθ.
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