Q#18
A thin circular ring of mass M and radius r is rotating about its axis with an angular speed ω. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become(a) ωM/(M + m) (b) ωM/(M + 2m) (c) ω(M – 2m)/(M + 2m) (d) ω(M + 2m)/M
Answer: (b)
Moment of inertia of the ring I = Mr²
Angular Momentum = I⍵
When the masses are attached, the moment of Inertia I' = Mr² + 2mr²
= (M + 2m)r²
Let the new angular speed be ⍵'. So the angular momentum = I'⍵'.
Since the angular momentum is conserved
I'⍵'= I⍵
⍵' = I⍵/I' = ⍵Mr²/(M + 2m)r² = ⍵M/(M + 2m)
Q#19
A person sitting firmly over a rotating stool has his arms stretched. If he folds his arms, his angular momentum about the axis of rotation
(a) increases (b) decreases (c) remains unchanged (d) doubles
Answer: (c)
If there is no external torque, the angular momentum is conserved. It is true if the moment of inertia of the rotating part of the stool is negligible with respect to the man.
Note: As informed by one of the students in the comment section below, the answer in the latest edition is changed to (b) i.e. decreases. It is true if the moment of inertia of the stool is considered and it is considerable with respect to the moment of inertia of the man. The angular momentum of the man with the stool will remain unchanged. If the initial angular speed = ⍵ and the final angular speed = ⍵', then
(Iₘ+ Iₛ)⍵ = (Iₘ'+ Iₛ)⍵'
Iₘ⍵ + Iₛ⍵ = Iₘ'⍵' + Iₛ⍵' = const. ------- (i)
{Iₘ and Iₘ' are initial and final M.I. of the man, Iₛ = M.I. of the stool}
When the man folds his hands, his moment of inertia decreases and hence the sum of M.I of man and the stool decreases (even though M.I. of the stool remains constant). To conserve the angular momentum the angular speed increases i.e. ⍵' > ⍵. So the angular momentum of the stool increases Iₛ⍵' > Iₛ⍵. Since the sum of the angular momentum of the stool and the man is constant the angular momentum of the man must decrease after folding his hands. As in (i), Iₘ'⍵' < Iₘ⍵.
Q#20
The Center of a wheel rolling on a plane surface moves with a speed v0. A particle on the rim of the wheel at the same level as the center will be moving at a speed
(a) zero (b) v0 (c) √2v0 (d) 2v0.
Answer: (c)
Explanation: Let the radius of the wheel = r.
The angular speed of the wheel ⍵ = v0/r
A particle on the rim of the wheel at the same level as the center will have a vertical speed =⍵r = (v0/r)r = v0 and a horizontal speed = v0
So the resultant speed = √(v0² + v0²) = √(2v0²) = v0√2
Q#21
A wheel of radius 20 cm is pushed to move it on a rough horizontal surface. It is found to move through a distance of 60 cm on the road during the time it completes one revolution about the center. Assume that the linear and the angular accelerations are uniform. The frictional force acting on the wheel by the surface is
(a) along the velocity of the wheel
(b) opposite to the velocity of the wheel
(c) perpendicular to the velocity of the wheel
(d) zero
Answer: (a)
Explanation: Perimeter of the wheel = 2π(0.20) = 0.4π = 1.26 m
But it moves only 0.60 m.
So the wheel slips on the road. At the point of contact with the road, the part of the wheel is sliding backward. Hence the force of friction acts forward i.e. along the velocity of the wheel.
Q#22
The angular velocity of the engine (and hence of the wheel) of a scooter is proportional to the petrol input per second. The scooter is moving at a frictionless road with uniform velocity. If the petrol input is increased by 10%, the linear velocity of the scooter is increased by
(a) 50% (b) 10% (c) 20% (d) 0%
Answer: (d)
With the increase of petrol input the angular velocity of the wheel increases but in the absence of friction on the road, no external force on the scooter is applied. Hence the linear velocity of the scooter is not increased.
Q#23
A solid sphere, a hollow sphere and a disc, all having same mass and radius, are placed at the top of a smooth incline and released. Least time will be taken in reaching the bottom by
(a) the solid sphere (b) the hollow sphere
(c) the disc (d) all will take same time.
Answer: (d)
Explanation: Since there is no friction on the incline, the only force on each of these objects along the incline is their weight component mg.sinθ.
So each of them have same initial velocity = 0 and same acceleration = (mgsinθ)/m = gsinθ
So the time taken by each of them t is given by
S = 0(t) + (½gsinθ)t²
t² = 2S/g.sinθ
t = √{2S/g.sinθ}
where S is the length of the incline.
Q#24
A solid sphere, a hollow sphere and a disc, all having same mass and radius, are placed at the top of an incline and released. The friction coefficients between the objects and the incline are same and not sufficient to allow pure rolling. Least time will be taken in reaching the bottom by
(a) the solid sphere (b) the hollow sphere
(c) the disc (d) all will take same time.
Answer: (d)
Since pure rolling does not occur, the objects slide. Therefore on each of them, a frictional force acts against the motion. Now, the normal force on each of the objects = mgcosθ.
Frictional force = µmg.cosθ
Net force along the incline = mgsinθ – µmgcosθ
= m(gsinθ – µgcosθ)
Acceleration of each of the block = m(gsinθ – µgcosθ)/m
= (gsinθ – µgcosθ)
Since the accelerations and the initial velocities are same the time taken to reach the bottom will be the same for all objects.
Time can be calculated as in the previous explanation.
In the previous question, the smallest kinetic energy at the bottom of the incline will be achieved by
(a) the solid sphere (b) the hollow sphere
(c) the disc (d) all will achieve same kinetic energy.
Answer: (b)
Since all objects reach the bottom at the same time, so their velocities are also the same hence same K.E. due to linear velocity.
Since each has the same radius and acted upon by equal forces, the torque on each of them is also the same. But the moment of Inertia is different for them. Since the mass is the most away from the center in the hollow sphere (or may say that the radius of gyration is greater for the hollow sphere) its M.I. is greater. Therefore it will have the smallest angular speed.
Rotational K.E. ∝ ω²
Hence it will have the smallest K.E.
Q#26
A string of negligible thickness is wrapped several times around a cylinder kept on a rough horizontal surface. A man standing at a distance l from the cylinder holds one end of the string and pulls the cylinder towards him (figure 10-Q3). There is no slipping anywhere. The length of the string passed through the hand of the man while the cylinder reaches his hands is
(a) l (b) 2l (c) 3l (d) 4l
Consider a vertical diameter of the cylinder. While rolling, at any instant, the lowest point of the diameter is in contact with the surface and does not move, The center moves a certain distance say x, and the top-most point of the diameter moves twice that = 2x. So when the center of the cylinder moves a distance l, the string length that unfolds through the top = 2l
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