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A puck is moving in a circle of radius r 0 with a constant speed V 0 on a level frictionless table. A string is attached to the puck, which holds it in the circle; the string passes through a frictionless hole and is attached on the other end to a hanging object of mass M. (See Fig.) (a) The puck is now made to move with a speed v 1 = 2 v0 , but still in a circle. The mass of the hanging object is left unchanged. The acceleration a’ of the puch and the radius r’ of the circle are now given by (A) a’ = 4a 0 and r’ = r 0 , (B) a’ = 2a­ 0 and r’ = r 0 , (C) a’ = 2a 0 and r’ = 2r 0 . (D) a’ = a 0 and r’ = 4r 0 . (b) The puck continues to move at speed v' = 2 v0 in a circle, but now the mass of the hanging object is doubled. The acceleration a' of the puck and the radius r' of the circle are now given by (A) a' = 4a 0 and r’ = r 0 . (B) a' = 2a 0 and r’ = r 0 . (C) a' = 2a 0 and r’ = 2r 0 . (D) a' = a 0 and r’ = 4r 0 .

A puck is moving in a circle of radius r0 with a constant speed V0 on a level frictionless table. A string is attached to the puck, which holds it in the circle; the string passes through a frictionless hole and is attached on the other end to a hanging object of mass M. (See Fig.)
(a) The puck is now made to move with a speed v1 = 2v0, but still in a circle. The mass of the hanging object is left unchanged. The acceleration a’ of the puch and the radius r’ of the circle are now given by
(A) a’ = 4a0 and r’ = r0,                            (B) a’ = 2a­0 and r’ = r0,
(C) a’ = 2a0 and r’ = 2r0.                         (D) a’ = a0 and r’ = 4r0.
(b) The puck continues to move at speed v' = 2v0 in a circle, but now the mass of the hanging object is doubled. The acceleration a' of the puck and the radius r' of the circle are now given by
(A) a' = 4a0 and r’ = r0.                       
(B) a' = 2a0 and r’ = r0.
(C) a' = 2a0 and r’ = 2r0.                     
(D) a' = a0 and r’ = 4r0.

Grade:11

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
7 years ago
The correct option is:
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(a) The magnitude of net force acting on the puck is:
236-939_54.PNG
The tension in the string provides the centripetal to the puck and therefore the force F should be equal to tension say T in the string, that is
236-466_55.PNG
Also, the hanging object is in vertical equilibrium, therefore the tension T in the string must be equal in magnitude to its weight and act in the opposite direction such that
T = Mg
Here g is the acceleration due to gravity and M is the mass of the hanging object.
236-629_56.PNG
236-1318_57.PNG
r’ = 4r0
Therefore, the new radius of the circle traced by the puck is 4r0.
It is important to note that the acceleration of the puck does not change with increase in velocity, because by doing so one has also increased the radius by the same factor. As a result there is no net change in the centripetal force experienced by the puck.
Hence, (D) is the correct option and the rest are ruled out.
The radius of the circle traced by the puck when it moves with velocity 2v0 is:


236-922_58.PNG
236-455_59.PNG
Therefore, upon doubling the mass of the hanging object one has reduced the radius of the circle traced by the puck by a factor of 2.
The force say F’ experienced by the puck is given as:

236-230_60.PNG
The mass of the puck being constant, the acceleration of the puck will increase by the same amount as the centripetal force, which is 2.
Therefore the acceleration say a’ of puck will be 2 times its acceleration when it moves with velocity v0 and was supported against the object of mass M , that is a’ = 2a0.
Hence (C) is the correct option and the rest are ruled out.

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