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Question

A toy car travels in a horizontal circle of radius 2a. It is kept on the track by a radial elastic string of unstretched length a. The period of rotation of the toy car is T0. Now, the toy car is speeded up until it is moving in a circle of radius 3a. Assuming that the string obeys Hooke's law, find the new period T in terms of T0.

A
T=T0
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B
T=32T0
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C
T=2T03
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D
T=34T0
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Solution

The correct option is B T=32T0
As the string obeys Hooke's law, we can say that F=k x
Also the toy-car attached to the string moves in a circular path
spring force provides the necessary centripetal force.
mv2r=kx
mω2r=kx
Given, unstretched length of the string is a. Initially, the string will be stretched by a length a and the angular speed of the toy car is given by
(ω)2=kxmr=kam(2a)=k2m .....(1)
After speeding up, the toy car moves in a circle of radius 3a, so the string stretches by length 2a. Then, the angular speed of the toy car will be given by
(ω)2=k(2a)m(3a)=2k3m .....(2)
From (1) and (2), we get
(ωω)2=2k3mk2m=43
⎜ ⎜ ⎜2πT2πT0⎟ ⎟ ⎟2=43
or T0T=43
T=32T0
Thus, option (b) is the correct answer.

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