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 $T_0$. 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 $T_0$.

• A. $T=T_0$
• B. $T=\frac{\sqrt{3}}{2}{T}_0$
• C. $T=\frac{2T_0}{\sqrt{3}}$
• D. $T=\frac{3}{4}{T}_0$

Answer 1

The correct option is B $T=\frac{\sqrt{3}}{2}{T}_0$

As the string obeys Hooke's law, we can say that $F=kx$
Also the toy-car attached to the string moves in a circular path
$\therefore$ spring force provides the necessary centripetal force.
$\frac{m{v}^2}{r}=kx$
$\Rightarrow m{\omega }^{2}r=kx$
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
$(\omega {)}^2=\frac{kx}{mr}=\frac{ka}{m\left(2a\right)}=\frac{k}{2m}.....(1)$
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
$({\omega }^{\prime }{)}^2=\frac{k\left(2a\right)}{m\left(3a\right)}=\frac{2k}{3m}.....(2)$
From $\left(1\right)$ and $\left(2\right)$, we get
${\left(\frac{{\omega }^{\prime }}{\omega }\right)}^2=\frac{\frac{2k}{3m}}{\frac{k}{2m}}=\frac{4}{3}$
$\Rightarrow {\left(\frac{\frac{2\pi }{T}}{\frac{2\pi }{T_0}}\right)}^2=\frac{4}{3}$
or $\frac{T_0}{T}=\sqrt{\frac{4}{3}}$
$\Rightarrow T=\frac{\sqrt{3}}{2}{T}_0$
Thus, option (b) is the correct answer.