Question

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

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

Answer 1

The correct option is C $\sqrt{\frac{3}{2}}T$

$T=\frac{m{v}^2}{r}=k.x$
$m{\omega }^{2}r=kx$
Initially, ${\omega }^2=\frac{kx}{mr}=\frac{ka}{m\left(2a\right)}=\frac{k}{2m}$ ..... (i)
Secondly, ${\omega }^2=\frac{k\left(a\right)}{m\left(3a\right)}=\frac{k}{3m}$ ...... (ii)
$\therefore {\left(\frac{{\omega }^{\prime }}{\omega }\right)}^2=\frac{k/3m}{k/2m}=\frac{2}{3}$
$\Rightarrow {\left(\frac{2\pi /T^{\prime }}{2\pi /T}\right)}^2=\frac{2}{3}$
or $\frac{T}{T^{\prime }}=\sqrt{\frac{2}{3}}$
$\Rightarrow T^{\prime }=\sqrt{\frac{3}{2}}T$