Question

A toy-car travels in a horizontal circle of radius $2a$, kept on the track by radial elastic string of unstretched lenght $a$. 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 obey Hooke's law, the new period $T$ is given as

• 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

Answer: (B)

$T=\frac{\sqrt{3}}{2}{T}_0$

Time Period, $T=\frac{2\pi }{\omega }$
Centripetal Force $=m{\omega }^{2}r$
For radius 2a:
Time period is $T_O=\frac{2\pi }{{\omega }_o}\therefore {\omega }_O=\frac{2\pi }{T_o}$
Centripetal force=Spring force
$(2a-a)=ka=m{{\omega }_O}^2\left(2a\right)-\left(1\right)$
For radius 3a:
Centripetal force=Spring force
$(3a-a)=2ka=m{\omega }^{2}r-\left(2\right)$
From (1) & (2)
$4m{{\omega }_O}^{2}a=3m{\omega }^{2}a\Rightarrow {\omega }^2=\frac{4}{3}{{\omega }_O}^2\Rightarrow \omega =\frac{2}{\sqrt{3}}{\omega }_O\Rightarrow \frac{2\pi }{T}=\frac{2}{\sqrt{3}}\frac{2\pi }{T_o}\Rightarrow T=\frac{\sqrt{3}}{2}{T}_o$