Question

A toy cart is tied to the end of an unstretch string of length $a$. When revolved the toy cart moves in a horizontal of radius $2a$ with time period $T$. Now they toy cart is speeded up until in a horizontal circle of radius $3a$ with period of $T$. If Hooke's law is obeyed throughout then

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

Answer 1

The correct option is D $T^{\prime }=\frac{\sqrt{3}}{2}T$

Centripetal force for a body in circular motion is given by

$F=mr{\omega }^2=mr(\frac{2\pi }{T}{)}^2$

Elastic force of spring = ka , where, k = force constant of the spring

$\therefore ka=m\left(2a\right)(\frac{2\pi }{T}{)}^2$

In the second case, radius of circle = 3a-a = 2a

$\therefore k\left(2a\right)=m\left(3a\right)(\frac{2\pi }{T^{\prime }}{)}^2$

So,

$\therefore k\left(2a\right)=m\left(3a\right)(\frac{2\pi }{T^{\prime }}{)}^2$/$ka=m\left(2a\right)(\frac{2\pi }{T}{)}^2$

$\therefore ka=m\left(2a\right)(\frac{2\pi }{T}{)}^2$

$T^{\prime }=\frac{\sqrt{3}}{2}T$