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 Hookes law then the new period will be

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 Hookes law then the new period will be (A) √(4/3) T (B) (32/42) T (C) (√3/2) T (D)  (3/4)T

Tardigrade Admin · Accepted Answer

For a body executing circular motion, centripetal force is given by

F = m r ω^{2} = m r (\frac{2 π}{T})^{2}

...(i)
If

k

is force constant of string, then elastic force

F = ka

...(ii)
From Eqs. (i) and (ii), we get

ka = m (2 a) (\frac{2 π}{T})^{2}

...(iii)
In second case New length of string = new radius of circle

= 3 a

Stretching of string

= 3 a - a = 2 a

Hence, elastic force

= k \cdot 2 a

So,

k \cdot 2 a = m (3 a) (\frac{2 π}{T ^{'}})^{2}

...(iv)
Dividing Eq. (iv) by Eq. (iii), we get

2 = \frac{3}{2} (\frac{T}{T ^{'}})^{2}

\Rightarrow (\frac{T}{T ^{'}})^{2} = \frac{4}{3}

\Rightarrow T^{'} = \frac{3}{2} T

$\frac{4}{3} T$

$\frac{3 ^{2}}{4 ^{2}} T$

$\frac{3}{2} T$

$\frac{3}{4} T$

Q. 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 Hookes law then the new period will be

34​​T

4232​T

23​​T

43​T

$\frac{4}{3} T$

$\frac{3 ^{2}}{4 ^{2}} T$

$\frac{3}{2} T$

$\frac{3}{4} T$