Question

A particle is kept fixed on a turntable rotating uniformly. As seen from the ground, the particle goes in the circle, its speed is $20cm/s$ and acceleration is $20cm/{{s}^{2}}$. The particle is shifted to a new position to make the radius half of the original value. The new values of the speed and the acceleration will be:
A) $10cm/s,10cm/{{s}^{2}}$
B) $10cm/s,80cm/{{s}^{2}}$
C) $40cm/s,10cm/{{s}^{2}}$
D) $40cm/s,40cm/{{s}^{2}}$

Answer 1

Hint: Since, there is no effect on the rotation of the turntable and it is unchanged, the angular frequency $\left( \omega \right)$ does not change. The velocity depends directly on the angular velocity and radius of the particle. The acceleration depends on the angular velocity squared and the radius.

Formula used:
$\text{ velocity }\left( v \right)\text{ = Angular velocity}\left( \omega \right)\times \text{Radius}\left( R \right)$
$\therefore v=\omega R$
$\therefore v\propto R$
$\text{ acceleration }\left( a \right)={{\left( \text{Angular velocity} \right)}^{2}}\times \text{Radius}\left( R \right)$
$\therefore a={{\omega }^{2}}R$
$\therefore a\propto {{R}^{2}}$

Complete step-by-step answer:

Let us first analyse the information given to us.
The initial velocity $\left( {{v}_{1}} \right)=20cm/s$ --(1)
The initial acceleration $\left( {{a}_{1}} \right)=20cm/{{s}^{2}}$ --(2)
The initial radius of the particle is $R$ --(3)

After the position of the particle is changed such that the radius is halved.
The final velocity is ${{v}_{2}}$ --(4)
The final acceleration is ${{a}_{2}}$ --(5)
Final radius is $\dfrac{R}{2}$ ---(6)
Now, since there is no change in the rotation of the turntable and there is no effect on it, the angular frequency $\left( \omega \right)$ remains the same.
The velocity depends directly on the angular velocity and radius of the particle. The acceleration depends on the angular velocity squared and the radius.
$\text{ velocity }\left( v \right)\text{ = Angular velocity}\left( \omega \right)\times \text{Radius}\left( R \right)$
$\Rightarrow v=\omega R$
$\Rightarrow v\propto R$ --(7)
$\text{ acceleration }\left( a \right)={{\left( \text{Angular velocity} \right)}^{2}}\times \text{Radius}\left( R \right)$
$\Rightarrow a={{\omega }^{2}}R$
$\Rightarrow a\propto R$ --(8)
Using (7), we get
$\dfrac{{{v}_{2}}}{{{v}_{1}}}=\dfrac{\dfrac{R}{2}}{R}=\dfrac{1}{2}$
$\Rightarrow {{v}_{2}}=\dfrac{{{v}_{1}}}{2}=\dfrac{20}{2}=10cm/s$
Hence, the new velocity is $10cm.{{s}^{-1}}$.
Using (8) we get,
$\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{\left( \dfrac{R}{2} \right)}{{{R}^{{}}}}=\dfrac{\dfrac{{{R}^{{}}}}{2}}{{{R}^{{}}}}=\dfrac{1}{2}$

$\therefore {{a}_{2}}=\dfrac{{{a}_{1}}}{2}=\dfrac{20}{2}=10cm/{{s}^{2}}$
Hence, the new acceleration is $10cm/{{s}^{2}}$
Therefore, the correct option is A) $10cm/s,10cm/{{s}^{2}}$.

Note: Students must keep in mind that the angular velocity (or angular frequency of the particle does not change. It can only change if there is a change in the speed of rotation of the turntable itself. Students often get confused at this point and commit silly mistakes.