#### Question

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

10 cm/s, 10 cm/s

^{2 }10 cm/s, 80 cm/s

^{2}40 cm/s, 10 cm/s

^{2}40 cm/s, 40 cm/s

^{2}

#### Solution

(a) 10 cm/s, 10 cm/s^{2}

It is given that the turntable is rotating with uniform angular velocity. Let the velocity be \[\omega\] .

We have:

\[\text{v = r}\omega\]

\[ \Rightarrow \text{v} \propto \text{r} (\text{P for constant }\omega)\]

\[\frac{v}{v'} = \frac{r}{r'}\]

\[ \Rightarrow \text{v}' = \frac{\text{v}}{2} = 10 \text{ cm/s}\]

Similarly, we have: \[a' = \frac{a}{2} = 10\text{ cm/ s}^2\]