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/s2. 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
(a) 10 cm/s, 10 cm/s2
(b) 10 cm/s, 80 cm/s2
(c) 40 cm/s, 10 cm/s2
(d) 40 cm/s, 40 cm/s2

Answer 1

The new value of the speed and acceleration is  10 cm/s, 10 cm/s2.

Explanation:

Step 1:

The turntable is given to be moving at uniform angular velocity. Let the speed be changed.

We have:  

$v=r \omega$

Where  

V is the velocity measured in meter per second  

r  is the circle radius  

$v \propto r$  ( for constant ω )

$\frac{v}{v^{\prime}}=\frac{r^{2}}{r^{\prime}}$

Step 2:

Given data in the question

$\mathrm{v}=20 \mathrm{cm} / \mathrm{second}$

$a=20 \mathrm{cm} / \mathrm{s}^{2}$

$v^{\prime}=\frac{v}{2}=\frac{20}{2}=10 \frac{\mathrm{cm}}{\mathrm{s}}$

Step 3:

Similarly,we have:  

$a^{\prime}=\frac{a}{2}=\frac{20}{2}=10 \mathrm{cm} / \mathrm{s}^{2}$