Q. 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 values of the speed and acceleration will be :

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Solution:

The angular velocity and angular acceleration remains constant :
$\Rightarrow \frac{v}{r} =\frac{v^{\prime}}{r^{\prime}}\,\,\,\,(v=\omega r)$
$\frac{20}{r} =\frac{v^{\prime}}{r / 2}$
$v^{\prime} =10\, cm / s$
Similarly, $ \frac{a}{r} =\frac{a^{\prime}}{r^{\prime}} $
$\frac{20}{r} =\frac{a^{\prime}}{r / 2}$
$a^{\prime}=10\, cm / s ^{2}$