A particle is securely placed on a uniformly rotating turntable. When observed from the ground, the particle travels in a circle with a speed of 20 cm/s and an acceleration of 20 cm/s^2 . The particle is subsequently

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Correct answer: C. 10 cm/s , 10 cm/s^2

For a particle on a uniformly rotating turntable, the angular velocity

\omega

remains constant.

The initial speed is given by

v_1 = \omega r_1 = 20 \text{ cm/s}

.

The initial centripetal acceleration is given by

a_1 = \omega^2 r_1 = 20 \text{ cm/s}^2

.

When the particle is relocated such that the new radius is

r_2 = \dfrac{r_1}{2}

, the new speed

v_2

is:

v_2 = \omega r_2 = \omega \left(\dfrac{r_1}{2}\right) = \dfrac{v_1}{2} = \dfrac{20}{2} = 10 \text{ cm/s}

The new acceleration

a_2

is:

a_2 = \omega^2 r_2 = \omega^2 \left(\dfrac{r_1}{2}\right) = \dfrac{a_1}{2} = \dfrac{20}{2} = 10 \text{ cm/s}^2

A particle is securely placed on a uniformly rotating turntable....

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