Question

A point on a spinning record experiences a centripetal acceleration of .75 cm/s^2 . How far is the point from center of the record if the record spins at 0.5 revolutions per second?

Answer 1

0.076 cm

Explanation:

Given:

a = 0.75 cm/s^2

angular velocity = 0.5 rps

`\begin{gathered} \omega=0.5\cdot2\pi\text{ rad/s} \\ \omega=\pi\text{ rad/s} \end{gathered}`

If the distance of record from the center be r, then as we know:

`\begin{gathered} a=\omega^2\cdot r \\ \text{ we solve for r} \\ r=(a)/(\omega^2) \\ r=(0.75)/((3.14)^2) \\ r=0.076cm \end{gathered}`

Therefore, this distance would be 0.076 centimeters