Question

A horizontal disk with a radius of 10 cm rotates about a vertical axis through its center. The disk starts from rest at t = 0 and has a constant angular acceleration of 2.1 rad/s^2. At what value of t will the radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude?

Answer 1

Answer:

0.69s

Explanation:

10 cm = 0.1 m

Let t be the time that radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude. At that time we have the angular velocity would be

$\omega = \alpha t = 2.1 t$

And so the radial acceleration is

$a_r = \omega^2 r = (2.1t)^2 r = 2.1^2 t^2 * 0.1= 0.441 t^2 m/s^2$

The tangential acceleration is always the same since angular acceleration is constant:

$a_t = \alpha * r = 2.1 * 0.1 = 0.21 m/s^2$

For these 2 quantities to be the same

$a_r = a_t$

$0.441 t^2 = 0.21$

$t^2 = 0.21/0.441 = 0.4762$

$t = \sqrt{0.4762} = 0.69 s$