Question

A rotating object starts from rest and has a constant angular acceleration. three secondslater the centripedal acceleration of a point on the object is 2m/s^2 . what is the magnitude of the centripedall acceleration of this point six seconds after the motion beings?

Answer 1

To solve this problem we will apply the kinematic equations of angular motion, we will begin by defining the centripetal acceleration which is given by,

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

Here,

v = Tangential velocity

r = Radius

The tangential velocity is also,

$v = r\omega$

Where $\omega$ is the angular velocity, then the expression of centripetal acceleration is,

$a = r\omega^2$

At the same time the angular velocity can be expressed in terms of angular acceleration as

$\omega = \alpha t$

At t = 3s

$\omega = 3\alpha$

Then at t = 6s

$\omega = 3\alpha + \alpha (6-3)$

$\omega = 6\alpha$

Since r is constant, we have proportionally that the centripetal acceleration is proportional to the square of the angular velocity

$a \propto \omega^2$

Then the relation between two acceleration is

$\frac{a_2}{a_1} =(\frac{\omega_2}{\omega_1})^2$

At 6s the centripetal acceleration would be

$a_2 = (\frac{6\alpha}{3\alpha})^2 (2)$

$a_2 = 8m/s^2$

Therefore the magnitude of the centripedal acceleration of this point six seconds after the motion beings is $8m/s^2$