Question

An electric drill starts from rest and rotates with a constant angular acceleration. After the drill has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the drill is 8.2 times the magnitude of the tangential acceleration. What is the angle?

Answer 1

Answer:

The angle is 4.1 rad.

         

Explanation:

The centripetal acceleration (α) is given by:

$\alpha = \omega^{2} r$    (1)                  

Where:

ω: is the angular velocity  

r: is the radius

And the tangential acceleration (a) is:                      

$a = \alpha r$      (2)

Since the magnitude of "α" is 8.2 times the magnitude of "a" (equating (2) and (1)) we have:

$\omega^{2} r = 8.2\alpha r  $

$\omega^{2} = 8.2\alpha$    (3)      

Now, we can find the angle with the following equation:

$\omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \Delta \theta$

Where:

$\omega_{f}$: is the final angular velocity                                                                              $\omega_{0}$: is the initial angular velocity = 0 (it starts from rest)

$\Delta \theta$: is the angle

$\omega^{2} = 2\alpha \Delta \theta$     (4)    

By entering equation (3) into (4) we can calculate the angle:

$8.2\alpha = 2\alpha \Delta \theta$

$\Delta \theta = 4.1 rad$

Therefore, the angle is 4.1 rad.

I hope it helps you!