Question

A disk is rotating around an axis located at its center. The angular velocity is 0.6 rad/s and the angular acceleration is 0.3 rad/s^2. The radius of the disk is 0.2 m. What is the magnitude of the acceleration at a point located on the outer edge of the disk, in units of m/s?

Answer 1

`a=0.0937 \ m/s^2`

Step-by-step explanation:

Given that

Angular velocity (ω)= 0.6 rad/s

Angular acceleration (α)= 0.3
`rad/s^2`

Radius (r)= 0.2 m

We know that is disc is rotating and having angular acceleration then it will have two acceleration .one is radial acceleration and other one is tangential acceleration.

So

`Radial\ acceleration(a_r)=\omega ^2r\ m/s^2`

`Radial\ acceleration(a_r)=0.6^2 *0.2 \ m/s^2`

`(a_r)=0.072 \ m/s^2`

`Tangential\ acceleration(a_t)=\alpha r\ m/s^2`

`Tangential\ acceleration(a_t)=0.3* 0.2\ m/s^2`

`(a_t)=0.06 \ m/s^2`

So the total acceleration ,a

`a=√(a_t^2+a_r^2)\ m/s^2`

`a=√(0.06^2+0.072^2)\ m/s^2`

`a=0.0937 \ m/s^2`