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

Answer:

$a=0.0937 \ m/s^2$

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

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

$Radial\ acceleration(a_r)=0.6^2 \times0.2 \ m/s^2$

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

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

$Tangential\ acceleration(a_t)=0.3\times 0.2\ m/s^2$

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

So the total acceleration ,a

$a=\sqrt{a_t^2+a_r^2}\ m/s^2$

$a=\sqrt{0.06^2+0.072^2}\ m/s^2$

$a=0.0937 \ m/s^2$