# A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of -4.60...

## Question:

A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of -4.60 rad/s{eq}^2 {/eq}. During a 2.20-s time interval, the wheel rotates through 52.9 rad. What is the angular speed of the wheel at the end of the 2.20-s interval?

## Rotational motion

The motion of a rigid body which rotates about its own axis is called Rotational motion.

All the formula and derivation in rotational motion is the same as in case of linear motion for rigid bodies.

The equations of linear motion are-

{eq}(1)\ \ \ \theta \ =\ \omega_0 t+\dfrac{1}{2}\alpha t^2\\ (2)\ \ \ \omega_{\theta}\ =\ \omega_0+\alpha t\\ {/eq}

Given:-

• {eq}\alpha \ =\ -4.60\ rad/s^2 {/eq} = angular acceleration.
• {eq}t \ =\ 2.20\ s {/eq} = time interval.
• {eq}\theta \ =\ 52.9^0. {/eq}

From formula,

We can findout the initial velocity.

The initial angular velocity is-

{eq}\theta \ =\ \omega_0 t+\dfrac{1}{2}\alpha t^2\\ \Rightarrow 52.9\ =\ \omega _0(2.20)+\dfrac{1}{2}(-4.60)(2.20)^2\\ \Rightarrow \omega_0\ =\ \dfrac{52.9+11.132}{2.20}\ =\ 29.1054\ rad/s.\\ {/eq}

Now,

From another formula,

The final angular velocity is

{eq}\omega_{\theta}\ =\ \omega_0+\alpha t\\ \Rightarrow \omega_{\theta }\ =\ 29.1054-4.60(2.20)\ =\ 29.1054-10.12\ =\ 18.9854\ rad/s.\\ {/eq}