Rotational Motion

Rotational motion is described as the turning motion of the object caused by a net torque applied to the object. It is measured by the angular displacement, angular velocity, and angular acceleration of the object.

The angular displacement is the angle \(\theta\) or the distance covered by the object as it turns. The change in the distance covered by the object over the change in time is the angular velocity. The angular acceleration is the change in angular velocity per change in time. The base unit for rotational motion is radians.

When an object is rotating, the angular velocity is uniform all throughout the motion. However, at some point of the object, there’s difference in the linear velocity. This depends on the radius from each point in the object.

Here is a list of equations in measuring the rotational motion of an object.

angular displacement: \(\theta\)

angular velocity: \(\omega=\frac{\Delta \theta}{\Delta t}\)

angular acceleration: \(\alpha=\frac{\Delta \omega}{\Delta t}\)

linear velocity: \(v=\omega r\)


 

Example 1.

A disc rotates at \(45\;rpm\). What is its angular velocity? If it takes \(3.5\;s\) to change its velocity, what is the angular acceleration of the disc?

Solution 1.

Given 45 revolutions per minute. This this the change in the angle \(\theta\) over the change in time \(t\). Thus, this gives the angular velocity of the disc. To solve, we have to convert \(rpm\) to \(rad/s\) since radians is the base unit of rotational motion. We use the conversion factors: \(1\; revolution = 2\pi\;radians\) and \(1\;min=60\;s\). So,

\(\omega=\frac{45\;revs}{min}\times \frac{2\pi\; rad}{1\;rev}\times \frac{1\;min}{60\;s}=4.7\;rad/s\).

Therefore, the angular velocity of the disc is \(4.7\;rad/s\).


Solution 2.

The disc’s angular acceleration is the change in velocity over the change in time, \(\alpha=\frac {\Delta \omega}{\Delta t}\).

\(\alpha=\frac{4.7\;rad/s}{3.5\;s}=1.34\;rad/s^2\)

Thus, the disc accelerates at a rate of \(1.34\;rad/s^2\).


Example 2.

What is the linear velocity at each point in the disc.

\(r_1=0\\ r_2=0.02\;m\\ r_3=0.06\;m\)

Solution 1:

At \(r=0\), the linear velocity \(v=\omega r\) is,

\(v=(4.7\;rad/s)(0\;m)=0\; m/s\)


Solution 2:

At \(r=0.02\;m\),

\(v=(4.7\;rad/s)(0.02\;m/s)=0.094\;m/s\)


Solution 3:

At \(r=0.06\;m\),

\(v=(4.7\;rad/s)(0.06\;m)=0.28\;m/s\)