A record turntable rotating at $\mathbf{33}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{}\mathbf{}\mathbf{\text{rev}}\mathbf{/}\mathbf{\text{min}}$slows down and stops in $\mathbf{30}\mathbf{}\mathbf{}\mathbf{\text{s}}\mathbf{}$after the motor is turned off.

(a) Find its (constant) angular acceleration in revolutions per minute-squared.

(b) How many revolutions does it make in this time?

We use the concept of angular kinematics. Using a kinematic equation, we can find the angular acceleration of the turntable for the given time. Also we can find the number of revolutions, that is, angular displacement for the given time using the displacement equation.

Formulae:

1. ${\omega }_f={\omega }_i+\alpha \text{t}$
   
2. $\theta ={\omega }_i\text{t}+\left(\frac{1}{2}\right)\alpha {\text{t}}^2$
   

1. Initial angular velocity is${\omega }_i=33.33\text{rev}/\text{min}$
2. Time is given as$\text{t}=30\text{s}$

We know ${\omega }_f=0$as turntable stops after$30\text{s}=0.5\text{min}.$We get

${\omega }_f={\omega }_i+\alpha \text{t}$

$0=33.33+\alpha \left(0.5\right)$

$\alpha =-\frac{33.33}{0.5}=-66.7\text{rev}/{\text{min}}^2$

$\alpha =-67\text{rev}/{\text{min}}^2$

We got the value of angular acceleration; we can find the angular displacement.

$\theta ={\omega }_i\text{t}+\left(\frac{1}{2}\right)\alpha {\text{t}}^2$

$\theta =\left(33.33\right)\left(0.5\right)+0.5\left(-67\right){\left(0.5\right)}^2$

$\theta =8.33\text{rev}$

Final statement:

We can use the concept of angular kinematics. Using angular kinematic equations, we can find the angular acceleration and angular displacement.