Question

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?

Short answer

1. Angular acceleration is $\alpha =-67\text{rev}/{\text{min}}^2$
2. Number of revolutions in the given time is$\theta =8.33\text{rev}$

Step-by-step solution

Step1: Explanation of Concept

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

Given

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

Calculations

a. Angular acceleration:

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

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

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

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

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

b. Number of revolutions in the given time:

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.