Angular Acceleration Equations (AQA A Level Physics): Revision Note

The turntable of a record player is spinning at an angular velocity of 45 RPM just before it is turned off. It then decelerates at a constant rate of 0.8 rad s⁻².

Determine the number of rotations the turntable completes before coming to a stop.

[4]

Answer:

Step 1: List the known quantities

• Initial angular velocity, ${\omega }_1=45\mathrm{RPM}$

• Final angular velocity, ${\omega }_2=0$

• Angular acceleration, $\alpha =-0.8\mathrm{rad}{\mathrm{s}}^{-2}$

• Angular displacement, $∆\theta$ = ?

Step 2: Convert the angular velocity from RPM to rad s⁻¹

• One revolution corresponds to 2π radians, and RPM = revolutions per minute, so

$\omega =2\mathrm{\pi }f$ and $f=\frac{RPM}{60}$(to convert to seconds)

${\omega }_1=\frac{2\mathrm{\pi }\times \mathrm{RPM}}{60}=\frac{2\mathrm{\pi }\times 45}{60}=\frac{3\mathrm{\pi }}{2}\mathrm{rad}{\mathrm{s}}^{-1}$ [1 mark]

Step 3: Select the most appropriate kinematic equation

• The values of ${\omega }_1$, ${\omega }_2$ and $\alpha$ are known, and the angular displacement $\theta$ is needed, so the best equation to use is

${{\omega }_2}^2={{\omega }_1}^2+2\alpha ∆\theta$

Step 4: Rearrange and calculate the angular displacement $∆\theta$

$0={{\omega }_1}^2+2\alpha ∆\theta$

$-2\alpha ∆\theta ={{\omega }_1}^2$

$∆\theta =\frac{{{\omega }_1}^2}{-2\alpha }=\frac{{\left({\displaystyle \frac{3\mathrm{\pi }}{2}}\right)}^2}{-2\times -0.8}$ [1 mark]

Angular displacement, $∆\theta =13.88\mathrm{rad}$ [1 mark]

Step 5: Determine the number of rotations in $∆\theta$

• There are 2π radians in one rotation

• Therefore, $\mathrm{number}\mathrm{of}\mathrm{rotations}=\frac{13.88}{2\mathrm{\pi }}=2.2$

• This means the turntable spins 2.2 times before coming to a stop [1 mark]