Angular Acceleration Formula (DP IB Physics: HL): 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.

Answer:

Step 1: List the known quantities

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

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

• Angular acceleration, $\alpha$ = 0.8 rad 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)

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

Step 3: Select the most appropriate kinematic equation

• We know the values of ${\omega }_i$, ${\omega }_f$ and $\alpha$, and we are looking for angular displacement $\theta$, so the best equation to use would be

${{\omega }_f}^2={{\omega }_i}^2+2\alpha ∆\theta$

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

$0={{\omega }_i}^2-2\alpha ∆\theta$

$∆\theta =\frac{{{\omega }_i}^2}{2\alpha }=\frac{{\left({\displaystyle \frac{3\mathrm{\pi }}{2}}\right)}^2}{2\times 0.8}$

Angular displacement, $∆\theta$ = 13.88 rad

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

• There are 2π radians in 1 rotation

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

• This means the turntable spins 2.2 times before coming to a stop