Question

an old 59 rpm record rotates through an angle of 300° as it slows down uniformly from 59.0 rpm to 17.0 rpm. what is the magnitude of the angular acceleration of the record?

Answer 1

3.34

Step-by-step explanation:

The relationship between final angular speed, initial angular speed, angular acceleration, and change in radians is given by

`\omega^2_f=\omega^2_i+2\alpha(\theta_f-\theta_i)`

Now in our case

`\begin{gathered} \omega_i=59rpm \\ \omega_i=59*(2\pi)/(60)\text{rad}/s \\ \end{gathered}`
`\omega_f=17*(2\pi)/(60)\text{rad}/s`
`\theta_f-\theta_i=300^o=(5)/(3)\pi(radians)`

Therefore, the above equation gives

`(17*(2\pi)/(60))^2=(59*(2\pi)/(60))^2+2((5)/(3)\pi)\alpha`

subtracting (17 2pi/ 60 )^2 from both sides gives

`(17*(2\pi)/(60))^2-(59*(2\pi)/(60))^2=2((5)/(3)\pi)\alpha`

dividing both sides by 2 * 5 / 3 pi gives

`((17*(2\pi)/(60))^2-(59*(2\pi)/(60))^2)/(2((5)/(3)\pi))=\alpha`

simplifying the above gives

`\alpha=3.34`

which is our answer!