Question

A phonograph record accelerates from rest to 28.0 rpm in 5.73 s.

Answer 1

Answer:

a) $\alpha=0.5117\ rad.s^{-2}$

b) $\theta=8.4\ rad$

Explanation:

Given:

• initial rotational speed of phonograph, $\omega_i=0\ rad.s^{-1}$

• final rotational speed of phonograph, $N_f=28\ rpm \Rightarrow \omega_f=2\pi\times\frac{28}{60} =2.932\ rad.s^{-1}$

• time taken for the acceleration, $t=5.73\ s$

a)

Now angular acceleration:

$\alpha=\frac{\omega_f-\omega_i}{t}$

$\alpha=\frac{2.932-0}{5.73}$

$\alpha=0.5117\ rad.s^{-2}$

b)

Using eq. of motion:

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$\theta=0+\frac{1}{2}\times 0.5117\times 5.73^2$

$\theta=8.4\ rad$