Question

The angular speed of the motor wheel is increased from 3120 rpm to 6240 rpm in 16 seconds

Answer 1

Explanation:

correct question :

The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16 seconds.

(i) What is its angular acceleration to be uniform? (ii) How many revolutions does the engine make during this time?

Answer :

(i) We shall use $\sf \omega = \omega_0 + \alpha \: t$

$\omega_0$=initial anugular speed in rad/s

$\omega_0$=2π× angular speed in rev/s

$\omega_0 \sf = \frac{2\pi \times angular \: speed \: in\:rev/min}{60s/min}$

$\sf\omega_0 \: = \frac{2\pi \times 1200}{60} = 40\pi \: rad/s$

Similarly,

$\sf \: \omega = \frac{2\pi \times 3120}{60} = 104 \: rad /s$

∴ Angular acceleration

$\sf \alpha = \frac{ \omega - \omega_0}{t} = 4\pi \: rad /{s}^{2}$

The angular acceleration of the engine = 4πrad/s²

(ii) The angular displacement in time t is given by

$\sf\theta = \omega_0t + \frac{1}{2} \alpha {t}^{2} </p><p></p><p>$

$\sf = 40\pi \times 16 + \frac{1}{2} \times 4\pi \times {16}^{2}$

$\sf = (640\pi + 512\pi)rad$

$\sf = 1152\pi \: rad$

Number of revolutions $\sf \: = \frac{1152\pi}{2\pi} = 576$