Question

)The angular speed of the motor wheel is increased from 3120 rpm to 6240 rpm in 16 seconds
(i) what is the angular acceleration (ii) how many revolutions does the engine make during this time.

Answer 1

Explanation:

i) We shall use ω=ω

0

+αt

ω

0

= initial anugular speed in rad/s

=2π× angular speed in rev/s

=

60s/min

2π×angular speed in rev/min

=

2π×3120/60

rad/s

=104

Similarly ω= final angular speed in rad/s

=

2π×6240/60

rad/s

=208πrad/s

∴ Angular acceleration

α=

t

ω−ω

0

=4πrad/s

2

The angular acceleration of the engine =4πrad/s

2

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

θ=ω

0

t+

2

1

αt

2

=(40π×16+

2

1

×4π×16

2

) rad

=(640π+512π)rad

=1152πrad

Number of revolutions =

2π

1152π

=576

Answer 2

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

Number of revolution = 576