Solveeit Logo
Login

Question

Physics Question on System of Particles & Rotational Motion

A rotating wheel changes angular speed from 1800 rpm1800\,rpm to 3000 rpm3000\, rpm in 20 s20\,s. What is the angular acceleration assuming to be uniform ?

A

60 π rad s−260\,\pi\, rad\,s^{-2}

B

90 π rad s−290\,\pi\, rad\,s^{-2}

C

2 π rad s−22\,\pi\, rad\,s^{-2}

D

40 π rad s−240\,\pi\, rad\,s^{-2}

Answer

2 π rad s−22\,\pi\, rad\,s^{-2}

Explanation

Solution

We know that
ω=2πn\omega =2 \pi n
∴ω1=2πn1\therefore \omega_{1}=2 \pi n_{1}
where n1=1800 rpmn_{1}=1800\, rpm
n2=3000 rpmn_{2}=3000\, rpm
Δt=20 s\Delta t=20\, s
ω1=2π×180060=2π×30=60π\omega_{1}=2 \pi \times \frac{1800}{60}=2 \pi \times 30=60 \pi
Similarly, ω2=2πn2=2π×300060\omega_{2}=2 \pi n_{2}=2 \pi \times \frac{3000}{60}
=2π×50=2 \pi \times 50
=100π=100 \pi
If the angular velocity of a rotating wheel about on axis changes by change in angular velocity in a time interval Δt\Delta t, then the angular acceleration of rotating wheel about that axis is
α= Change in angular velocity  Time interval \alpha =\frac{\text { Change in angular velocity }}{\text { Time interval }}
α=ω2−ω1Δt\alpha =\frac{\omega_{2}-\omega_{1}}{\Delta t}
=100π−60π20=\frac{100 \pi-60 \pi}{20}
=40π20=\frac{40 \pi}{20}
=2π rad/s2=2 \pi\, rad / s ^{2}