A wheel undergoes a constant angular acceleration from time t=0 to t=20s and thereafter angular acceleration is zero. If angular velocity at t=2s is found to be 5rad‌/‌s, then the number of revolutions made by the wheel in time interval t=0s to t=50s is
Let ω1 be the initial angular velocity i.e. ω1=0 rads ‌−1 at time t1=0s ω2 be the angular velocity at t2=2s i.e. ω2=5rads−1 Let α be the angular acceleration at t=2s and n be the number of revolution. As we know that, α=‌
∆ω
∆t
∴‌‌α=‌
ω2−ω1
t2−t1
=‌
5−0
2−0
=2.5rads−2 ω3=ω1+αt3 where, ω3 is angular speed at time, t3=20s ∴‌‌ω3=0+2.5×20=50rads−1 If angle swept by body in 1 st 20s be θ. then from equation ω32−ω12=2αθ ∴‌‌502−02‌‌=2×2.5θ θ‌‌=‌
2500
5
=500rad and θ′ be the angle swept by body from time t3=20s ‌ to ‌t4=50s ∴ Remaining time be t=50−20=30s Since, θ′=ω3t+‌
1
2
α′t2 where, α′=0 rads ‌−1 ∴‌‌θ′=ω3t=50×30=1500rad Total angle covered from time 0 to 50s θ‌net ‌‌‌=θ+θ′ ‌‌=500+1500=2000rad ∵‌ Number of revolution ‌‌‌=‌