The moment of inertia of a wheel is 1000 kg m2. Its rotation is uniformly accelerated.
At some instant of time, its angular speed is 10 rad s1. After the wheel has rotatedthrough an angle of 100 radians, the angular velocity of the wheel becomes100 rad s1. Calculate the torque applied to the wheel and the change in its kineticenergy.

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Solution

Given information
Initial angular speed of the wheel $ω_{o} = 10 r a d / s f i n a l a n g u l a r v e l o c i t y o f t h e w h e e l ω = 100 r a d / s a n d a n g u l a r d i s p l a c e m e n t θ = 100 r a d i a n f r o m t h e g i v e n i n f o r m a t i o n w e c a n c a l c u l a t e t h e a n g u l a r a c c e l e r a t i o n b y u \sin g t h e e q u a t i o n ω^{2} = ω_{o}^{2} + 2 α θ (100)^{2} = (10)^{2} + 2 \times 100 \times α \Rightarrow α = \frac{10000 - 100}{200} = \frac{9900}{200} = 49.5 m / s^{2} t o r q u e a c t i n g o n t h e w h e e l = I α = 1000 \times 49.5 = 49500 N m^{2} C h a n g e i n r o t a t i o n a l k i n e t i c e n e r g y = \frac{1}{2} I [ω^{2} - {ω_{O}}^{2}] = \frac{1}{2} \times 1000 (10000 - 100) = 500 \times 9900 = 4950000 J$

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