A flywheel whose moment of inertia about its axis of rotation is 16kg-m^(2) is rotating freely in its own plane about a smooth axis through its centre. Its angular velocity is 9rads^(-1) when a torque is applied to bring it to rest in t_(0) seconds find t_(0) if (a). The torque is constant and of magnitude 4N-m (b). The magnitude of the torque after t second is given by kt.
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Given: Moment of Inertia = 16kg m²
Angular velocity = 9 rads
To Find: The torque and the magnitude of torque.
Solution:
a) Angular impulse = Change in angular momentum
Therefore, t0=Iω
t0 = Iω/τ
= 16 × 9/5
= 36
b) ∫rdt = Iω
∫kdt = Iω
= kt²/2 = Iω
= t0 = √Iω√2/k
= √ 16 × 9 √2/k
= 12√2/k
Answer: The magnitude of torque is 12 √2/k
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