Physics · Equilibrium of rigid bodies, rigid body rotation and equations of rotational motion, comparison of linear and rotational motions
The moment of inertia of a flywheel is which is initially stationary. constant e
The moment of inertia of a flywheel is \( 0.2 k g m^{2} \) which is initially stationary. constant external torque \( 5 N m \) acts on the wheel. The work done by this torque during 10 sec is:
- A. \( 1250 J \)
- B. \( 2500 J \)
- C. \( 5000 J \)
- D. \( 6250 J \)
Step-by-step solution
Given I=0.2 kg m², τ=5 N m, t=10 s. Angular acceleration α=τ/I=25 rad/s². Initial ω=0, so after t, ω=αt=250 rad/s. Angular displacement θ=½αt²=1250 rad. Work done by torque = τθ = 5×1250 = 6250 J. Alternatively, change in rotational KE = ½Iω² = 0.5×0.2×250² = 6250 J.
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