Rotating Particle With Time Dependent Torque

Question:

An object of negligible size and 10kg mass is attached to a rod of negligible mass and 1.5 meter length. The rod is pivoted at the end opposite the object and a torque T=3*t²+5*t+2 Nm is applied, where t is in seconds. At t=0 the object has a velocity of 2m/s. Determine the velocity of the object at t=2.

Answer:

The moment of inertia, I, of a single particle is m*r², in this case I=22.5 kgm². The angular acceleration of the object Aa is the torque divided by the moment of inertia,
Aa=(3*t²+5*t+2)/22.5.
The angular velocity at time t is the integral of the angular acceleration from time zero to time t, plus any initial angular velocity. The initial angular velocity is the initial tangential velocity divided by the radius, 2/1.5=1.333 radians/s.

The integral of Aa is
(t^3+5/2*t²+2*t)/22.5.
Integrated from t=0 to t=2 this is
(2^3+2.5*2²+2*2)/22.5 =
(22)/22.5 = 0.978 radians per second.

To this we must add the initial angular velocity of 1.333 radians per second to get 2.311 radians per second. To convert this to the tangential velocity we must multiply by the 1.5 meter radius to get v=3.47m/s

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