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*t^{2}+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*r^{2}, in this case I=22.5
kgm^{2}. The angular acceleration of the object Aa
is the torque divided by the moment of inertia,

Aa=(3*t^{2}+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*t^{2}+2*t)/22.5.

Integrated from t=0 to t=2 this is

(2^3+2.5*2^{2}+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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