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Wewaii [24]
3 years ago
9

Consider a motor that exerts a constant torque of 25.0 nâ‹…m to a horizontal platform whose moment of inertia is 50.0 kgâ‹…m2. A

ssume that the platform is initially at rest and the torque is applied for 12.0 rotations. Neglect friction.
Physics
1 answer:
Crazy boy [7]3 years ago
3 0

Answer:

W = 1884J

Explanation:

This question is incomplete. The original question was:

<em>Consider a motor that exerts a constant torque of 25.0 N.m to a horizontal platform whose moment of inertia is 50.0kg.m^2 . Assume that the platform is initially at rest and the torque is applied for 12.0rotations . Neglect friction. </em>

<em> How much work W does the motor do on the platform during this process?  Enter your answer in joules to four significant figures.</em>

The amount of work done by the motor is given by:

W=\Delta K

W= 1/2*I*\omega f^2-1/2*I*\omega o^2

Where I = 50kg.m^2 and ωo = rad/s. We need to calculate ωf.

By using kinematics:

\omega f^2=\omega o^2+2*\alpha*\theta

But we don't have the acceleration yet. So, we have to calculate it by making a sum of torque:

\tau=I*\alpha

\alpha=\tau/I     =>     \alpha = 0.5rad/s^2

Now we can calculate the final velocity:

\omega f = 8.68rad/s

Finally, we calculate the total work:

W= 1/2*I*\omega f^2 = 1883.56J

Since the question asked to "<em>Enter your answer in joules to four significant figures.</em>":

W = 1884J

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