# A simple torque question

A 1000 lb uniform platform is rotated by a green shaft (centered) as shown. The ends of the green shaft are each connected to a motor - one on either side of the platform. If the platform starts out from rest in horizontal position, how much torque is need to rotate this platform 180 degrees in 3 seconds (assume gravity is present)?

*see attachment for picture

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

A 1000 lb uniform platform is rotated by a green shaft (centered) as shown. The ends of the green shaft are each connected to a motor - one on either side of the platform. If the platform starts out from rest in horizontal position, how much torque is need to rotate this platform 180 degrees in 3 seconds (assume gravity is present)?

Solution:

Some new features were added on the figure.

If the platform starts from rest and assuming that a constant torque is applied, then an uniform angular acceleration is obtained, denoted as .

This can be computed using the formula:

( 1)

where t = the time needed to reach the angle , which in our case is t = 3 s for

( 2)

Since the shaft is centered on the platform as in the figure, the gravity will have no effect on the dynamics of the considered system.

By applying formula (1), the angular acceleration will be found:

( 3)

Next, the second Newton's low for the circular motion will be applied:

( 4)

where Jyy = the moment of inertia about y axis (the platform rotates about y axis according to the figure)

Using the notations introduced in the figure, for such a rectangular platform the moment of inertia is given by the formula:

( 5)

The given mass is m = 1000 lb = 453.6 kg.

The sizes a and b are not given, but they can be chosen according to the concrete application.

Finally, the expression of the needed torque will be:

( 6)

where both a and b need to be taken in meters.

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