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Arslan89 · Feb 7, 2012, 11:45 PM

I want to rotate a 500 kg mass supported by a hollow shaft at 10 rps, what is the required amount of torque?

ebaines · Feb 8, 2012, 07:12 AM

If the mecahnism is frictionless, the answer is: zero. Once the mass is rotating torque is needed only to overcome friction. However, if the mass is initially at rest torque is required to get it spinning, and the mount of torque needed is dependent on (a) the rotational moment of inertia of the mass and (b) how quickly you want it to get up to speed. For (a) we need to know the shape and mass distribution of the 500 Kg object. For (b) you need to tell us the amount of time you want to give the motor to reach 10 rps. And then finally we need an estimate from you of the amount of torque presented by friction.

Arslan89 · Feb 8, 2012, 10:45 PM

The 500kg mass is actually a dish type antenna, how can I calculate its rotational moment of inertia(don't know the mass distribution of antenna), and lets say the rotational velocity of 10 rps is required in 2 seconds,the antenna will be mounted on the shaft and the shaft will be supported by bearings so I guess the friction will be negligible.

ebaines · Feb 9, 2012, 06:33 AM

We can make an estimate if you tell us:

a) the diameter of the dish antenna, and
b) the position of the antenna relative to the shaft? In other words - is it centered above the shaft (like an umbrella), or is it hanging off to one side? And if it's hanging off to the side - how far from the shaft?

Arslan89 · Feb 10, 2012, 12:15 AM

The antenna(dia 5ft) is like an umbrella, with its cloth dish inverted,the antenna is centered(welded) at one end of a arm(1m high) and the other end of the arm is mounted on the shaft,such that the load causes the shaft to bend and shear. The shaft has to support the load(500kg dish) and it also has to rotate the load with the help of a motor at required speed for dish orientation.

ebaines · Feb 10, 2012, 06:46 AM

OK - we can do a rough calculation that assumes that the worst case is the antenna starting at rest in the horizontal position, so that all its weight is suspended 1 meter from the axis of rotation. The torque needed can be calculated as follows:

$ \tau = I \alpha $

where I = moment of inertia of the antenna and $\alpha$ = angular acceleration of the system. We can roughly model the antenna as a 500Kg point mass at the end of a 1m arm, so its moment of inertia is
$ I = mr^2 = 500Kg-m^2 $

The acceleration you want is
$ \alpha = 5 \frac { Rev} {s^2} = 10 \pi s^{-2} $

So:

$ \tau = 500 \times 10 \pi Kg m^2/s^2 = 15700 N-m $

That's a lot of toque. You may want to relax the requirement that the system spin up to 10rps in just 2 seconds. Also if the antenna is positioned directly above the axis of rotation instead of hanging off the side the amount of torque needed is cut by about a factor of 4.

asaleem89 · Feb 12, 2012, 10:21 PM

Thanks a lot ebaines for your time and suggestions it really helped me a lot!!