I have a large boom of a radio antenna weighing $3,500 kg$ which is rotating at $0.5 RPM$. The antenna is a T shape and the boom radius is $14m$. The power to the motor is cut off suddenly and the motor acts as a rigid brake. The boom is not rigid and it continues to move/bend in the same direction for 1 second (at which points it bounces back and fourth until the forces damper out). I am trying to work out the max torque this would apply to the gearbox shaft, assuming there is no friction.

For angular movement, I have designated the centroid of each boom radius as a point load (so a point load at a radius of 7 m on both ends of the boom).

I can work out the angular moment which is $m \cdot r \cdot v$, which works out to $1750 kg \cdot 7m \cdot 0.367m/s$ or $4489.86kg\frac{m^2}{s}$

This is where I am not sure about though. I have read that torque is simply the change in angular momentum over time. So would that be $4489.86kg\cdot m $ $(44.1kN \cdot m)$?

Would this be an accurate way to determine the torque in the shaft? This is for a real scenario so If I have missed any big considerations please let me know.


2 Answers 2


You make 1 major false assumption, and that is that the torque is uniform over the 1 second period of deceleration of the beam. If the motor is truly a rigid break (doesn't back drive at all) then the bottom of the beam is stopping instantaneous (infinite torque) and the far end of the beam stops 1 second later.

To get the shock load right as you stop may require testing or detailed analysis of the rigidity of your gear box, etc.

Even if we assume your idealized scenario where all the mass is at a radius of 7 meters attached to the gear box by a massless beam in bending varies with the deflection. So zero deflection (time t=0) means 0 force on the beam, so 0 torque and maximum deflection would mean maximum force and maximum torque. I suggest using beam in bending calculations to give you the torque curve vs time.



Assuming your antenna's boom is solid then its moment of inertia along the long axis is$$ I = 1/12ml^2 $$ approximately which is less than you attributing the mass to the two ends. The torque this boom causes while stopping and wobbling is $$ \tau = I\alpha $$

In this case is $$ 1/2-0 = 1/2 \space = \pi $$ So your torque is $$1/12\times 3500\times28^2\times \pi = \pi/12\times 28^2\times 3500 \space Nm $$

But We should consider the vibration of the boom and its dynamic whiplash effect two, which is dependant on the stiffness of the boom and its detailed construction such as bolted together or welded together or what not.

codes in similar situations have a dynamic load factor of 2.8.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.