Rotational Dynamics

Does anyone know how to work out the solutions to the following?

An electric motor is used to drive the drum that raises a cage up a mineshaft. The pulley on the motor has a diameter of 160mm, and the drum has a diameter of 800mm. When the motor is turning at a speed of 720 rpm, the tension F1 on the tight side of the drive belt is 1400N, and on the slack side of the belt the tension F2 is 400N.

For this system calculate:

1. The angular velocity of the motor pulley
2. The angular velocity of the drum (hint find the gear ratio of the system)
3. The net torque developed by the motor
4. The net torque exerted by the drum
5. The power delivered by the drum as it raises its load.

Reply 1

Euclid

danette

Not absolutely sure about this one, but may have the following structure.

To find the angular velocity we use: v = rω (v = angular velocity (m/s), r = radius (m) and ω is the angular frequency (rad/s))

First convert the turning speed from rpm to radians per second.

720 rpm = 720 x 2π rad per min
= 24π rad /s
» ω = 24 π rad/s

Using this go straight to the angular velocity equation and input the radius of the motor (in METRES) and the angular frequency above.

2) This time do exactly the same as in 1) but plug in the radius of the drum.

3) Torque (moment) = magnitude of force x perpendicular dist. from pivot.

The 'net' torque would be the overall resultant torque:

net torque = torque (tight belt) - torque (slack belt)

= 0.8(1400 - 400) - perpend. dist. from pivot = 0.8m (half diam.)

= 800 Nm

4) Not sure, could be same as above, but replace distance from pivot.

5) Power = Force x angular velocity.

Hope this helps a bit.