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December 20, 2014

December 20, 2014

Posted by **BIGEYE** on Saturday, March 24, 2007 at 7:46am.

A hoist lifts a 3.5kg load from rest, and has a final velocity of 2 m/s after 4 seconds. The motor can develop a torque of 4 Nm. Determine:

i) The linear acceleration of the mass.

ii) The angular acceleration of the mass.

I have calculated the linear acceleration to be 0.5 m/s^-2, how can I calculate the angular acceleration from this.

TIA

Your answer for the linear acceleration is correct. The question should have asked what is the angular acceleration of the MOTOR. The mass that is being lifted is not roatating. What rotates is the cylindrical drum of the hoisting pulley.

The work done in 4 seconds, ignoring the moment of inertia of the hoist and friction, is (1/2)(3.5)^2^2 = 7 J

You could assume a radius for that drum, but it will cancel out in the end. Here is another way to use the torque:

(Torque) x (angle of rotation of the hoist pulley wheel) = Work done

Angle of rotation = 7/3.5 = 2 radians

2 = (1/2)*(angular acceleration)* t^2

Angular acceleration = 1.0 rad/s^2

Actually, the angular acceleration I computed is for the motor shaft where the torque is specified. To compute the angular acceleration anywhere else will require gear transmission ratios. If you were provided with a diagram, one could figure out angular acceleration from the radius of the pulley hoist and the linear acceleration

Thanks for that, but by transposing, shouldn't it be 2 = 0.5 * a * t^2

a = 2/8

a = 0.25 r/s^2

That's what I'm after, the motor shaft angular acceleration. Can you recheck please, as you got 1 rad/s^2, but my calcs give

2 = 0.5 * a * t^2

2 = 0.5 * 4^2 * a

a = 2/8

a = 0.25 r/s^2

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