What torque must be applied?

Question

A motor is connected to a solid cylindrical drum with a diameter of 1.2 m and a mass 51 kg. A massless rope is attached to the drum and tied at the other end to a 38 kg weight, so the rope will wind onto the drum as it turns. What toque must be applied if the weight is to be lifted with an acceleration of 1.1 m/s^2?

I know the drum deals with rotational motion and the weight simply with linear motion.

Fnet of the weight is: mg- T(tension)

Torque = I a

I'm not sure where to go from here.

Answer 1

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Answer 2

The moment of inertia of a solid cylinder is

I = 1/2 m r^2 = 1/2 ( 51 kg ) ( 1.2 m )^2 = 37 kg m^2

Linear and angular acceleration are related as

a = r α

so α = ( 1.1 m/s^2 ) / ( 1.2 m ) = .92 rad / s^2

The net torque required to accelerate the drum is then found using Newton's Second Law:

Г = I α = ( 37 kg m^2 ) ( .92 rad / s^2 ) = 34 N m.

Solve for the net torque Г. You already have one torque, ( 38 kg ) ( 9.8 m/s^2 ) ( 1.2 m ) = 450 N m . The applied torque is thus 490 N m.

Answer 3

Well, here's how you do it. First you need to get the tension the rope needs to lift the weight at 1.1 m/s^2. From a free body diagram you get that

-W+F=ma

where W is the weight F is the tension, m is the mass and a is the acceleration you want, so

F = ma + W

F = ma + mg

F = m (a+g)

F = 38kg(1.1m/s^2+9.81m/s^2) = 414.6N

now you do the same for ratational movement, in this case

Sum of Torques = I*a

in this case, I=1/2mr^2

so

I = (51kg*(0.6m)^2)/2 = 9.18 kg*m^2

in this case a is going to be the angular acceleration, so

a=la/r

where la is the linear acceleration

a = 1.1m/s^2/0.6m = 1.8333rad/s^2

now, the sum of torques would be

-Fr+T=Ia

where T is the torque the cylinder needs, so

T = Ia+Fr

T = (9.18kg*m^2)(1.8333rad/s^2)+(414.6N)(0.6...

T = 266N*m

Hope this helps