Torque Calculation for Suspended Mass and Cylinder System

[Richard Dumfry]

Homework Statement

A 16 kg mass is suspended from the end of a rope around a 22 kg cylinder with a 6.0 m diameter. What torque applied to the drive axle of the cylinder is required to accelerate the 16 kg mass 2.0 m/s^2 upwards?

Homework Equations

Torque = Radius cross Force

The Attempt at a Solution

Since the problem asks for the an acceleration 2 m/s^2 upwards, it has to counteract gravity. Therefore, the acceleration upwards (which also happens to be the tangential acceleration since it's tangent to the pulley) is 11.81 m/s^2.

From there, I multipled by mass (f = ma) to get force, and multiplied that by the radius, 3. However, my for torque was incorrect. I believe I left out the rotational energy of the pulley, but I am not sure where it goes in my calculations.

 

[haruspex]

I left out the rotational energy of the pulley, but I am not sure where it goes in my calculations.

What's the angular acceleration of the pulley?

 

[PhanthomJay]

Homework Statement

A 16 kg mass is suspended from the end of a rope around a 22 kg cylinder with a 6.0 m diameter. What torque applied to the drive axle of the cylinder is required to accelerate the 16 kg mass 2.0 m/s^2 upwards?

Homework Equations

Torque = Radius cross Force

The Attempt at a Solution

Since the problem asks for the an acceleration 2 m/s^2 upwards, it has to counteract gravity. Therefore, the acceleration upwards (which also happens to be the tangential acceleration since it's tangent to the pulley) is 11.81 m/s^2.

From there, I multipled by mass (f = ma) to get force, and multiplied that by the radius, 3. However, my for torque was incorrect. I believe I left out the rotational energy of the pulley, but I am not sure where it goes in my calculations.

The acceleration of the hanging mass is 2 not 11.81. Draw a free body diagram of the hanging mass, identify forces acting, and apply Newton 2 to find the rope tension. The draw a free body diagram of the pulley, find the net torque and use Newton 2 for rotational acceleration ...you need to determine the mass moment of inertia of the cylinder...

 

[Richard Dumfry]

What's the angular acceleration of the pulley?

Err...should be 11.81 / 3m, since angular acceleration is just tangential acceleration / radius, right? So 3.93?

I tried something like that, and then plugging into torque = I * a, but couldn't get the answer from that either.

 

[Richard Dumfry]

The acceleration of the hanging mass is 2 not 11.81. Draw a free body diagram of the hanging mass, identify forces acting, and apply Newton 2 to find the rope tension. The draw a free body diagram of the pulley, find the net torque and use Newton 2 for rotational acceleration ...you need to determine the mass moment of inertia of the cylinder...

Right, but I thought the tangential acceleration you'd actually have to apply would be 11.81, since gravity is pulling 9.81 in the opposite direction.

 

[PhanthomJay]

Right, but I thought the tangential acceleration you'd actually have to apply would be 11.81, since gravity is pulling 9.81 in the opposite direction.

Stick to the basics of kinematics and Newton's laws. Use tangential acceleration = 2 m/s^2, then angular acceleration is 2/3 radians/sec^2.

Note in a FBD of the hanging mass, the tension pulls up and the weight acts down, hence, T -mg =ma, or T = m(a+g), and a+g equals 11.81. But when you look at the pulley, you must use the tangential acceleration with respect to the ground of 2 m/s^2, not the a+g figure you calculated. Gets you into trouble.

 

[Richard Dumfry]

Stick to the basics of kinematics and Newton's laws. Use tangential acceleration = 2 m/s^2, then angular acceleration is 2/3 radians/sec^2.

Note in a FBD of the hanging mass, the tension pulls up and the weight acts down, hence, T -mg =ma, or T = m(a+g), and a+g equals 11.81. But when you look at the pulley, you must use the tangential acceleration with respect to the ground of 2 m/s^2, not the a+g figure you calculated. Gets you into trouble.

So... 2/3 angular acceleration, and 99 Inertia (Since 1/2 * 22 * 3^2 = 99), then torsion = Inertia * acceleration = 99 * 2/3 = 66?

 

[haruspex]

Right, but I thought the tangential acceleration you'd actually have to apply would be 11.81, since gravity is pulling 9.81 in the opposite direction.

The tension would be the same as for an 11.81 acceleration in a gravity-free environment, but that does not mean the acceleration is actually 11.81 (so it was wrong to use 11.81 for calculating the acceleration of the pulley).

 

[haruspex]

So... 2/3 angular acceleration, and 99 Inertia (Since 1/2 * 22 * 3^2 = 99), then torsion = Inertia * acceleration = 99 * 2/3 = 66?

With the right units, yes. But that that's just the axial torque to accelerate the pulley. You need the total axial torque.

 

[Richard Dumfry]

With the right units, yes. But that that's just the axial torque to accelerate the pulley. You need the total axial torque.

Huh? Aren't they the same thing?

 

[haruspex]

Huh? Aren't they the same thing?

If the pulley were massless, there would stil be axial torque to raise the suspended mass. If there were no suspended mass it would still take axial torque to accelerate the pulley. So, no, they're not the same thing.