Solving Rotational Mechanics Homework: \theta, \alpha, Torque

[crybllrd]

Homework Statement

A block hangs by a massless string that is wound around a solid cylindrical pulley of radius 0.20m. The pulley is free to rotate on frictionless bearings about an axis through its center. The block is released from rest. The pulley completes 3 rotations in 4 seconds.

A) Through what angle \theta (in radians) does the pulley rotate during this time?

B) What is the angular acceleration \alpha of the pulley (magnitude and direction)?

C)How far does the block drop in this time?

A hand now touches the pulley and exerts a connstant frictional force such that the downward speed of the block is decreasing.

D)What is the direction of the pulley's angular velocity?

E)What is the direction of the pulley's angular acceleration?

f)Is the magnitude of the torque exerted by the hand on the pulley less than, greater than or equal to the magnitude exerted by the string on the pulley?

Homework Equations

The Attempt at a Solution

A) Through what angle \theta (in radians) does the pulley rotate during this time?

My book shows an example leading me to this:

\Delta\theta=3revolutions(\frac{2\pi rad}{1revolution})=19radiansB) What is the angular acceleration \alpha of the pulley (magnitude and direction)?

I should insert \Delta\theta in this equation:

\Delta\theta=\omega_{i}\Deltat+\frac{1}{2}\alpha\Deltat^{2}

\alpha=2.4rad/s/s

C)How far does the block drop in this time?

Here I think I should solve for \Delta\theta, making me realize I did part B wrong.

D)What is the direction of the pulley's angular velocity?

Hard to explain this, it would be pointed out the back of my monitor (away from the viewer)

E)What is the direction of the pulley's angular acceleration?

It is slowing down, so it is in the opposite direction of the velocity, so pointing toward the viewer, out of the monitor.

f)Is the magnitude of the torque exerted by the hand on the pulley less than, greater than or equal to the magnitude exerted by the string on the pulley?

Greater than, it is slowing the pulley down, and it will eventually stop.

 

[rock.freak667]

The Attempt at a Solution

A) Through what angle \theta (in radians) does the pulley rotate during this time?

My book shows an example leading me to this:

\Delta\theta=3revolutions(\frac{2\pi rad}{1revolution})=19radians

Good.

B) What is the angular acceleration \alpha of the pulley (magnitude and direction)?

I should insert \Delta\theta in this equation:

\Delta\theta=\omega_{i}\Deltat+\frac{1}{2}\alpha\Deltat^{2}

\alpha=2.4rad/s/s

Didn't check the answer, but your equation should be θ=ω_it+(1/2)αt²

C)How far does the block drop in this time?

Here I think I should solve for \Delta\theta, making me realize I did part B wrong.

Well remember the distance the pulley rotates would dictate the amount the mass drops by. So what distance did the pulley rotate?

D)What is the direction of the pulley's angular velocity?

Hard to explain this, it would be pointed out the back of my monitor (away from the viewer)

Yes, using the right hand grip rule.

E)What is the direction of the pulley's angular acceleration?

It is slowing down, so it is in the opposite direction of the velocity, so pointing toward the viewer, out of the monitor.

I think this is correct as well.

f)Is the magnitude of the torque exerted by the hand on the pulley less than, greater than or equal to the magnitude exerted by the string on the pulley?

Greater than, it is slowing the pulley down, and it will eventually stop.

Assuming the hand is stopping it, yes.