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

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SUMMARY

The discussion focuses on solving a rotational mechanics homework problem involving a solid cylindrical pulley and a block. The pulley, with a radius of 0.20m, completes 3 rotations in 4 seconds, leading to an angular displacement of 19 radians and an angular acceleration of 2.4 rad/s². The direction of the pulley's angular velocity is away from the viewer, while the angular acceleration points toward the viewer due to deceleration caused by a frictional force. The torque exerted by the hand on the pulley is greater than that exerted by the string, as it is responsible for slowing down the pulley.

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crybllrd
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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.

213kcnc.jpg


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

B) What is the angular acceleration [tex]\alpha[/tex] 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 [tex]\theta[/tex] (in radians) does the pulley rotate during this time?

My book shows an example leading me to this:

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

I should insert [tex]\Delta\theta[/tex] in this equation:

[tex]\Delta\theta=\omega_{i}\Deltat+\frac{1}{2}\alpha\Deltat^{2}[/tex]

[tex]\alpha=2.4rad/s/s[/tex]

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

Here I think I should solve for [tex]\Delta\theta[/tex], 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.
 
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crybllrd said:

The Attempt at a Solution



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

My book shows an example leading me to this:

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

Good.

crybllrd said:
B) What is the angular acceleration [tex]\alpha[/tex] of the pulley (magnitude and direction)?

I should insert [tex]\Delta\theta[/tex] in this equation:

[tex]\Delta\theta=\omega_{i}\Deltat+\frac{1}{2}\alpha\Deltat^{2}[/tex]

[tex]\alpha=2.4rad/s/s[/tex]

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

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

Here I think I should solve for [tex]\Delta\theta[/tex], 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?


crybllrd said:
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.

crybllrd said:
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.

crybllrd said:
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.
 

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