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

In summary: If the hand is just slowing it down, then it would be equal to the magnitude of the torque exerted by the string. In summary, a block hangs by a massless string around a solid cylindrical pulley that can rotate freely. After being released from rest, the pulley completes 3 rotations in 4 seconds. The pulley rotates through 19 radians during this time and has an angular acceleration of 2.4 rad/s/s. The distance the block drops is dependent on the distance the pulley rotates. The direction of the pulley's angular velocity is pointed out the back of the monitor, while the angular acceleration is pointed towards the viewer. The magnitude of the torque exerted by the hand on the pulley is
  • #1
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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  • #2
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.
 

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

What is rotational mechanics?

Rotational mechanics is a branch of classical mechanics that studies the motion of objects that rotate around a fixed axis. It involves concepts such as angular displacement, angular velocity, and angular acceleration.

What is the difference between \theta and \alpha in rotational mechanics?

\theta represents the angular displacement, which is the change in angle of an object rotating around a fixed axis. \alpha represents the angular acceleration, which is the rate of change of the angular velocity of an object.

How do you calculate torque in rotational mechanics?

Torque is calculated by multiplying the applied force by the distance from the axis of rotation to the point where the force is applied. The formula for torque is T = F * r * sin(\theta), where F is the applied force, r is the distance, and \theta is the angle between the force and the lever arm.

What is the unit of torque in rotational mechanics?

The unit of torque in rotational mechanics is Newton-meters (N*m) in the SI system. In the imperial system, it is measured in foot-pounds (ft*lb).

How can I apply rotational mechanics to real-life situations?

Rotational mechanics is applicable to a wide range of real-life situations, such as the motion of planets around the sun, the rotation of wheels on a car, and the movement of a spinning top. It is also used in engineering and design to understand and optimize the motion of machines and structures.

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