Need help with hw problem (torque question)

  • Thread starter bigbuck
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In summary, the question is asking for the torque required to turn a solid disk with a mass of 60.0kg and a diameter of 48.0cm from rest through 12.0 revolutions in 6.00 seconds. To solve this, we can use the equation Tau=I*alpha (torque equals moment of inertia times angular acceleration). The moment of inertia, I, can be calculated by taking half of the mass, 60.0kg, and multiplying it by the square of the radius, which is half of the diameter, 24.0cm. This gives us a moment of inertia of 1.728kg*m^2. To find the angular acceleration, alpha, we can use the
  • #1
bigbuck
6
0
The question is:

A solid disk with m=60.0kg and a diameter of 48.0cm is to be turned from rest through 12.0 revolutions in 6.00s. Calculate the torque required to accomplish this.

What I've got is this:

Tau=I*al(angular acceleration)

I=.5*m*r^2 .5*60.0kg*.240m^2=1.728kg*m^2

Al = om/t (convert to om and then to al)12revs/6sec=2.00revs/sec = 12.56rads/s / 6

sec = 2.09 rads/s^2=al

1.728kg*m^2 * 2.09rads/s^2= 3.61N*m ...is this right?

This is an online physics course and I am getting a little lost at this point. The course notes are vague at best and I am just not sure if i have the irght answers or not.
ANy help would be appreciated.
 
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  • #2
bigbuck said:
The question is:

A solid disk with m=60.0kg and a diameter of 48.0cm is to be turned from rest through 12.0 revolutions in 6.00s. Calculate the torque required to accomplish this.

What I've got is this:

Tau=I*al(angular acceleration)

I=.5*m*r^2 .5*60.0kg*.240m^2=1.728kg*m^2

Al = om/t (convert to om and then to al)12revs/6sec=2.00revs/sec = 12.56rads/s / 6

sec = 2.09 rads/s^2=al
If a disk has a constant angular acceleration, a rad/s2, then in t seconds, it will have reached angular speed at rad/s and will have moved (1/2)at2 rad or (1/2)[\pi]at2 revolutions. Another way of looking at that is that if it has constant acceleration, a, it will have average speed (1/2)at over time t. Moving through 12.0 revolutions in 6 seconds means it had an angular acceleration of 12/72= 1/6 rev/s2= [itex]\pi[/itex]/6= 0.523 rad/sec2, not 2.09.

1.728kg*m^2 * 2.09rads/s^2= 3.61N*m ...is this right?

This is an online physics course and I am getting a little lost at this point. The course notes are vague at best and I am just not sure if i have the irght answers or not.
ANy help would be appreciated.
 
  • #3
still confused

I am sorry.
One of the problems is that this proffesor, while a good and helpful person, has used his own characters for the variables..ie "om" instead of w for omega, and "al" for angular acceleration. While i know he thinks he's helping, it is causing me great problems when i refer to the text or get help from others, I am getting "lost in translation" because I don't understand the language. I am very frustrated because i know this is simple.

Is "at" angular acceleration?

I don't see how to get the acceleration from the given info.
 
  • #4
anyone?
 
  • #5
It is unfortunate that your professor does that. I would suggest buying a used textbook and create a key or something so you can translate. Anyway, you are looking for angular acceleration. Heres the equation in the proper language:

a(avg)=w2-w1/t2-t1 rad/s^2

average angular acceleration = change in angular speed divided by change in time
Units: radians per second squared

Once you find this, you can easily derive the torque. Good luck!
 

1. What is torque?

Torque is a measure of the twisting force applied to an object. It is calculated by multiplying the force applied to the object by the perpendicular distance from the point of application to the axis of rotation.

2. How is torque related to angular acceleration?

Torque is directly proportional to angular acceleration. This means that the greater the torque applied to an object, the greater the angular acceleration it will experience.

3. What is the equation for calculating torque?

The equation for torque is torque = force x distance, where force is measured in newtons (N) and distance is measured in meters (m).

4. How does the direction of the force affect the torque?

The direction of the force determines the direction of the torque. If the force is applied perpendicular to the object's axis of rotation, the torque will be at its maximum. If the force is applied parallel to the axis of rotation, the torque will be zero.

5. Can torque be negative?

Yes, torque can be negative. This occurs when the force is applied in the opposite direction to the object's motion or rotation. Negative torque is often referred to as "counterclockwise" torque, while positive torque is referred to as "clockwise" torque.

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