How to Calculate Speed, Spin, and Unwrapped String in a Rotary Motion Problem?

In summary, the disk is at rest on a frictionless surface and a string is wrapped around the rim. When the disk has moved a distance of 5.0 m, the disk is moving at a speed of 6.8 m/s and it is spinning at a rate of 1200 rpm.
  • #1
faizulhu
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Homework Statement



A solid uniform disk of mass 21.0 and radius 85.0 is at rest flat on a frictionless surface. The figure shows a view from above. A string is wrapped around the rim of the disk and a constant force of 35.0 is applied to the string. The string does not slip on the rim.
(i) When the disk has moved a distance of 5.0 m, determine how fast it is moving.
(ii) How fast it is spinning (in radians per second)
(iii) How much string has unwrapped from around the rim.

Homework Equations



Work = tau * theta
rotational energy = 0.5I (omega)^2
translational KE = 0.5Mv^2

The Attempt at a Solution


I have tried using energy conservation but for some reason the answer isn't right. tau = 35*0.85. But I'm having trouble finding theta. I used theta = 5 / 0.85 but that isn't working since that would mean that the answer to question (iii) is 5 which it isn't. The work done equals the translational and kinetic energy. But without a correct theta, I can't work out any of the problem. The problem is from Physics for Scientists and Engineers by Giancoli Chapter 10, Problem 108
 

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  • #2
Text of the problem says that the disk is at rest on a flat frictionless surface. It means that the disk is not rolling. String acts on the disk with the force equal to the force applied to the string and it's the only relevant force. It means that the disk is accelerating translationally because of that force (2. Newton's law). There is also a rotation present. It is caused by the torque of the force acting on the disk.

Using [tex]F=ma[/tex] we can get the translational acceleration and also the speed of the disk.

Then, using [tex]\tau=I\alpha=rF[/tex] we can also get angular acceleration and from there the angular velocity.

Since the body is rotating, string unwrapps from the rim. Length of the unwrapped part is equal to [tex]l=r\varphi[/tex] where [tex]\varphi[/tex] is the angle traversed because of the rotation ([tex]\varphi[/tex] can be calculated using angular acceleration).
 
  • #3
Thank you very much. I have been at this problem since lunch and didn't manage to consider doing the translational and rotational motion individually. You are a life saver! Thanks again
 

1. What is the concept behind a disc pulled by a string?

The concept behind a disc pulled by a string is that the string, when pulled, provides a force that causes the disc to move. This force is known as tension and is transmitted along the string to the disc, causing it to accelerate and move in the direction of the pull.

2. How does the tension in the string affect the motion of the disc?

The tension in the string determines the magnitude and direction of the force acting on the disc. This force causes the disc to accelerate and move in the direction of the pull. The greater the tension, the greater the force and the faster the disc will move.

3. Are there any other factors that affect the motion of the disc?

Yes, there are other factors that can affect the motion of the disc. These include the mass of the disc, the angle at which the string is pulled, and any external forces acting on the disc, such as friction or air resistance. These factors can impact the speed and direction of the disc's motion.

4. What happens if the string is pulled at an angle instead of straight?

If the string is pulled at an angle instead of straight, the tension in the string will be split into two components: one parallel to the direction of the pull and one perpendicular to it. The parallel component will cause the disc to accelerate and move in the direction of the pull, while the perpendicular component will cause the disc to change direction.

5. Can the disc continue moving without being pulled by the string?

No, the disc will eventually come to a stop if it is not being pulled by the string. This is because the tension in the string provides the force that keeps the disc in motion. Without this force, the disc will eventually slow down and come to a stop due to the effects of friction and air resistance.

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