Rotational Dynamics Questions (2)

In summary, for the first problem, use the work-energy theorem and rotational kinematic equations to solve for the elapsed time and number of revolutions before the wheel comes to a stop. For the second problem, use the work-energy theorem and consider both linear and rotational energies to find the tension in each cord and the linear acceleration of the cylinder as it falls. Keep in mind that the radius of gyration of an object is the radius at which all the mass would need to be concentrated to give the same moment of inertia.
  • #1
csnsc14320
57
1

1. A wheel of mass M and radius of gyration "k" spins on a fixed horizontal axle passing through its hub. Assume that the hub rubs the axle of radius "a" at only the top-most point, the coefficient of kinetic friction being μk. The wheel is given an initial angular velocity Wo. Assume uniform deceleration and find (a) the elapsed time and (b) the number of revolutions before the wheel comes to a stop.

2. A solid cylinder of length L and radius R has a weight W. Two cords are wrapped around the cylinder, one near each end, and the cord ends are attached to hooks on the ceiling. The cylinder is held horizontally with the two cords exactly vertical and is then released. Find (a) the tension in each cord as they unwind and (b) the linear acceleration of the cylinder as it falls.





For the first question, I'm not really sure where to start, and we didn't cover radius of gyration. I think if I just get a set up of what information to work with I can get this.

The second one, I drew a FBD and I see that the two tensions will be exerting a torque on the cylinder, but I don't think that the weight gives any torque given where it's acting from - but then I don't know how to incorporate it into getting my answer. I have that 2TR + ?? =I*a/r so far =\


Thanks for any help to get nudge me in the right direction :D
 
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  • #2
The radius of gyration of an object is the radius if all the mass were concentrated at that radius to give the same moment of inertia of the object

I = Mk2

To start Problem 1 use the work-energy theorem, the rotational kinematic relation

Work = Torque*theta

and the fact that the wheel will initially and very quickly move off the high point of the axle due to friction and reach an equilibrium position. Utilize the angle of repose to determine the equilibrium position.

For Problem 2, the weight, W, does impart a torque since T = r x Mg = r x W . If the strings were attached over the center of the cylinder then no torque would exist.
 
  • #3
chrisk said:
The radius of gyration of an object is the radius if all the mass were concentrated at that radius to give the same moment of inertia of the object

I = Mk2

To start Problem 1 use the work-energy theorem, the rotational kinematic relation

Work = Torque*theta

and the fact that the wheel will initially and very quickly move off the high point of the axle due to friction and reach an equilibrium position. Utilize the angle of repose to determine the equilibrium position.

For Problem 2, the weight, W, does impart a torque since T = r x Mg = r x W . If the strings were attached over the center of the cylinder then no torque would exist.

For problem one, we haven't yet covered the work-energy theorem for rotational motion (unless we are suppose to infer that for ourselves), is there any other way to go about solving it? Or should I just look ahead and try it that way?
Also, I'm not sure if the axle fits perfectly through the wheel's hub, but if it did, why would the wheel move from the high point due to friction? What I imagine seeing is the hub being slightly larger and the inner radius of it to move around the circumference of the axle? (but then the wheel would be wobbling =\)

For problem 2, I used Iα=2TR, and using α=a/R, I=(MR^2)/2 got that T=Ma/4
For the acceleration I used Newtons 2nd law, and got Mg - 2T = Ma --> Mg -Ma/2 = MA --> a = 2g/3 (which makes T = Mg/6
 
  • #4
The work-energy theorem for linear motion should have been previously covered. Just use the analagous rotational kinematic equations to establish the equations. The hub is a little larger than the axle diameter so a slight shift off of vertical will occur.

Use the work-energy theorem for Problem 2. Solving Problem 1 first will give you the insight to solve Problem 2. Just keep in mind that two types of energy exist for the second problem; linear and rotational.
 

1. What is rotational dynamics?

Rotational dynamics is the study of the motion of objects that are spinning or rotating, and the forces that affect their movement. It is a branch of physics that focuses on the relationship between rotational motion, angular velocity, and torque.

2. What is the difference between linear and rotational motion?

Linear motion is the movement of an object along a straight line, while rotational motion is the movement of an object around an axis or point. In linear motion, the velocity and acceleration vectors are in the same direction, while in rotational motion, the velocity and acceleration vectors are perpendicular to each other.

3. How is angular velocity related to linear velocity?

Angular velocity is a measure of how quickly an object is rotating, while linear velocity is a measure of how quickly an object is moving in a straight line. The two are related through the radius of rotation, with linear velocity being equal to the angular velocity multiplied by the radius of rotation.

4. What is torque and how does it affect rotational motion?

Torque is a measure of the rotational force applied to an object, and it is calculated by multiplying the force applied by the distance from the axis of rotation. Torque can cause rotational motion by changing an object's angular velocity, and it is also responsible for maintaining the stability of rotating objects.

5. How do you calculate moment of inertia?

Moment of inertia is a measure of an object's resistance to changes in its rotational motion and is calculated by summing the product of each infinitesimal mass element and the square of its distance from the axis of rotation. The formula for moment of inertia varies depending on the shape and distribution of mass of the object.

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