What is the Angular Momentum of a Spinning Disk Subject to Applied Forces?

In summary, at t=0, a disc of mass 5.0 kg and radius 0.45m is spinning clockwise about the axis passing through the center of the disc. Three forces are applied to the disc: a net torque of 5.4 N, and two counterclockwise torques of -1.2 N and 1.2 N, respectively. Immediately after the forces are applied, the disc's angular momentum is increasing by 6.48 N, decreasing by -0.92 N, and staying the same. The change in angular momentum between t=0 and t=1.2 s is 7.62 N. The disc's angular speed between t=0 and the moment described in Part 3 is
  • #1
postfan
259
0

Homework Statement



A solid disc of mass m = 5.0 kg and radius r = 0.45 m is spinning clockwise about the axis passing through the center of the disc. At t = 0, three forces are applied to the disc as shown.

Part 1. What is the magnitude of the net torque acting on the disc about its center?
Part 2. Immediately after the three forces are applied, is the magnitude of the angular momentum of the disc increasing, decreasing or staying the same?
Part 3. Assuming that the net torque stays constant, what is the change in the magnitude of the angular momentum of the disc between t = 0 and t = 1.2 s?
Part 4. What is the change in the magnitude of the angular speed between t = 0 and the moment described in Part 3?

Homework Equations





The Attempt at a Solution



For part 1 (taking counterclockwise to be positive) I found the torque by calculating 14*r/2+9*r-sin(30)*r*8=7*r+9r-4r=12r=12*.45=5.4

For part 2, since the disk is rotating clockwise, adding counterclockwise torques will cause it to reduce its angular momentum.

For part 3, I multiplied the time interval by the change in torque: 5.4*1.2=6.48

What am I doing wrong?
 

Attachments

  • BK107.PNG
    BK107.PNG
    4.9 KB · Views: 552
Physics news on Phys.org
  • #2
postfan said:
1.

The Attempt at a Solution



For part 1 (taking counterclockwise to be positive) I found the torque by calculating 14*r/2+9*r-sin(30)*r*8=7*r+9r-4r=12r=12*.45=5.4


Check the signs of the 9r and 8rsin30 terms?
 
  • #3
So is it 7r-9r+4r=2r? Why?
 
  • #4
postfan said:
So is it 7r-9r+4r=2r? Why?

Yes. Look at the way the forces tend to turn the disc!
 
  • #5
OK I figured out parts 1 and 2 from above and common sense, on part 3 I just took the integral over the time period of the torque, but I am not sure about part 4.

I used the equation 1.08L=MR^2/2*omega, m=5, r=.45 and I got omega=2.13, is that right?
 
  • #6
postfan said:
OK I figured out parts 1 and 2 from above and common sense, on part 3 I just took the integral over the time period of the torque, but I am not sure about part 4.

I used the equation 1.08L=MR^2/2*omega, m=5, r=.45 and I got omega=2.13, is that right?

For part 1 they ask for magnitude so your answer must be +.
For part 2, the disc is spinning cw and your net trque is also in that direction. Again, it asks for magnitude of change.
3. part 3 your approach is right but you need to make the correction in 1. I frankly don't remember the formula for I for a disc. BTW it's not 'change in torque' but just 'torque'.
part 4: you know how much angular momentum has changed (= I*Δω), surely deriving Δω is a snap?
 
  • #7
The part of part 4 that I don't understand is on the LHS of the equation do I use 1.08L or L+1.08?
 
  • #8
postfan said:
The part of part 4 that I don't understand is on the LHS of the equation do I use 1.08L or L+1.08?

Where does 1.08 come from? What is L?

You have determined the change in angular momentum in part 3. Angular momentum change = I*Δω. Solve for Δω.
 
  • #9
1.08 is the change of angular momentum, and L is the symbol of angular momentum.
 
  • #10
postfan said:
So is it 7r-9r+4r=2r? Why?
You said you were taking counterclockwise as positive. The above is taking clockwise as positive. In the OP it was the 14r/2 term that had the wrong sign. Anyway, you got 2r clockwise, which is correct.
on the LHS of the equation do I use 1.08L or L+1.08?
As rude man asks, what is L and where does 1.08 come from? I will add, what did you get for part 3? (It is not 1.08.)
 
  • #11
postfan said:
1.08 is the change of angular momentum, and L is the symbol of angular momentum.

OK, so what is L(t=0)? BTW you have I = mR^2/2 right, I looked it up.
Then ΔL = L(t=1.2) - L(t=0). And I*Δω = ΔL.

@haruspex, direction is irrelevant since they're asking for magnitudes thruout.
 

Related to What is the Angular Momentum of a Spinning Disk Subject to Applied Forces?

1. What is angular momentum?

Angular momentum is a physical quantity that measures the amount of rotational motion an object has around a specific axis. It is a vector quantity, meaning it has both magnitude and direction.

2. How is angular momentum calculated for a disk?

The angular momentum of a disk can be calculated by multiplying the moment of inertia (a measure of an object's resistance to rotational motion) by the angular velocity (the rate at which the disk is rotating).

3. What factors affect the angular momentum of a disk?

The angular momentum of a disk can be affected by its moment of inertia, its angular velocity, and any external torque acting on the disk. Additionally, the distribution of mass and the axis of rotation can also impact the angular momentum.

4. Is angular momentum conserved for a disk?

Yes, according to the law of conservation of angular momentum, the total angular momentum of a system remains constant as long as there is no external torque acting on the system. This means that the angular momentum of a disk will remain constant unless an external force or torque is applied to it.

5. How is angular momentum related to rotational kinetic energy?

Angular momentum and rotational kinetic energy are related through the equation L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity. This equation shows that as the angular momentum increases, so does the rotational kinetic energy.

Similar threads

  • Introductory Physics Homework Help
Replies
26
Views
318
  • Introductory Physics Homework Help
Replies
5
Views
320
  • Introductory Physics Homework Help
2
Replies
45
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
30
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
1K
Replies
6
Views
495
  • Introductory Physics Homework Help
Replies
10
Views
943
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top