# Homework Help: Twisting Disk - angular speed

1. May 2, 2010

### mybrohshi5

1. The problem statement, all variables and given/known data
A thin metal disk of mass m=4.50 g and radius R=2.70 cm is attached at its center to a long fiber. When the disk is turned from the relaxed state through a small angle theta, the torque exerted by the fiber on the disk is proportional to theta.

Suppose the disk is twisted away from equilibrium by an angle of 27 degrees and released from rest. What is the maximum angular speed of the disk as it twists back and forth?

2. Relevant equations

w = sqrt(k/I)
t = -k(theta)

3. The attempt at a solution

So i found the first two parts of the problem fine but i am having trouble with this part. I am not sure where i can go with the given information

I converted the degrees to radians

not sure what to do now

2. May 2, 2010

### Staff: Mentor

Are you given additional information that would allow you to calculate the spring constant of the fiber?

3. May 2, 2010

### mybrohshi5

That is all the information i was given.

I didnt think i could find the spring constant with what is given...

4. May 2, 2010

You can't.

5. May 2, 2010

### mybrohshi5

so is there any other way i can solve this without finding the spring constant?

Its a mastering physics question and it asks for an answer in rad/s so i guess there has to be a way to find it :(

6. May 2, 2010

### mybrohshi5

Not sure if this will help but this was the question it asked before this one.

The disk, when twisted and released, oscillates with a period T=1.05 s. Find the torsional constant kappa of the fiber.

i found k to be 5.87 x 10^-5

7. May 2, 2010

### Staff: Mentor

Of course it helps! Assuming the question refers to the same disk, you now have the spring constant.

8. May 2, 2010

### mybrohshi5

How do i have the spring constant now?

The torsional constant kappa (k) is the same as the spring constant?

Last edited: May 2, 2010
9. May 2, 2010

### Staff: Mentor

Yes, it's the torsional spring constant (measured in Newton-meters per radian).

10. May 2, 2010

### mybrohshi5

oh it is ok.

so then can't i just use w = sqrt(k/m)

11. May 2, 2010

### Staff: Mentor

No. You have the correct equation in your first post. (One is for linear motion; the other for rotational motion.)

12. May 2, 2010

### mybrohshi5

Thats what i tried earlier and got

w = sqrt (k/I)

w = sqrt (5.87*10^-5 / (.5*.0045kg*.027m^2))

w = 5.98 rad/s but it says that is wrong?

do i have to factor the 27 degrees in some how?

13. May 2, 2010

### Staff: Mentor

You've found the angular frequency (ω) of the torsional pendulum. How does that relate to the angular speed?

Of course.

14. May 2, 2010

### mybrohshi5

It doesnt i was getting my symbols confused :(

I cant seem to get the angular speed though. i know

angular speed = V/r

also V = (angular frequency)(A)

so i found V

V = sqrt (k/I) * A

V = 2.819

then i used angular speed = V/r

angular speed = 104 rad/s but that cant be right?

am i using a wrong equation?

15. May 2, 2010

### Staff: Mentor

In that equation, V is the tangential speed of something in circular motion. Not what you want here.

Hint: Write the angle as a function of time like this:

θ = θmax cos(ωt)
where θmax is the maximum angle (in radians)

How would you write the angular speed (dθ/dt) as a function of time? What's the maximum angular speed?

16. May 2, 2010

### mybrohshi5

would (dθ/dt) = -θmax*ω*sin(ωt)

17. May 2, 2010

### Staff: Mentor

Exactly! So what is the maximum value of dθ/dt?

18. May 2, 2010

### mybrohshi5

I am a little confused on what to plug into where now.

for ω would i use (2pi)/1.05 ------ 1.05 is the Period given for the earlier part.

and then i am not sure what to use for t?

is t maybe 1/4*(1.05) = .2625 s

then for theta max just use 27*pi/180 = .471239 rad right?

19. May 3, 2010

### Staff: Mentor

That's correct, but it's easier than that. What's the maximum value of sinωt?

Right.

20. May 3, 2010

### mybrohshi5

the max value of sinωt is always 1. That comes in handy haha :)

Thank you for all the help Doc Al. I really appreciate it.