What is the maximum angular speed of a twisted disk?

Click For Summary

Homework Help Overview

The problem involves a thin metal disk attached to a fiber, which is twisted and released from a small angle. The objective is to determine the maximum angular speed of the disk as it oscillates back and forth. The context includes concepts from rotational dynamics and oscillatory motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between torque, angular speed, and the spring constant. There are attempts to relate angular frequency to angular speed and questions about the relevance of the angle of twist.

Discussion Status

Participants are exploring different methods to calculate the maximum angular speed, including the use of angular frequency and the relationship between angular displacement and time. Some have provided hints and clarifications regarding the equations involved, while others express confusion about the application of certain formulas.

Contextual Notes

There is a mention of a previous question regarding the period of oscillation, which may influence the current problem. Participants are also grappling with the definitions and relationships between angular speed, angular frequency, and the parameters of the system.

mybrohshi5
Messages
365
Reaction score
0

Homework Statement


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?


Homework Equations



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

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

27 degrees = .471239 rad

not sure what to do now
 
Physics news on Phys.org
Are you given additional information that would allow you to calculate the spring constant of the fiber?
 
That is all the information i was given.

I didnt think i could find the spring constant with what is given...
 
mybrohshi5 said:
I didnt think i could find the spring constant with what is given...
You can't.
 
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 :(
 
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
 
mybrohshi5 said:
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
Of course it helps! Assuming the question refers to the same disk, you now have the spring constant.
 
How do i have the spring constant now?

The torsional constant kappa (k) is the same as the spring constant?
 
Last edited:
mybrohshi5 said:
So the torsional constant kappa (k) is the same as the spring constant?
Yes, it's the torsional spring constant (measured in Newton-meters per radian).
 
  • #10
oh it is ok.

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

so w = 0.1142 rad/s
 
  • #11
mybrohshi5 said:
so then can't i just use w = sqrt(k/m)
No. You have the correct equation in your first post. (One is for linear motion; the other for rotational motion.)
 
  • #12
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
mybrohshi5 said:
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?
You've found the angular frequency (ω) of the torsional pendulum. How does that relate to the angular speed?

do i have to factor the 27 degrees in some how?
Of course.
 
  • #14
It doesn't i was getting my symbols confused :(

I can't 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 = 5.98 rad/s * ( .4712 rad)

V = 2.819

then i used angular speed = V/r

angular speed = 104 rad/s but that can't be right?

am i using a wrong equation?
 
  • #15
mybrohshi5 said:
It doesn't i was getting my symbols confused :(

I can't seem to get the angular speed though. i know

angular speed = V/r
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
would (dθ/dt) = -θmax*ω*sin(ωt)
 
  • #17
mybrohshi5 said:
would (dθ/dt) = -θmax*ω*sin(ωt)
Exactly! So what is the maximum value of dθ/dt?
 
  • #18
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.

ω = 5.984 rad/s

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
mybrohshi5 said:
for ω would i use (2pi)/1.05 ------ 1.05 is the Period given for the earlier part.

ω = 5.984 rad/s
Yes, you've already found ω.

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

is t maybe 1/4*(1.05) = .2625 s
That's correct, but it's easier than that. What's the maximum value of sinωt?

then for theta max just use 27*pi/180 = .471239 rad right?
Right.
 
  • #20
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.
 

Similar threads

Replies
30
Views
4K
Replies
8
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 20 ·
Replies
20
Views
3K
Replies
3
Views
2K
Replies
3
Views
2K
Replies
28
Views
5K
Replies
9
Views
8K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
18
Views
3K