Force Required to stop a Rotating Disc

Click For Summary

Homework Help Overview

The discussion revolves around calculating the force required to stop a rotating disc within a specified time frame. The disc's specifications include a rotation speed of 5800 rpm, a radius of 5 inches, and a material density of titanium. The original poster is seeking assistance in determining the necessary force to halt the disc's rotation in 0.005 seconds.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of angular acceleration and its relationship to the force needed to stop the disc. There is an exploration of relevant equations, including those for torque and angular motion.

Discussion Status

The discussion has progressed with participants providing guidance on how to relate angular acceleration to torque and subsequently to force. There is an ongoing exploration of the necessary equations to finalize the calculations.

Contextual Notes

Participants note the original poster's uncertainty regarding the integration of concepts from dynamics and the specific equations needed to complete the problem. There is an acknowledgment of the original poster's previous coursework and the challenges of recalling relevant formulas.

rjm3db
Messages
3
Reaction score
0

Homework Statement


Essentially, I am trying to determine the force that must be applied to a rotating disc to that stop that disc from rotating in a certain time period.

The disc is rotating at 5800 rpm, the disc has a radius of 5 inches, and a thickness of 0.087 inches. The disk is made of titanium (density of 4507 kg/m^3). The time required to stop the disk is 0.005 seconds.


Homework Equations


K = 1/2*I*ω^2 ; I = 1/2*m*r^2 ; -------> K = 1/2(1/2*m*r^2)*ω^2

The Attempt at a Solution


Knowing ω = (2∏) / T , I was able to calculate ω to be ω= 607.375 rad/sec

Volume of the disc = V = ∏r^2h, which leads to V = 1.1197 E-4 m^3

Using the density, I found the mass of the disk to be m = 0.50465 kg.

Plugging all of this into equation above for kinetic energy, I found K to be K = 750.7 joules

It's been a while since I took a dynamics course. How can I determine the force necessary (I guess I'm kind of assuming a friction-like force such as a break being applied) to stop the disc from spinning in 0.005 seconds? I know I'll need an integral, but I forget the exact formula and steps necessary to finish up this problem.

Thanks!
 
Physics news on Phys.org
Hi rjm3db, welcome to Physics Forums.

You've done most of the background calculations for the problem. Consider that in order to go from a speed ω to zero in a given time t that there must be an acceleration \alpha. How might you find that?
 
Lets see. So, the angular acceleration is equal to α = dω / dt

So, α = (-607.375 rad / sec) / 0.005 sec

→ α = -121475 rad / sec^2

Alright, so now I've calculated both the angular acceleration.

Any other hints to getting me closer to finishing this one up?

It's been a while since I took a dynamics course, but does it involve the equation a = (alpha x r) + (omega x omega x r) , and later on the equation ƩF = ma ?
 
All you need now is the angular equivalent to F = MA to find the torque, then convert the torque to a force applied (presumably) to the rim of the disk.
 
Great, so something along the lines of

torque = I * α

torque = r x F → Force = torque / r ?

I really appreciate all of your help. Thanks a bunch
 
rjm3db said:
Great, so something along the lines of

torque = I * α

torque = r x F → Force = torque / r ?
Yup. Looks good.
I really appreciate all of your help. Thanks a bunch
Glad to be of help.
 

Similar threads

  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
879
  • · Replies 17 ·
Replies
17
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
26
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
4K