Force Required to stop a Rotating Disc

Click For Summary
SUMMARY

The discussion focuses on calculating the force required to stop a rotating titanium disc with a radius of 5 inches and a thickness of 0.087 inches, rotating at 5800 RPM. The calculations reveal that the disc has a mass of 0.50465 kg and a kinetic energy of 750.7 joules. The angular acceleration needed to stop the disc in 0.005 seconds is determined to be -121475 rad/sec². The final steps involve using torque equations to convert angular acceleration into a linear force applied at the rim of the disc.

PREREQUISITES
  • Understanding of angular motion and acceleration
  • Familiarity with kinetic energy calculations
  • Knowledge of torque and its relationship to force
  • Basic principles of dynamics and rotational mechanics
NEXT STEPS
  • Study the relationship between torque and angular acceleration in rotational dynamics
  • Learn about the equations of motion for rotating bodies
  • Explore the concept of friction and its role in stopping rotating objects
  • Investigate practical applications of angular deceleration in engineering
USEFUL FOR

Students in physics or engineering, mechanical engineers, and anyone interested in the dynamics of rotating systems and force calculations.

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
831
  • · Replies 17 ·
Replies
17
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
26
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
4K