# Force Required to stop a Rotating Disc

• rjm3db
Remember to always check your units and make sure everything is consistent. Good luck with the rest of your problem!In summary, the problem involves determining the force needed to stop a rotating disc with a radius of 5 inches and a thickness of 0.087 inches, made of titanium with a density of 4507 kg/m^3, rotating at 5800 rpm in 0.005 seconds. Using the equations K = 1/2*I*ω^2 and V = ∏r^2h, the kinetic energy of the disc is found to be 750.7 joules. To find the necessary force, the angular acceleration is calculated to be -121475 rad/sec^2 and then converted to a torque

## 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!

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