# Flywheel & Disk Brake System

1. Jun 2, 2015

### Mika-999

1. The problem statement, all variables and given/known data
Hey y'all. Heres the problem:

The flywheel to the right is spinning at 300 radians per second in a counter clockwise direction. The
brake (to the left) is applied with a force F of 5kN.

How long will it take the flywheel to stop rotating?

Assume:
 The flywheel is made of steel
 The units above are in cm
 The brake disc uses all the friction surface area on the flywheel

2. Relevant equations

Here are the relevant equations:

Friction: 2500 N
Area of friction plate: 0.1018 m^2
Torque (brake): 2250 N.m
Inertia (don't have mass):

3. The attempt at a solution

I'm fairly stumped at which direction to take. At the moment my attempt goes something like this:

1. working out the area that the disk brake will affect
2. working out the friction force that will be applied from the disk brake to the flywheel
3. the amount of torque the brake has

How i imagined the problem:

speed of the flywheel --> disk brake is applied --> the friction of the brake to the flywheel will slow it down --> result is it will stop moving.

How to get time from all of this baffles me.

Any suggestions will be greatly appreciated

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2. Jun 2, 2015

### SammyS

Staff Emeritus
Finding the torque may be your biggest problem .

How are torque and angular acceleration related ?

3. Jun 3, 2015

### SteamKing

Staff Emeritus
Why can't you calculate the mass & inertia of the flywheel?

You are given the dimensions of the flywheel and its material. Not everything will be handed to you on a platter.

4. Jun 4, 2015

### Mika-999

Okay, following the lead given by SteamKing (thank you for the point out).

The mass calculation went as follows:

mass = density x area x width
= 169 kg

The inertia of the flywheel is then given by:

I =(1/2)mr^2
= 17.111 kg.m^2

Then i was thinking of using Newtons second law of F = ma but for a rotational setting, so it becomes τ=Iα
and as α = ω/t , i can substitute that in and solve for t, which ends up giving me 2.28 sec.

Using the torque given by the brake:

τ = Fr
= 5 kN x 0.45 m
= 2250 N.m

However i was thinking that the value I'm using for F doesn't include the friction and r is in reference to the whole flywheel. So from this i decided to take F = 5kN + 2500N and r as 0.18m.

Substituting this into the procedure above and i get t = 3.802 sec

5. Jun 4, 2015

### SteamKing

Staff Emeritus
I've tried to confirm your mass calculation for the flywheel, but I'm not getting your result.

Can you show all your calculation details for the mass of the flywheel?

6. Jun 4, 2015

### Mika-999

My mistake. Following the below procedure i get 835kg not 169kg

This is what i assumed you can do to calculate mass.

mass = density x area x width
= 8050 x (pi x 0.45^2) x 0.163
=835 kg

(mass seems a little too big for the actual size of this flywheel though)

Anyways, calculating the time with this mass gives 18.787 sec

7. Jun 5, 2015

### SteamKing

Staff Emeritus
I agree with your calculation of the mass, except the density of steel seems a little high.

This table:

http://www.engineeringtoolbox.com/metal-alloys-densities-d_50.html

gives a density of 7850 kg / m3 for both iron and mild steel. I think you have used a density value for some sort of stainless or other alloy steel, which you probably wouldn't find used making a simple flywheel.

8. Jun 5, 2015

### Mika-999

That was pretty careless of me, i took the first answer google gave me and took that as an absolute.

So, now using a density of 7850 kg / m3 and plugging it in, gives me 18.315 sec

- would you say this is a hit or a miss value?

9. Jun 5, 2015

### SteamKing

Staff Emeritus
The force applied to the flywheel brake, 5 kN, can't be assumed to brake the flywheel directly. You must calculate the actual amount of friction force which opposes the rotary motion of the flywheel. The force applied to the brake is applied perpendicular to the motion of the flywheel and cannot retard its motion except by creating the friction between the brake and the brake pad attached to the flywheel.

In any event, you couldn't assume that the braking force is applied at the outer radius of the flywheel.

10. Jun 5, 2015

### Mika-999

No of course, the brake only applies to the friction surface which it is applied to.

With that in mind then, the retarding force is 2 500 N and the brake torque must only be applicable to 0.30-0.12 = 0.18m

The time i now get is 54.945 sec

11. Jun 5, 2015

### SteamKing

Staff Emeritus
I think you should show your braking calculations as revised after correcting the mass of the flywheel. It's hard to follow these calculations piecemeal.