# Torque required to decelarate a disc

1. Oct 8, 2015

Hello everyone if someone can helo me out here
I have a rotating shaft wiht a disc mounted on it.The moment of inertia of the total system is 170 kgm2.It is rotating with a veolicty of 20 deg /sec.i want to bring it to stand still within 80 deg travel from the application of brakes

I=170 kgm2
wf=0
Δt=4 sec(taken from the fact that it is moving at speed of 20 deg/sec,80 deg/sec will be covered in 4 sec.Please correct as i m not sure abt this value)

So
T=Iα
T=I(Δω/Δt)
T=(170) (0-0.34/4)
T= -14.45Nm

2. Oct 8, 2015

### Daeho Ro

You cannot simply set it as 4 seconds because when it slow down, the disc stop before it rotate 80 deg after the application of brakes.

Just leave it as variable and consider
$$T = \dfrac{d\omega}{dt} = \dfrac{d\theta}{dt} \dfrac{d\omega}{d\theta} = \omega \dfrac{d\omega}{d\theta}.$$
When you integrate it as $\theta$, then
$$\int_{\theta_i}^{\theta^f} T d\theta = T(\theta_f - \theta_i)$$
and also it is same as
$$\int_{\theta_i}^{\theta^f} T d\theta = \int_{\omega_i}^{\omega_f} \omega d\omega = \dfrac{1}{2} \left(\omega_f^2 - \omega_i^2\right).$$

3. Oct 8, 2015

hey thnks for the reply.....can you elaborate it more....possibly solve it

4. Oct 8, 2015

### Daeho Ro

Use the last relation,
$$T(\theta_f - \theta_i) = \dfrac{1}{2} (\omega_f^2-\omega_i^2),$$
for your situation. You already know each variable without $T$, so it is elementary calculation

5. Oct 8, 2015

ok thnks....but cant i determine a stopping distance?

6. Oct 8, 2015

### Daeho Ro

You mention that you want to stop the disc within 80 deg,
so just set $\theta_i = 80^\circ$ and $\theta_f = 0$. It's up to you.

7. Oct 8, 2015

And what abt the inertia....i see you have ignored inertia in your equations

8. Oct 8, 2015

### Daeho Ro

Oh! Sorry. I used $T$ for the angular acceleration. The equation should be
$$\alpha (\theta_f - \theta_i) = \dfrac{1}{2} (\omega_f^2-\omega_i^2),$$
and then
$$T = I \alpha$$.
It's my mistake.