Torque required to decelarate a disc

In summary, the disc will stop within 80 deg of the desired position when the applied brakes cause a change in the rotational velocity of the system of 170 kgm2.
  • #1
Shady99
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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
wi=20 deg/sec=0.34 rad/sec
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

Please correct and help
 
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  • #2
Shady99 said:
Δt=4 sec
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
[tex] T = \dfrac{d\omega}{dt} = \dfrac{d\theta}{dt} \dfrac{d\omega}{d\theta} = \omega \dfrac{d\omega}{d\theta}. [/tex]
When you integrate it as [itex] \theta[/itex], then
[tex] \int_{\theta_i}^{\theta^f} T d\theta = T(\theta_f - \theta_i) [/tex]
and also it is same as
[tex] \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). [/tex]
 
  • #3
hey thnks for the reply...can you elaborate it more...possibly solve it
 
  • #4
Use the last relation,
[tex] T(\theta_f - \theta_i) = \dfrac{1}{2} (\omega_f^2-\omega_i^2), [/tex]
for your situation. You already know each variable without [itex] T [/itex], so it is elementary calculation
 
  • #5
ok thnks...but can't i determine a stopping distance?
 
  • #6
Shady99 said:
ok thnks...but can't i determine a stopping distance?
You mention that you want to stop the disc within 80 deg,
Shady99 said:
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
so just set [itex] \theta_i = 80^\circ [/itex] and [itex] \theta_f = 0 [/itex]. It's up to you.
 
  • #7
And what abt the inertia...i see you have ignored inertia in your equations
 
  • #8
Oh! Sorry. I used [itex] T [/itex] for the angular acceleration. The equation should be
[tex] \alpha (\theta_f - \theta_i) = \dfrac{1}{2} (\omega_f^2-\omega_i^2), [/tex]
and then
[tex] T = I \alpha [/tex].
It's my mistake.
 

FAQ: Torque required to decelarate a disc

1. What is torque and how does it relate to deceleration?

Torque is a measure of the force that causes an object to rotate around an axis. In the context of deceleration, torque is the force that causes a rotating disc to slow down or come to a stop.

2. How is torque required to decelerate a disc calculated?

The torque required to decelerate a disc can be calculated using the formula T = I * α, where T is the torque, I is the moment of inertia of the disc, and α is the angular acceleration.

3. What factors affect the torque required to decelerate a disc?

The torque required to decelerate a disc is affected by the mass and shape of the disc, as well as the speed and direction of its rotation.

4. Can the torque required to decelerate a disc be reduced?

Yes, the torque required to decelerate a disc can be reduced by decreasing its moment of inertia, increasing the friction between the disc and the surface it is rotating on, or reducing its rotational speed.

5. How is torque applied to a disc to decelerate it?

Torque is typically applied to a disc to decelerate it through the use of brakes, such as caliper brakes or drum brakes, which create friction and slow down the rotation of the disc.

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