Torque required to decelarate a disc

  • Thread starter Thread starter Shady99
  • Start date Start date
  • Tags Tags
    Disc Torque
Click For Summary

Discussion Overview

The discussion revolves around calculating the torque required to decelerate a rotating disc mounted on a shaft. Participants explore the relationship between angular velocity, moment of inertia, and the stopping distance of the disc, with a focus on the dynamics involved in bringing the disc to a stop within a specified angular travel of 80 degrees.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the initial calculation for torque using the moment of inertia and angular deceleration, but expresses uncertainty about the time variable used in the calculation.
  • Another participant challenges the assumption that the time to stop can be simply set to 4 seconds, suggesting that the disc will stop before completing the 80 degrees of rotation.
  • A different approach is proposed involving integration of torque with respect to angular displacement and velocity, indicating a more complex relationship than initially considered.
  • Participants discuss the possibility of determining a stopping distance, with one suggesting to set initial and final angular positions for the calculation.
  • Concerns are raised about the treatment of inertia in the equations, prompting a correction regarding the relationship between torque, angular acceleration, and moment of inertia.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to calculate the torque or the stopping distance, with multiple competing views and methods presented throughout the discussion.

Contextual Notes

Some participants express uncertainty about the time variable and its implications for the stopping distance, while others highlight the need to consider inertia in the calculations. The discussion reflects various interpretations of the equations involved and their application to the problem at hand.

Shady99
Messages
71
Reaction score
0
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
 
Engineering news on Phys.org
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]
 
hey thnks for the reply...can you elaborate it more...possibly solve it
 
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
 
ok thnks...but can't i determine a stopping distance?
 
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.
 
And what abt the inertia...i see you have ignored inertia in your equations
 
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.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
9K
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
10K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 5 ·
Replies
5
Views
22K
  • · Replies 1 ·
Replies
1
Views
6K