Braking Torque Needed to Stop Shaft w/ 34kg m Inertia @ 600rpm

[Louis Harriss]

A tyre on a shaft with a moment of inertia of 34kg m is initially running at 600rpm. It is brought to rest in 18 complete revolutions by a braking torque; reversed, and accelerated in the opposite direction by a driving torque of 675Nm. The friction couple throughout Is 160Nm.
Find the braking torque required on the tyre to stop the shaft:
The only formula I can find is Braking torque= total inertia x RPM / constant x stopping time( which I don't know)?
I should be expecting an answer of 423
Secondly, the revolutions turned through in attaining full speed again?
All equations refer to time not by the amount of rotations
Thanks

 

[BvU]

Hello Louis,

This is clearly homework and as such should be posted in the homework forum, where there is a most useful template waiting to be filled in.
In exchange for your effort: the formulas for angular motion with constant angular acceleration are analog to those for linear motion with constant acceleration. So ##\tau = I\alpha## and ##\alpha = {\Delta \omega\over \Delta t}## (not times delta t !).
You don't know ##\Delta t## but you do know it takes 18 revolutions...

 

[CWatters]

There are direct equivalents between the equations for linear and rotational motion. For example compare..

Linear case...
Force = mass * acceleration

Rotational case..
Torque = Moment of inertia * angular acceleration

If you are familiar with the equations of motion (eg SUVAT for constant acceleration) then you can convert those to their rotational equivalents. It's almost as simple as changing the units. eg using angular velocity (radians/sec) in place of linear velocity (meters/second). Likewise for angular displacement (Rads), angular acceleration (rads/S^2) etc.