- #1
Louis Harriss
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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
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