The flywheel of an engine has a moment of inertia 2.10{\rm kg} \cdot {\rm m}^{2} about its rotation axis.

What constant torque is required to bring it up to an angular speed of 450rev/min in a time of 8.50s , starting from rest?

What is its final kinetic energy?

Question

The flywheel of an engine has a moment of inertia 2.10{\rm kg} \cdot {\rm m}^{2} about its rotation axis.


What constant torque is required to bring it up to an angular speed of 450rev/min in a time of 8.50s , starting from rest?


What is its final kinetic energy?

anonymous · Answer

$$\Sigma \tau=I\alpha \ \alpha=\frac{\Delta \omega}{\Delta t} $$
Strategy:
1) Find alpha first
2) Find torque later :)

anonymous · Answer

thank you :)

anonymous · Answer

how to find the Wo?

anonymous · Answer

It's zero (from rest)

anonymous · Answer

oh okay thanks ill try to solve now

anonymous · Answer

i got $$\alpha=47.123$$

anonymous · Answer

next is?

anonymous · Answer

Plug in this alpha to $$\Sigma \tau=I\alpha  $$ I=2.10

anonymous · Answer

i did it. but when i entered the answer in mastering physics its worng -_- N.m

anonymous · Answer

Omega must be in radians per second

anonymous · Answer

oh now i know whats wrong. i didnt divide it with time. waiiit :) ill see

anonymous · Answer

now correct :D how about the second question? how to do it :)

anonymous · Answer

$$K=\frac12 I\omega^2$$

anonymous · Answer

same value from the first? the w?

anonymous · Answer

Yes, final omega in rad/s

anonymous · Answer

okay 2x

anonymous · Answer

got new question. ill post now :)

anonymous · Answer

Keep posting :) Lot of people will help you here.