Question #123777

The angular momentum of a flywheel having a Moment of inertia 0.125kgm^3 decreases from 3.0 to 2.0kgm^2/s^2 in a period of 1.5s

(i) Assuming a uniform angular acceleration, through how many revolutions will the flywheel have tuned?

(ii) how much work is done?

(i) Assuming a uniform angular acceleration, through how many revolutions will the flywheel have tuned?

(ii) how much work is done?

Expert's answer

Solution.

"I=0.125kgm^2;"

"L1=3.0kgm^2\/s;"

"L_2=2.0kgm^2\/s;"

"t=1.5s;"

i)"M=\\dfrac{\\Delta L}{\\Delta t};"

"M=\\dfrac{1kgm^2\/s}{1.5s}=0.7kgm^2\/s;"

"M=I\\epsilon\\implies \\epsilon=\\dfrac{M}{I};"

"\\epsilon=\\dfrac{0.7kgm^2\/s}{0.125kgm^2}=5.6rad\/s^2;"

"\\epsilon=\\dfrac{\\Delta\\omega}{\\Delta t};" "\\Delta \\omega=\\dfrac{\\Delta\\phi}{\\Delta t};"

"\\Delta\\phi=\\epsilon(\\Delta t)^2;"

"\\Delta\\phi=5.6rad\/s^2\\sdot(1.5s)^2=12.6rad;"

"n=\\dfrac{12.6rad}{2\\pi rad}=8;"

ii )"A=\\dfrac{I\\omega_1^2}{2}-\\dfrac{Iw_2^2}{2};"

"L_1=I\\omega_1;"

"\\omega_1=\\dfrac{3.0kgm^2\/s}{0.125kgm^2}=24rad\/s;"

"\\omega_2=\\dfrac{2.0kgm^2\/s}{0.125kgm^2}=16rad\/s;"

"A=\\dfrac{0.125kgm^2\\sdot(24rad\/s)^2}{2}-\\dfrac{0.125kgm^2\\sdot(16rad\/s)^2}{2}=20J;"

**Answer: **"i)n=8;"

"ii)A=20J."

Learn more about our help with Assignments: MechanicsRelativity

## Comments

## Leave a comment