Question

A torque of 77.7 Nm causes a wheel to start from rest, completes 5.55 revolutions and attains a final angular velocity of 88.8 rad/sec. What is the moment of inertia of the wheel?

I am trying to find angular acceleration first. How do I find it given this information? I need to know time...Also, I don't see how that information about revolutions will be used. What do I do?

Thank you!

Answer 1

torque = I alpha
where alpha = d omega/dt = angular acceleration

now
velocity proportional to time so
average angular velocity = 88.8/2 = 44.4 rad/s
does 5.55 *2pi radians
so
time = 5.55 * 2 pi /44.4 rad/s
= .0765 seconds to accelerate
acceleration = alpha = change in omega/ change in time
= 88.8/.0765 = 1161 rad/s^2

and of course
77.7 = I (1161)

do not trust my arithmetic.

Answer 2

time = 5.55 * 2 pi /44.4 rad/s
= .785 seconds to accelerate
acceleration = alpha = change in omega/ change in time
= 88.8/.785 = 113 rad/s^2

and of course
77.7 = I (113)

Answer 3

the answer is 0.687

Answer 4

Convert 5.55 revolutions into radians. 1 rev = 2pi rad, so 5.55rev = 34.87 rad.

Use the equation ωf^2 = ωi^2 + 2α∆θ.

ωf = 88.8rad/sec, ωi = 0, and ∆θ = 34.87 rad.

Knowing those variables, you can calculate α = 113.069.

Finally, you can use the equation τ = Iα (where τ = torque, I = moment of inertia, and α = angular acceleration) to calculate the moment of inertia.

τ = 77.7Nm and α = 113.069.

I = τ/α = 77.7/113.069 = 0.687

Hope that helps