Calculating Applied Torque for a Grinding Wheel

[PeachBanana]

Homework Statement

A grinding wheel is a uniform cylinder with a radius of 10.0 cm and a mass of 0.570 kg.

Calculate its moment of inertia about its center.
2.85 x 10^-3 kg * m^2

Calculate the applied torque needed to accelerate it from rest to 1700 rpm in 4.80s if it is known to slow down from 1700 rpm to rest in 57.0s .

Homework Equations

I = 1/2 Mr^2
ω final - ω initial / t = α
τ = Iα

The Attempt at a Solution

1700 rpm = 178.02 rad./s
τ = Iα

τ = (2.85 * 10^-3 kg * m^2)(37.0815 rad./s^2) = 0.1056 N*m (Accelerating from rest to 1700 rpm in 4.80 s)

τ = (2.85*10^-3 kg * m^2)(-3.123 rad./s^2) = -0.00890 N*m (Decelerating from 1700 rpm to rest in 57.0s)

I have these two torques and I'm unsure of what to do with them.

 

[cepheid]

The second bit of info about the slowing down tells you what the frictional torque is. The first bit of info about speeding up tells you what NET torque you need to achieve that acceleration. So, given that frictional torque, what must the *applied* torque be in order to produce the net torque that is required?

 

[PeachBanana]

net τ = the sum of the magnitudes of the two torques I listed. I added them together and got the correct answer of 0.1145 N * m. Thank you!

 

[cepheid]

net τ = the sum of the magnitudes of the two torques I listed. I added them together and got the correct answer of 0.1145 N * m. Thank you!

Actually, it's

τ_net = τ_app + τ_fric

The two that you have are net and frictional. It's applied that you're trying to solve for:

τ_net - τ_fric = τ_app

However, since the frictional torque is negative, when you subtract it, it's the same as adding its magnitude. That's why adding the two torques gave you the right answer.

So your arithmetic was right, it was just your interpretation that was a bit off.

 

[PeachBanana]

Oops. Thank you for correcting me.