Solving for Torque and Angular Speed in an Engine

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[SOLVED] Torque & Angular Speed

Homework Statement

The wheel of an engine has a moment of inertia 2.80kgm^2 about its rotation axis.

What constant torque is required to bring it up to an angular speed of 360 rev/min in a time of 7.60 s, starting from rest?

What is its final kinetic energy?

Homework Equations

torque = rF sin(theta)
k=.5mv^2

The Attempt at a Solution

First I converted 360 rev/min to seconds

so

360 (rev/min) ( 1 min/60s)( 2pi/rev) = 37.699 rad/s

t = 7.60

I = 2.80 kgm^2

so I decided to get angular acceleration, in case I need it is

alpha = 4.96

From what I spotted in my book Sum of torque = I(angular acceleration)

torque = (2.80)(4.96)
torque = 13.89

this is right, but is they way I have done it correct or did I just get lucky. I am afraid I just plugged into the "sum of torquez = I (angular accelerationz)"

without understanding it :O

part two of the question is

What is its final kinetic energy?

for this

K=.5mv^2

how do I implement this into the problem if I don't have mass?

omega = omega initial + alpha(time)

 

[Doc Al]

The Attempt at a Solution

First I converted 360 rev/min to seconds

so

360 (rev/min) ( 1 min/60s)( 2pi/rev) = 37.699 rad/s

t = 7.60

I = 2.80 kgm^2

so I decided to get angular acceleration, in case I need it is

alpha = 4.96

From what I spotted in my book Sum of torque = I(angular acceleration)

torque = (2.80)(4.96)
torque = 13.89

this is right, but is they way I have done it correct or did I just get lucky. I am afraid I just plugged into the "sum of torquez = I (angular accelerationz)"

without understanding it :O

That's fine. Realize that the equation "Torque = I*alpha" is just the rotational version of Newton's 2nd law; the translational version (which you know and love) is "Force = mass*acceleration".

part two of the question is

What is its final kinetic energy?

for this

K=.5mv^2

how do I implement this into the problem if I don't have mass?

You don't need mass. For rotational kinetic energy, use I. What's the rotational version of the kinetic energy formula?

Read this: Rotational-Linear Parallels

 

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Great! Thank you very much Doc Al, you are the greatest. :)