Moment of inertia and torque problem

Homework Statement

A grinding wheel is a uniform cylinder with a radius of 8.10 cm and a mass of 0.550 kg.

(a) Calculate its moment of inertia about its center.
kg·m2
(b) Calculate the applied torque needed to accelerate it from rest to 1700 rpm in 2.00 s if it is known to slow down from 1700 rpm to rest in 55.0 s.
m·N

m= .550 kg
r= .081 m

Homework Equations

I have tried I=mr^2, but that is not working. That is all I know how to do.

The Attempt at a Solution

Attempted I=(.550)(.081)^2 and got a very small decimal that was incorrect. I would love to know how this equation is done.

Thank you!

Yes it does! I had no idea to do that. :/ The book doesn't mention it! Now I'm working on the second part.

Yes it does! I had no idea to do that. :/ The book doesn't mention it! Now I'm working on the second part.

You are using the definition of the moment of inertia for a point particle $$m$$ at a distance $$r$$ from the axis of rotation.

For an object of general shape, an integral is needed.

The moment of inertia about some defined axis is:

$$I=\int r^2 dm$$

The physical meaning of that integral, would be to split up your object into small mass elements, $$dm$$, which are point-like in nature and to sum up all their moments of inertia.

Taking the integral over different shapes yields different moments of inertia of the form:

$$I=m(k_1 a^2+k_2 b^2...)$$ where $$k_n$$ is a numeric constant, and $$a,b...$$ are characteristic length elements of the shape.

You can look at the Wikipedia link posted to see some examples to help illustrate the concept.

Finding the moment of inertia of an object is just an exercise in volume integration, on tests, you'll usually be given the formula for the moment of inertia of the objects you'll be working with, but sometimes the point of the exercise is just that, to find the moment of inertia of a particular object.

As for (b), I've got a tip for you, assume that the frictional torque has a constant magnitude, and remember that torque is a vector quantity. You can have clockwise spinning torque, and counter-clockwise spinning torque, one you can define as + and the other as -
(Are you familiar with the right hand rule?)

No I'm not. I'm stuck again... Could I get the two torques and then use the difference as the answer?

No I'm not. I'm stuck again... Could I get the two torques and then use the difference as the answer?

The right hand rule is just a common convention for determining what is a positive cross product, what is a negative cross product and which points where.

http://en.wikipedia.org/wiki/Right_hand_rule

Remember that:

$$|\alpha|=\frac{\tau_{net}}{I}$$

An equivalent notation would be:

$$\vec \tau_{net}=I\vec \alpha$$

Where $$\tau_{net}$$ is the total moment about the axis of rotation (Remember, clockwise and counter-clockwise moments cancel each-other out. We assign a positive value to the counter-clockwise ones, and a negative value to the clockwise ones (Right hand convention), $$\alpha$$ is the angular acceleration about that axis of rotation and $$I$$ is the moment of inertia about that axis of rotation.

First, what is the angular acceleration the question asks for? Hint: $$\alpha\equiv \frac{d\omega}{dt}$$