Moment of inertia and torque problem

Isn't there supposed to be a 1/2 in that equation for a solid cylinder?
Check http://en.wikipedia.org/wiki/List_of_moments_of_inertia

 

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

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

The moment of inertia about some defined axis is:

[tex]I=\int r^2 dm[/tex]

The physical meaning of that integral, would be to split up your object into small mass elements, [tex]dm[/tex], 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:

[tex]I=m(k_1 a^2+k_2 b^2...)[/tex] where [tex]k_n[/tex] is a numeric constant, and [tex]a,b...[/tex] 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?)

 

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:

[tex]|\alpha|=\frac{\tau_{net}}{I}[/tex]

An equivalent notation would be:

[tex]\vec \tau_{net}=I\vec \alpha[/tex]

Where [tex]\tau_{net}[/tex] 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), [tex]\alpha[/tex] is the angular acceleration about that axis of rotation and [tex]I[/tex] is the moment of inertia about that axis of rotation.

I'll help you break the question down.

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

Knowing the moment of inertia of the grinding wheel, what kind of net torque would you need on it to achieve that angular acceleration?

The two torques that act on the grinding wheel are the frictional torque, and the one that you apply. How would you go about finding the frictional torque, and what does its existence tell you about the torque you need to apply? (Assume that the frictional torque is constant in magnitude)

 

You're off by a factor of a bit less than 10, I'm afraid. Please show your work so we can see where the mistake is (I got about 1.58 Nm, so it may just be a factor of 10+rounding errors, but I only used a calculator at the very end, so I doubt it).