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Homework Help: Moment of inertia and torque problem

  1. Oct 12, 2009 #1
    1. The problem statement, all variables and given/known data
    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.
    (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= .550 kg
    r= .081 m

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

    3. 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!
  2. jcsd
  3. Oct 12, 2009 #2


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  4. Oct 12, 2009 #3
    Yes it does! I had no idea to do that. :/ The book doesn't mention it! Now I'm working on the second part.
  5. Oct 12, 2009 #4
    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?)
  6. Oct 12, 2009 #5
    No I'm not. I'm stuck again... Could I get the two torques and then use the difference as the answer?
  7. Oct 12, 2009 #6
    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.


    Remember that:


    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)
  8. Oct 12, 2009 #7
    Okay I got the answer to part b as .1664418144. Does that sound correct?

    I have a couple more questions after this one... I swear there are at least three problems on every homework that kill me.
  9. Oct 12, 2009 #8
    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).
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