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Homework Help: Moment of inertia + torque

  1. Mar 15, 2010 #1
    1. The problem statement, all variables and given/known data

    A vehicle wheel has a mass 2.5 kg and diameter 310 mm. The vehicle moves from rest to a linear velocity of 13 kmh -1 in 60 s. Assuming that the wheel can be treated as a solid disc, find the torque applied to the wheel


    2. Relevant equations

    Not so sure on how to add equations that aren't basic...


    3. The attempt at a solution

    I started with

    0.5 x 2.5 x 0.1552 = 0.3003125

    then

    3.61ms-1 / 0.155 = 23.29 rads-1

    23.29-0/60=0.388172043 (angular acceleration)

    so then it should be

    0.388172043 x 0.3003125 = 0.1165729167 Nm right? :confused:
     
  2. jcsd
  3. Mar 15, 2010 #2

    rl.bhat

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    Your calculation appease to be correct.
     
  4. Mar 15, 2010 #3
    i'm on my last question, i'll put it here instead of starting numerous threads... :uhh:

    A hollow cylinder of mass 50 g and external and internal diameters of 50 mm and 44 mm respectively, rolls down a slope for a distance of 5 m without slipping. If the cylinder starts from rest and has a velocity of 3 ms-1 by the time it reaches the 5m point, what was the angle of incline of the slope

    i'm a bit lost on where to start... i know it's something to do with potential energy...

    btw thanks for looking at the other one
     
  5. Mar 15, 2010 #4
    use conservation of energy.
     
  6. Mar 15, 2010 #5
    ke = pe ?

    h = 1/2 x 32/9.81 ?

    can i then assume it's a right angle triangle? or am i off target...? :redface:
     
  7. Mar 15, 2010 #6

    rl.bhat

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    Initial velocity, final velocity and displacement of the cm of the cylinder is given. Find the acceleration of the center of mass.
    The equation of motion for translation a nd for rotation about the axis through cm is
    ΣFx = Mg*sinθ + (-f) = M*a(cm) ....(1)
    Στ = f*R = Iα = I*R*a(cm) .......(2) where f is the frictional force.
    Substitute the value of I in the eq.2 and find the expression for a(cm). From that find θ.
     
  8. Mar 15, 2010 #7
    ok went over some notes... was getting carried away

    i've got

    0.05x9.81x5Sinθ=0.225+0.198 (all sources of ke)

    sinθ=0.1724 radians right?

    180/pi x 0.1724 = 9.91 degrees?
     
  9. Mar 15, 2010 #8
    Well, what is its total kinetic energy? By the way, it would be helpful if you shown all your work. I was a little lost for a few minutes.
     
  10. Mar 15, 2010 #9
    yes apologies!! bit lazy of me :tongue2:
     
  11. Mar 15, 2010 #10
    ok, so you used conservation of energy instead of forces. But could you explain how you got "0.225+0.198 (all sources of ke)". It seems too small for me.
     
  12. Mar 15, 2010 #11
    Never mind, I got it. It's the mass 0.05.
    It looks good now.
     
  13. Mar 15, 2010 #12
    1/2x0.05x32=0.225

    1/2x1.1x10-4x602=0.198

    i've done 3/0.05=60 rads-1 used the radius of the first diameter, no idea if this is correct or not or whether there is a way of using them both.
     
  14. Mar 15, 2010 #13
    I believe 1.1*104 is the moment of inertia, could you show me how you did it as well.
     
  15. Mar 15, 2010 #14
    ok cool, cheers. does my answer seem ok in your opinion?

    having an open-book test in a couple of days :smile: thought i would do as many examples as possible for reference.
     
  16. Mar 15, 2010 #15
    moment of inertia for a hollow disc = 1/2xm(r1^2+r2^2) if that makes sense. sorry i'm not familiar with how to add fancy text for formulas
     
  17. Mar 15, 2010 #16
    It's ok for now, but it would be really helpful if you did know. Here's how to learn latex code:
    https://www.physicsforums.com/misc/howtolatex.pdf

    I thought the + should be -. But ya, you are right. Good luck with your review!
     
  18. Mar 18, 2010 #17
    Is it possible to work out an objects angular velocity when you're only give it's moment of inertia, mass and radius...?
     
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