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Homework Help: Rotational energy on an incline

  1. Mar 1, 2015 #1
    1. The problem statement, all variables and given/known data
    A cart whose body has mass M = 1.5 kg is set on four tires each of which has mass m = 0.3 kg and radius r = 0.1 meters. Each tire can be treated as a solid disk with rotational inertia mr2/2. The cart is set on an incline a height h = 1.2 meters high and released. At the bottom, the cart runs into a spring whose force constant is 600 N/m. What is the speed of the cart when it reaches the bottom of the incline? How much is the spring compressed when the cart comes to rest?

    2. Relevant equations
    Mgh = 1/2*MV^2 + 1/2*Iω^2
    U(s) = 1/2*Kx^2
    I = 4(m*r^2)
    3. The attempt at a solution
    I = 4(0.3*0.1^2)
    I = 0.006 Kg m^2

    2*Mgh = V^2 (M + I/r^2)
    V = √(2*Mgh)/(M + I/r^2)
    V = √(2*1.5*9.8*1.2)/(1.5 + 0.006/0.1^2)
    V = √(35.28)/(2.1)
    V = 4.09 m/s

    Mgh = 1/2*MV^2 + 1/2*Iw^2 + 1/2*Kx^2
    35.28 = 12.546 + 5.01843 + 250x^2
    x = √(17.71557)/(250)
    x = 0.26 m

    So what did I do wrong?
  2. jcsd
  3. Mar 1, 2015 #2


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    Hello Zach, welcome to PF :smile:

    A clear and concise post. Not all that verbose on the attempt at solution, but it looks as if you know what you are doing.
    So just a quick hint and you'll be fine, I hope:

    I see that you fill in 1.5 kg for M. But it's not just the cart body that's losing potential energy ! Nor is it just the cart body that has kinetic energy from linear motion: a moving wheel has kinetic energy also if it doesn't rotate !

    And in b) the problem formulation "when the cart comes to rest" should help you see what you can do to improve that answer :wink:

    I sense a little snag there: If your problem statement means the spring is along the incline, the cart will lose some more potential energy, so it becomes a bit nastier equation than when the spring is horizontal.
    Last edited: Mar 1, 2015
  4. Mar 1, 2015 #3
    so M is the mass of the moving system, which means M is 1.5 + 4(0.3) = 2.7 Kg
    V = √(2*2.7*9.8*1.2)/(2.7 + 0.006/0.1^2)
    V = √(63.504)/(3.3)
    V = 4.38 m/s
    And for part b I think you should see the diagram for better understanding

    Attached Files:

  5. Mar 2, 2015 #4


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    Sure helps ! So it's the easy way. Now think of ##v_{\rm final}## and ##\omega_{\rm final}##
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