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Torque and rotational kinetic energy

  1. Jun 9, 2012 #1
    This is problem from a text book whose solution I don't understand.

    A small rock of mass 0.5kg is attached to a 1.5m string, then it is whirled for 5 seconds until it achieves a near horizontal orbit at 120 rpm. What is the torque required?

    I used the equation : τ=Iα (moment of inertia x angular acceleration) giving an answer of 2.8 Nm or Joules. This is the book's answer.

    My question is, why I don't get the same answer with the formula for rotational kinetic energy
    E_k=1/2 Iω^2 which gives 177 Nm.

    Both formulas have the same units of Nm or energy. Why the energy used to impart the motion is not the same as the kinetic energy stored in the moving mass?
  2. jcsd
  3. Jun 9, 2012 #2
    From the point of view of units, both torque and kinetic energic carry the Nm tag. However, their meanings are vastly different; that´s why an energy is given in Joules and a torque in Nm. You've just compared apples to oranges
  4. Jun 9, 2012 #3
    Torque isn't exactly a measure of energy. While the two quantities have the same dimensions, they aren't obtained in the same way: radians are treated as dimensionless, so ω2 has dimensions [T]-2, which happens to coincide with the dimensions of dω/dt = α

    To illustrate the difference, consider a wheel spinning at a constant rate. The rotational kinetic energy is non-zero, but there is no torque on the wheel (so the torque is zero).
  5. Jun 9, 2012 #4
    Thank you Cauchyam and Gordianus for your help.

    I still have some questions. In the problem in question supposing the mass continues swinging around with no friction. If I want to calculate its energy, is it correct to apply the rotational kinetic energy formula, and would also be valid to say that that was the same amount of energy used to bring it from rest to its current rotation?
  6. Jun 10, 2012 #5


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    And here you can see the relation between kinetic energy and angular momentum: The 2.8Nm were applied while the rock made 5 rotations* with a total angle of 10pi. And 2.8Nm*10pi=88J. I get the same value for the kinetic energy.

    * 0/s at t=0, 2/s at t=5s, with constant acceleration, gives 5 rotations
  7. Jun 10, 2012 #6
    Thank you mfb.
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