I'm taking an introduction to modern physics course and would appreciate some help on this problem:

A yo-yo is attached to a string, whose end is held tightly. It is released from rest.

  1. After the yo-yo has fallen a distance h, find the total kinetic energy of the yo-yo.

  2. Find the speed of the yo-yo after it has fallen a distance h.

  3. Find the magnitude of the tension in the string while the yo-yo is falling.


Attempt at a solution:
$$\Delta KE = KE_f-KE_i=KE_f$$ Going from a point particle system, $$\Delta KE_{trans} = W_{gravity}=mgh$$ But I am not sure this is true because the string also does work on the yo-yo, but I don't know that force.


1 Answer 1


Since energy is conserved, the total kinetic energy after the yo-yo has fallen a distance $h$ is simply the initial energy, $E_1=mgh$. However, gravitational potential energy is converted into two forms of kinetic energy: translational kinetic energy ($\frac{1}{2}mv^2$) and rotational kinetic energy ($\frac{1}{2}I\omega^2$). Here, $I$ is the moment of inertia of the yo-yo, and $\omega$ is the angular velocity.

Now, since the rope is always tangent to the yo-yo, the speed of the rope is simply the tangential velocity of the yo-yo, $v=\omega r$. Thus you can use conservation of energy $E_1=E_2$, where $E_2=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2$ to determine $v$.

Finally, to find the tension, apply Newton's laws $F=ma$, and $\tau=I\alpha,$ with $\alpha$ the angular acceleration. Both the linear force equation and the torque equation apply! This will allow you to determine the tension in the rope in terms of the mass, radius of the yo-yo, etc.

  • 3
    $\begingroup$ Correct but you should also add the point that tension does no work as point of application of force of tension is always stationary. $\endgroup$
    – user600016
    Dec 10, 2019 at 6:35
  • $\begingroup$ That is a good point. Consider another scenario where the yo-yo is not accelerating in the vertical direction (i.e. someone is pulling up on the rope with a force exactly equal to $mg.$) The yo-yo still undergoes angular acceleration. Is work being done by the tension in this case? $\endgroup$
    – aRockStr
    Dec 10, 2019 at 14:48
  • $\begingroup$ @user600016 That is what I'm confused about. Why does the fact the point of application is stationary mean no work is done even though the force still acts through a distance? $\endgroup$
    – elbecker
    Dec 10, 2019 at 20:26
  • $\begingroup$ @elbecker Another example which may help is a wheel which is rolling without slipping along flat ground. As the wheel rolls, the single point along the circumference of the wheel which is instantaneously in contact with the floor is stationary relative to the floor. Because of this, friction does no work (friction requires the object to slide along the floor, which isn't happening here). $\endgroup$
    – aRockStr
    Dec 10, 2019 at 20:56

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