I have a conceputal question regarding the following problem:

A round massive stone disk with diameter $0.600 m$ has a mass of $50.0 kg$. The stone rotates at an angular velocity of $115.2 rad/s$, around an axis located at the center of the disk. An axe pressed against the disk applies a tangential frictional force, $F_f$ at $50.0 N$. Assume that we turn off the rotating power of the disk, so that the frictional force is the only force acting on the disk. What is the magnitude of the angular acceleration, $\alpha$, of the disk?

OK, so I have calculated the torque ($-15 Nm$), and the moment of inertia ($I = 2.25 kg \cdot m^2$). By using the formula:

$$\tau = I \alpha$$

It is easy to show that $\alpha = -6.67 rad/s^2$, which is also the correct answer.

However, say I want to try to solve this in another way, without using the torque and moment of inertia. Since the frictional force is the only force acting on the disk, I would assume that we would have:

$$F_f = ma_T$$

$$-50 = ma_T$$

$$-50 = 50 \cdot a_T$$

$$a_T = - 1 m/s^2$$

And since we have:

$$a_T = r \alpha$$

This gives:

$$\alpha = \frac{-1}{0.3} = -3.33 rad/s^2$$

Which is not the same answer. So why doesn't this second approach work? If anyone can explain this to me, I would really appreciate it!

  • 1
    $\begingroup$ Well, I interpreted the problem such that $F$ is the actual frictional force. After all, I used this value to calculate the torque, and then my answer became correct. $\endgroup$
    – user12277
    Commented Oct 9, 2012 at 19:44
  • $\begingroup$ If you managed to figure it out, feel free to post an answer to your own question explaining what you did. $\endgroup$
    – David Z
    Commented Oct 9, 2012 at 19:57
  • $\begingroup$ @David Zaslavsky: The answer above was to a comment from someone else (the comment is deleted now I see). So I still don't understand why my second approach above gives the incorrect answer. $\endgroup$
    – user12277
    Commented Oct 9, 2012 at 20:12
  • $\begingroup$ Oh, OK, I didn't see the earlier deleted comment. $\endgroup$
    – David Z
    Commented Oct 9, 2012 at 20:18
  • $\begingroup$ @DavidZaslavsky: Hi there David, It was me who posed at the question... The user responded to the comment and instead of replying "Thanks", I just upvoted his & erased mine... $\endgroup$ Commented Oct 10, 2012 at 5:30

1 Answer 1


Your second approach is conceptually very wrong... The equation you wrote, $F = m\cdot a$, holds for the acceleration of the center of mass of the whole disk, not for individual points of it. And since your disk is not going anywhere when you apply $F_f$ with the axe, we can assume that the axis of rotation is applying a force $-F_f$ that exactly compensates the other one, so that the center of mass remains at rest.

Note that while this pair of forces are parallel, they are not co-linear, so they still generate the torque you calculated before. I don't think there is a way of solving your problem without dealing with the moment of inertia...


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