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!
Best Answer
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...