# Torque and Rotational Kinematics

1. Nov 3, 2007

### zbobet2012

1. The problem statement, all variables and given/known data
A grindstone in the shape of a solid disk with diameter 0.490 m and a mass of m = 50.0 kg is rotating at omega = 890 rev/min. You press an ax against the rim with a normal force of F = 170 N, and the grindstone comes to rest in 7.20 s.

2. Relevant equations

τ=Iα
ωz0zkαt
τ=r×F

3. The attempt at a solution
$$\tau =I\alpha$$

$${\omega }_z={\omega }_{0z}+{\mu }_k\alpha t$$

$$I=\frac{1}{2}MR^2$$

$$\tau =\frac{1}{2}MR^2\alpha$$

$${\omega }_z=0\alpha =\ \frac{{\omega }_{0z}}{{\mu }_kt}$$

$$\tau =-\frac{1}{2}Mr^2\frac{{\omega }_{0z}}{{\mu }_kt}$$

$$\tau =r\times F$$

$$r\times F=\frac{1}{2}Mr^2\frac{{\omega }_{0z}}{{\mu }_kt}$$

$${\mu }_k=\frac{Mr{\omega }_{0z}}{2tF}$$

$$F=f{\mu }_k$$

$${\mu }_k=\frac{Mr{\omega }_{0z}}{2tf{\mu }_k}$$

$${\mu }^2_k=\frac{Mr{\omega }_{0z}}{2tf}$$

$${\mu }_k=\sqrt{\frac{Mr{\omega }_{0z}}{2tf}}$$

Last edited: Nov 3, 2007
2. Nov 3, 2007

### saket

Okay, although not provided in the question, I am assuming that you are supposed to find coefficient of kintic friction. (Your attempt at the solution tells me that.)
You erred in this statement:
Note that, torque produced by F is not the one which is producing deceleration in the disc (It is in a perpendicular direction to ωz!), rather it is f, torque due to friction which is producing deceleration. Keep this in mind and I hope you would be able to solve it, as you know other things, it seems.

3. Nov 6, 2007

### zbobet2012

$$F=f{\mu }_k$$ <---isn't that the force due to friction? Or am I confused? Realizing I kind of messed up my notation, $$f = 170N$$ and $$F_{friction}=F$$

Last edited: Nov 6, 2007