Torque and Rotational Kinematics

[zbobet2012]

Homework Statement

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.

Homework Equations

τ=Iα
ω_z=ω_0z+μ_kαt
τ=r×F

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}}

 

[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:

\tau =r\times F

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

Note that, torque produced by F is not the one which is producing deceleration in the disc (It is in a perpendicular direction to ω_z!), 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.

 

[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