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

[tex]\tau =I\alpha[/tex]

[tex]{\omega }_z={\omega }_{0z}+{\mu }_k\alpha t[/tex]

[tex]I=\frac{1}{2}MR^2[/tex]

[tex]\tau =\frac{1}{2}MR^2\alpha[/tex]

[tex]{\omega }_z=0\alpha =\ \frac{{\omega }_{0z}}{{\mu }_kt}[/tex]

[tex]\tau =-\frac{1}{2}Mr^2\frac{{\omega }_{0z}}{{\mu }_kt} [/tex]

[tex]\tau =r\times F[/tex]

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

[tex]{\mu }_k=\frac{Mr{\omega }_{0z}}{2tF}[/tex]

[tex]F=f{\mu }_k[/tex]

[tex]{\mu }_k=\frac{Mr{\omega }_{0z}}{2tf{\mu }_k}[/tex]

[tex]{\mu }^2_k=\frac{Mr{\omega }_{0z}}{2tf}[/tex]

[tex]{\mu }_k=\sqrt{\frac{Mr{\omega }_{0z}}{2tf}} [/tex]

 

[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:

[tex]\tau =r\times F[/tex]

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

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]

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