- #1
kudoushinichi88
- 129
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A uniform horizontal disk of mass M and radius R is rotating about its vertical axis with an angular velocity [itex]\omega[/itex]. When it is placed on a horizontal surface, the coefficient of kinetic friction between the disk and the surface is [itex]\mu_k[/itex].Find
a)the torque [itex]\tau[/itex] exerted by the force of friction on a circular element of radius r with width dr.
Let the element has a mass of dm.
[tex]
\frac{dm}{M}=\frac{\pi(r+dr)^2-\pi r^2}{\piR^2}[/tex]
[tex]dm=\frac{2M}{R^2}rdr[/tex]
The torque is provided by the frictional force, therefore
[tex]
d\tau=f_{friction}r=\mudm\ gr[/tex]
[tex]d\tau=\frac{2M\mu g}{R^2}r^2dr[/tex]
b)the total torque exerted by friction on the disk
Integrating,
[tex]\tau=\frac{2M\mu g}{R^2}\int_{0}^{R}r^2dr[/tex]
gives us
[tex]\tau=\frac{2}{3}M\mu gR[/tex]
c)the total time required to bring the disk to a halt.
[tex]\tau=I\alpha[/tex]
Since disk is decelerating,
[tex]\alpha=-\frac{\tau}{I}=-\frac{4\mu g}{3R}[/tex]
[tex]\omega=\omega_0+\alpha t[/tex]
[tex]t=\frac{3R\omega}{4\mu g}[/tex]
Can someone check my work? I don't have solutions to this question...
a)the torque [itex]\tau[/itex] exerted by the force of friction on a circular element of radius r with width dr.
Let the element has a mass of dm.
[tex]
\frac{dm}{M}=\frac{\pi(r+dr)^2-\pi r^2}{\piR^2}[/tex]
[tex]dm=\frac{2M}{R^2}rdr[/tex]
The torque is provided by the frictional force, therefore
[tex]
d\tau=f_{friction}r=\mudm\ gr[/tex]
[tex]d\tau=\frac{2M\mu g}{R^2}r^2dr[/tex]
b)the total torque exerted by friction on the disk
Integrating,
[tex]\tau=\frac{2M\mu g}{R^2}\int_{0}^{R}r^2dr[/tex]
gives us
[tex]\tau=\frac{2}{3}M\mu gR[/tex]
c)the total time required to bring the disk to a halt.
[tex]\tau=I\alpha[/tex]
Since disk is decelerating,
[tex]\alpha=-\frac{\tau}{I}=-\frac{4\mu g}{3R}[/tex]
[tex]\omega=\omega_0+\alpha t[/tex]
[tex]t=\frac{3R\omega}{4\mu g}[/tex]
Can someone check my work? I don't have solutions to this question...