- #1
kudoushinichi88
- 125
- 2
A uniform horizontal disk of mass M and radius R is rotating about its vertical axis with an angular velocity \omega. When it is placed on a horizontal surface, the coefficient of kinetic friction between the disk and the surface is \mu_k.Find
a)the torque \tau exerted by the force of friction on a circular element of radius r with width dr.
Let the element has a mass of dm.
<br /> \frac{dm}{M}=\frac{\pi(r+dr)^2-\pi r^2}{\piR^2}
dm=\frac{2M}{R^2}rdr
The torque is provided by the frictional force, therefore
<br /> d\tau=f_{friction}r=\mudm\ gr
d\tau=\frac{2M\mu g}{R^2}r^2dr
b)the total torque exerted by friction on the disk
Integrating,
\tau=\frac{2M\mu g}{R^2}\int_{0}^{R}r^2dr
gives us
\tau=\frac{2}{3}M\mu gR
c)the total time required to bring the disk to a halt.
\tau=I\alpha
Since disk is decelerating,
\alpha=-\frac{\tau}{I}=-\frac{4\mu g}{3R}
\omega=\omega_0+\alpha t
t=\frac{3R\omega}{4\mu g}
Can someone check my work? I don't have solutions to this question...
a)the torque \tau exerted by the force of friction on a circular element of radius r with width dr.
Let the element has a mass of dm.
<br /> \frac{dm}{M}=\frac{\pi(r+dr)^2-\pi r^2}{\piR^2}
dm=\frac{2M}{R^2}rdr
The torque is provided by the frictional force, therefore
<br /> d\tau=f_{friction}r=\mudm\ gr
d\tau=\frac{2M\mu g}{R^2}r^2dr
b)the total torque exerted by friction on the disk
Integrating,
\tau=\frac{2M\mu g}{R^2}\int_{0}^{R}r^2dr
gives us
\tau=\frac{2}{3}M\mu gR
c)the total time required to bring the disk to a halt.
\tau=I\alpha
Since disk is decelerating,
\alpha=-\frac{\tau}{I}=-\frac{4\mu g}{3R}
\omega=\omega_0+\alpha t
t=\frac{3R\omega}{4\mu g}
Can someone check my work? I don't have solutions to this question...