Physics Help Forum Tumble dryer problem
 Apr 22nd 2019, 08:34 AM #11 Junior Member     Join Date: Jan 2019 Posts: 25 I looked at this problem to see if the book solution made sense. If the clothes make it all the way around without falling, then at the top of basket $F_c = mr \omega^2 = N + mg$ The minimum speed occurs when $N \to 0$ $mr \omega^2 = mg \implies \omega = \sqrt{\dfrac{g}{r}}$ measured in RPM, $\omega = \sqrt{\dfrac{g}{r}} \cdot \dfrac{30}{\pi}$ substituting the given values yields an $\omega$ of 52 RPM. So, an angular speed of 75 RPM is more than sufficient to keep the clothes from falling at any position in the basket. At any position in the upper half of the basket $F_c = mr \omega^2 = N + mg\cos{\theta}$, where $\theta$ is the radial angle of that position measured from the vertical. As $N \to 0$, $mr \omega^2 = mg\cos{\theta} \implies \omega = \sqrt{\dfrac{g\cos{\theta}}{r}}$ for RPM $\omega = \sqrt{\dfrac{g\cos{\theta}}{r}} \cdot \dfrac{30}{\pi}$ substituting $\theta = 60^\circ$ and $r = 0.325 \, m$ yields an $\omega$ of 37 RPM. Attached Thumbnails   Last edited by Cervesa; Apr 22nd 2019 at 10:04 AM.