In a home laundry dryer, a cylindrical tub containing wet clothes is rotated ste

Question

In a home laundry dryer, a cylindrical tub containing wet clothes is rotated steadily about a horizontal axis, as shown in the figure below. So that the clothes will dry uniformly, they are made to tumble. The rate of rotation of the smooth-walled tub is chosen so that a small piece of cloth will lose contact with the tub when the cloth is at an angle of ? = 61.0° above the horizontal. If the radius of the tub is r = 0.435 m, what rate of revolution is needed in revolutions per second?


________ rev/sec

Explanation / Answer

As the cloth rotates against the wall of the drier drum, there are essentially two forces acting on it, the normal force of the drier wall and the weight of the cloth (mg). These forces cause a centripetal acceleration and both act toward the center of the drum and therefore are both positive. So from Newton’s 2nd law: ?F = ma(c) = n + mg Of course centripetal acceleration is v²/r: mv²/r = n + mg At the 61.0° angle the piece of cloth falls from the wall so the normal force at that point vanishes. Also, the component of gravity acts at a 61.0° angle off the drier wall. The equation becomes: mv²/r = mgsin? v²/r = gsin? v = v[rgsin?] = v[(0.435m)(9.80m/s²)sin61.0°] = 1.93m/s This equates to an angular speed of: ? = v / r = 1.93m/s / 0.435m = 4.43rad/s The rate of revolution is: (v/2(pi)(.385)) * 60sec = 47.87 Revolutions/min Hope you find this helpful!

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