A vacuum cleaner belt is looped over a shaft of radius 0.44 and a wheel of radiu

Question

A vacuum cleaner belt is looped over a shaft of radius 0.44 and a wheel of radius 2.50 . The arrangement of the belt, shaft, and wheel is similar to that of the chain and sprockets in the figure (Figure 1) . The motor turns the shaft at rotational velocity 62.0 and the moving belt turns the wheel, that in turn is connected by another shaft to the roller that beats the dirt out of the rug being vacuumed. Assume that the belt doesn't slip on either the shaft or the wheel.

Explanation / Answer

When you have two wheels connected like that, you need to use the fact that the velocity of any point on the two wheels has to be the same because the belt or chain can only move at one speed. Use these relationships: 1 rpm = 2pi / 60 rad/s rotational speed 1 rpm = 2pi*R / 60 m/s point speed on the belt So multiply / divide by R to transform rotational speed (different for each wheel) to / from belt speed (which is uniform for the system) So if Wheel A is rotating at wa rad/s, this corresponds to the belt B moving at wa*0.073 m/s. For Wheel C, since belt is moving at wa*0.073 m/s. this corresponds to a rotational speed wc = wa*0.073/0.343 rad/s - since it is bigger, it is turning slower. Since 91.5 rpm = 91.5*2*pi/60 rad/s = 9.58 rad/s = final speed of wheel C in the above description. wa at any point in time is: wa = 8.9t So the rotational speed of wheel C is: wc = wa*0.073/0.343 = 8.9*0.073/0.343 * t = 1.89*t set wc equal to the final rotational speed and solve for t: wc = 9.58 1.89*t = 9.58 t = 9.58/1.89 = 5.06s Given n = 68/15 = 4.53 By omega = 2 x pi x n = 2 x 3.14 x 4.53 = 28.47 rad/sec By omega1 = omega2 + alfa x time =>28.47 = 0 + alfa x 15 =>alfa = 1.90 rad/sec^2 By a = r x alfa =>a = 39 x 10^-2 x 1.90 = 0.74 m/s By s = ut + 1/2at^2 =>s = 0 + 1/2 x (0.74) x 15 x 15 =83.25m

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