Question

In a manufacturing process, a large, cylindrical roller is used to flatten material fed beneath it. The diameter of the roller is 9.00 m, and, while being driven into rotation around a fixed axis, its angular position is expressed as theta = 2.90t^22 - 0.650t^3 where 6 is in radians and t is in seconds. Find the maximum angular speed of the roller. What is the maximum tangential speed of a point on the rim of the roller? At what time t should the driving force be removed from the roller so that the roller does not reverse its direction of rotation? I Through how many rotations has the roller turned between t = 0 and the time found in part (c)?

Answer 1

here,

diameter , d = 9 m

radius , r = 4.5 m

theta = 2.9 t^2 - 0.65 *t^3

(a)

angular speed w = dθ/dt = 5.8*t - 1.95 *t^2

dw/dt = 5.8 - 3.9 t

dw/dt = 0 for max w

so max w occurs at t = 5.8/3.9 s

t = 1.49 s

so w max = 5.8 * 1.49 - 1.95 * (1.49)^2

wmax = 4.31 rad/s

(b)

tangential speed v = r*w

so v = 0.5 * 4.31

v = 2.16 m/s

(c)

w is positive until 5.8 t = 1.95 *t^2

so t = 5.8/1.95

t = 2.97 s (or t = 0 invalid)

After t = 2.97s

, w is negative (starts reversing direction of rotation)

Driving force would actually have to be removed some time before t = 2.97 s because the roller can't stop instantaneously, but insufficient info to calculate this.

(d)

Up to t = 2.97s,

theta = 2.9 * 2.97^2 - 0.65 * 2.97^3

theta = 8.63 rad = 1.37 rotations