In a manufacturing process, a large, cylindrical roller is used to flatten mater
ID: 1510756 • Letter: I
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)?Explanation / Answer
here,
diameter , d = 9 m
radius , r = 4.5 m
theta = 2.9 t^2 - 0.65 *t^3
(a)
angular speed w = d/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
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