# In a manufacturing process, a large, cylindrical roller is used to flatten material fed beneath it. The diameter of the roller is 1.00 m and

Question

In a manufacturing process, a large, cylindrical roller is used to flatten material fed beneath it. The diameter of the roller is 1.00 m and while being driven into rotation around a fixed axis, its angular position is expressed as
θ = 2.50t2 – 0.600t3
where θ is in radians and t is in seconds.
a) Find the maximum angular speed of the roller
b) what is the maximum tangential speed of the point an the rim of the roller?
c) at what time t should the driven force be removed from the roller so that the roller does not reverse its direction of rotation?
d) Through how many rotations has the roller turned between t=0 and the time found in part c?

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1 year 2021-09-03T19:24:46+00:00 1 Answers 186 views 0

B) 1.47m/s

C) t = 2.78s

D) 5.4 rotation

Explanation:

(a) angular speed w = dθ/dt = 5t – 1.8t^2

dw/dt = 5 – 3.6t = 0 for max w

so max w occurs at t = 5/3.6 s = 1.39s

so w max = 5*1.39 – 1.8*(1.39)^2 = 3.47 rad/s

(b) tangential speed v = r*w

r = D/2 = 0.5m

so v = 0.5*w = 1.74 m/s

(c) w is positive until 5t = 1.8t^2

so t = 5/1.8s = 2.78s (or t = 0 invalid)

After t = 2.87s, w is negative (starts reversing direction of rotation)

Driving force would actually have to be removed some time before t=2.78s because the roller can’t stop instantaneously, but insufficient info to calculate this.

(d) Up to t = 2.78s, θ = 2.5*(2.78)^2 – 0.6*(2.78)^3 rad = 33.95 rad = 5.40 rotations