In a manufacturing process, a large, cylindrical roller is used to flatten mater
ID: 1600461 • 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 4.00 m, and, while being driven into rotation around a fixed axis, its angular position is expressed as
= 2.30t2 0.850t3
where is in radians and t is in seconds.
(a) Find the maximum angular speed of the roller.
rad/s
(b) What is the maximum tangential speed of a point on the rim of the roller?
m/s
(c) At what time t should the driving force be removed from the roller so that the roller does not reverse its direction of rotation?
s
(d) Through how many rotations has the roller turned between t = 0 and the time found in part (c)?
rotations
Explanation / Answer
w = dtheta/dt
theta = 2.30t^2 - 0.850t^3
w = 4.6t - 2.55t^2
it will be maximum
dw/dt = 4.6- 5.1t = 0
t = 4.6/5.10 = 0.9 s
w_max at 0.9 s
w_max = 4.6 * 0.9 - 2.55 * 0.9^2
w_max = 2.0745 rad/s
part b )
vmax = w_max *r
r = diameter/2 = 4/2 = 2 m
vmax = 4.149 m/s
part c )
The roller reverses its direction when the angular velocity is zero
w = 4.6t - 2.55t^2 = 0
t = 4.6/2.55
The driving force should be removed from the roller at t = 1.8 s
part d )
put t = 1.8s in theta
theta = 2.3 x 1.8^2 - 0.850*1.8^3
theta = 2.4948 rad
2.4948 * 1rotation/2pi rad = 0.397 rotation
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