The maximum angular speed of the roller is 3.47rad/a,
the maximum tangential speed of the point an the rim of the roller is 1.47m/s,
the time t should the driven force be removed from the roller so that the roller does not reverse its direction of rotation is t = 2.78s,
5.4 rotation has the roller turned between t=0 and the time found in part c
<h3>What does tangential speed refer to?</h3>
Any item moving along a circular path has a linear component to its speed called tangential velocity. An object's velocity is always pointed tangentially when it travels in a circle at a distance r from the center. The word for this is tangential velocity.
To calculate the tangential speed, divide the circumference by the time required to complete one spin. For instance, if it takes 12 seconds to complete one rotation, divide 18.84 by 12 to get 1.57 feet per second as the tangential velocity.
(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
Therefore,
A) 3.47rad/a
B) 1.47m/s
C) t = 2.78s
D) 5.4 rotation
To learn more about tangential speed, refer to:
brainly.com/question/19660334
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