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
θ = 2.40t2 − 0.650t3
where θ 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?

(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?

(d) Through how many rotations has the roller turned between
t = 0
and the time found in part (c)?

Answer 1

Given angular posistion

(a) angular speed

  

The maximum angular speed , occurs at t= 4.8/3.9 s= 1.23 s

Placing this value for t into the equation for angular velocity, we
find, Therefore

     

(b) Tangential speed

Given r = D/2 = 9/2 = 4.5 m

  Therefore the maximum tangential speed is v_max = 4.5 x 2.95 = 13.27 m/s

(c) The roller reverses its direction when the angular velocity is zero—recall an object moving vertically upward against gravity reverses its motion when its velocity reaches zero at the maximum height.

   is positive until 4.8t = 1.95t²

t= 4.8/1.95s = 2.46 s (or t = 0 invalid)

The driving force should be removed from the roller at t = 2.46 s.

(d) Upto t = 2.46 s,

   = 0.77 rotations