A roller in a printing press turns through an angle given by theta(t) = gamma t^

Question

A roller in a printing press turns through an angle given by theta(t) = gamma t^2 - beta t^2, where y = 3.15 rad/s^2 and p = 0.585 rad/s^3. (a) Calculate the angular velocity of the roller as a function of time. (Use any variable or symbol stated above as necessary. Do not substitute numerical values; use variables only.) omega (t) = (b) Calculate the angular acceleration of the roller as a function of time. (Use any variable or symbol stated above as necessary. Do not substitute numerical values; use variables only.) alpha (t) = (c) What is the maximum positive angular velocity? rad/s At what value of t does it occur? s

Explanation / Answer

PRINTING PRESS ROLLER

Given : (t) = t2 t3, where = 3.15 rad/s2 and = 0.595 rad/s3

A) To find the angular velocity of the roller as a function of time.
The angular velocity (t) of the roller can be obtained by differentiating the angle with respect to time t.

Therefore we get (t) = (d/dt)((t) = d/dt (t2 t3)

(t) = 2t 3t2

B) To find the angular acceleration of the roller as a function of time.
The angular acceleration () of the roller can be obtained by differentiating the angular velocity (d) with respect to time Therefore we get (t) = (d/dt) (d)

= (d/dt) (2t 3t2)

(t) = 2 6t

C) To find the maximum positive angular velocity
The maximum positive angular velocity (t) is obtained when d/dt = 0 . i.e., d/dt = 0

We know that d/dt = 2 6t   (from part B)
Therefore (2 6t) = 0
2 = 6t = 3t    t = ( / 3)

Given: = 3.15 rad/s2 and = 0.595 rad/s3 t = [3.15 / (3)(0.595)] t = 1.794 s

Max value of angular velocity (t) is given by

(t) max = 2t 3t2

= 2 (3.15) (1.794) - 3(0.595)(1.794)2

= 11.30 -5.64

(t) max = 5.66 rad/s

The maximum angular velocity of 5.66 rad/s occurs at time  t = 1.794 s

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