Question

An object rotates according to the following expression: θ(t) = 2.3t3 – 2t2 + 0.5t with θ in radians and time in seconds.

Find the average angular speed from t = 0 to t = 2 seconds.

Find the instantaneous angular speed of the object at t = 2 seconds.

Find the average angular acceleration from t = 0 to t = 2 seconds.

Find the instantaneous angular acceleration of the object at t = 2 seconds.

Answer 1

Part A:

(t) = 2.3*t^3 - 2*t^2 + 0.5*t

Average angular speed is given by:

w_avg = [(t2) - (t1)]/[t2 - t1]

from t1 = 0 to t2 = 2 sec

t2 - t1 = 2 - 0 = 2

(t1) = 2.3*0^3 - 2*0^2 + 0.5*0 = 0

(t2) = 2.3*2^3 - 2*2^2 + 0.5*2 = 11.4

So,

w_avg = [11.4 - 0]/[2 - 0] = 5.7 rad/sec

Part B.

Instantaneous angular velocity is given by:

w(t) = d/dt

w(t) = d[2.3*t^3 - 2*t^2 + 0.5*t]/dt

w(t) = 2.3*3*t^2 - 2*2*t + 0.5*1

w(t) = 6.9*t^2 - 4*t + 0.5

Now at t = 2 sec

w(2) = 6.9*2^2 - 4*2 + 0.5

w(2) = 20.1 rad/sec

Part C.

w(t) = 6.9*t^2 - 4*t + 0.5

Average angular Acceleration is given by:

_avg = [w(t2) - w(t1)]/[t2 - t1]

from t1 = 0 to t2 = 2 sec

t2 - t1 = 2 - 0 = 2

w(t1) = 6.9*0^2 - 4*0 + 0.5 = 0.5

w(t2) = 6.9*2^2 - 4*2 + 0.5 = 20.1

So,

_avg = [20.1 - 0.5]/[2 - 0] = 9.8 rad/sec^2

Part D.

Instantaneous angular acceleration is given by:

(t) = dw/dt

(t) = d[6.9*t^2 - 4*t + 0.5]/dt

(t) = 6.9*2*t - 4*1 + 0

(t) = 13.8*t - 4

Now at t = 2 sec

(2) = 13.8*2 - 4

(2) = 23.6 rad/sec^2

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