A child is pushing a merry-go-round. The angle through which the merry-go-round

Question

A child is pushing a merry-go-round. The angle through which the merry-go-round has turned varies with time according to (t)=t+t3, where = 0.439 rad/s and = 1.30×102rad/s3.

NEED PARTS A-D

PART A: Calculate the angular velocity of the merry-go-round as a function of time.

Express your answer in terms of the variables , , and t.

ANSWER: Wz(t) =______________ rad/s

PART B: What is the initial value of the angular velocity?

ANSWER Wz =_________________ rad/ s

PART C: Calculate the instantaneous value of the angular velocity

z at t= 4.50 s .

ANSWER Wz =_________________ rad/ s

PART D: Calculate the average angular velocity

avz for the time interval t=0 to t= 4.50 s .

ANSWER :

avz =_________________ rad/ s

Explanation / Answer

PART A:
angular velocity = omega(t)
in order to find angular velocity, differentiate theta(t) given in the question.
answ:
omega(t) = (gamma)+(3*beta*t^2) rad/s

PART B:
initial angular velocity occur when t=0
from equation in PART B: omega(t) = (gamma)+3(beta)*t^2
substitute gamma = 0.439, beta = 0.013 and t=0
answ:
initial angular velocity = 0.439 + (3*0.013*0^2) = 0.439 rad/s

PART C:
the working is as in PART B
omega(5) = (gamma)+3(beta)*t^2 = 0.439 + (3*0.013*5^2) = 1.414 rad/s

PART D:
the formula of average angular velocity for the time interval t = 0 to t = 5.00 seconds is : (theta(5) - theta(0)) / ( 5-0 )
answ:
(0.439(5) + (0.013*(5^3)) - 0) / (5-0) = 0.764 rad/s

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