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 = 0.450 rad/s and = 1.35×102 rad/s3.

(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.

(b)What is the initial value of the angular velocity?

(c)Calculate the instantaneous value of the angular velocity z at t= 5.20 s .

(d)Calculate the average angular velocity avz for the time interval t=0 to t= 5.20 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:
omega_0 occur when t=0
from equation in PART B: omega(t) = (gamma)+3(beta)*t^2
substitute gamma = 0.450, beta = 0.0135 and t=0
answ:
omega_0 : 0.450 + (3*0.0135^2) = 0.45 rad/s

PART C:
the working is as in PART B
omega(5) = (gamma)+3(beta)*t^2 = 0.450 + (3*0.0135*5.2^2) = 1.54 rad/s

PART D:
the formula of average angular velocity for the time interval t = 0 to t = 5.20 seconds is : (theta(5.2) - theta(0)) / ( 5.2-0 )
answ:
(0.45(5) + (0.0135*(5.2^3)) - 0) / (5.2-0) = 0.79 rad/s

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