1. A merry-go-round with diameter 4m rotates at angular speed 0.8rev/s. Suddenly

Question

1. A merry-go-round with diameter 4m rotates at angular speed 0.8rev/s. Suddenly a 30kg child sits on the edge of the merry-go-round. If the rotational inertia of the merry-go-round is 100kg.m2 find the new angular speed. (Hint: assume that the angular momentum is conserved for system and the initial angular momentum of the child be zero.) x=12cos(24t+4) 2. The displacement (X) of a simple harmonic oscillation is given by . Answer the following questions. (a) Find the amplitude, frequency and the phase constant of the motion. b) What is the maximum speed and maximum acceleration of the oscillator. (c) Find the dispalcement, velocity and acceleration when t-0.5 seconds.

Explanation / Answer

solution :

1) we have

moment of inertia, I = 100 k.g * m2

radius of marry-go-round, r = 4/2 = 2 m

angular speed of marry-go-round, w = 0.8 rev/s = 0.8*2*3.14 rad/s = 5.024 rad/s

angular momentum before child sit on marry-go-round = 100*5.024 = 502.4 K.g * m2/s

let w1 be the angular velocity after child sit on marry-go-round then

from conservation of angular momentum

total initial momentum = total final momentum

502.4 = moment of inertia of child * w1 + moment of inertia of marry-go-round * w1

502.4 = 30*2*2*w1 + 100*w1

w1 = 502.4/220 = 2.836 rad/s

2) we have

x= 12*cos(24*t + Pi/4)

and equation of simple harmonic motion is given by

x = A*cos(2*Pi*f*t + phase constant)

where A = amplitude, Pi= 3.14, f = frequency of oscillation,

therefore

a) amplitude,A = 12, frequency,f = 24/(2*31.4) = 3.8216 s-1, phase constant = 3.14/4 = 0.785 rad = 45 degree

b) In SHM, we know that maximum speed is given by

v = A*angular velocity = 12*24 = 288 m/s

and maximum acceleration is given by

a = A*(angular velocity)2 = 12*24*24 = 6912 m/s2

c) x at t = 0.5 s is

x = 12*cos(24*0.5 + Pi/4) = 12*0.707 = 8.484 m

velocity at t=0.5 s

v= A*(angular velocity)*sin((angular velocity)*t + Pi/4) = 12*24*0.707 = 203.616 m/s

acceleration at t=0.5 s

a = A*(angular velocity)2*sin((angular velocity)*t + Pi/4) = 12*24*24*0.707 = 4886.784 m/s2 ............ANSWER

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