A heavy, circular merry-go-round has a radius of 1.8m and moment of inertia (abo

Question

A heavy, circular merry-go-round has a radius of 1.8m and moment of inertia (about its central axis) of 500. kg*m^2.

a. Four children, standing on the ground, all pull simultaneously with four different forces on the edge of the merry-go-around:

Child A: 450 N, clockwise (CW), tangential

Child B: 620N, counter-clockwise (CCW), tangential

Child C: 350N, exactly away from center

Child D: 550N, counter-clockwise (CCW), at an angle 120 degrees away from the center

Find the angular acceleration of the merry-go-round at the moment described. Will it be CCW or CW? (Assume the merry-go-rounds's axle is frictionless.)

b. Suppose sometimes later that the same merry-go-round has no outside torques acting on it. A few children are riding on the merry-go-round near its edge, so that the whole system an initial total moment of inertia of 800. kg*m^2. The system is spinning with an initial angular speed of 1.20 rad/s. Then the system is spinning with an initial angular speed of 1.20 rad/s. Then the children migrate toward the ceneter of the merry-go-round, decresing the system's moment of inertia to a final value of 650 kg*m^2 . What will be the merry-go-round's final angular speed?

c. Find the system's change in kinetic energy during the transformation described in part (b).

Explanation / Answer

Answer questionnet torque = -Fa*R sin 90+Fc *R*sin0+Fb*Rsin90+Fd*R*sin60 = 1163 CCW...... net torque = I *alpha =======>alpha = 2.32 rad/s^2 CCW........... b) I1*W1=I2*W2===> W2 =1.47 rad /s. = c)delta KE = .5I1*W1^2 - .5*I2*W2^2 =1173.8 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.