contestada

*Urgent*


(Part 1)


A merry-go-round rotates at the rate of 0.4 rev/s with an 99 kg man standing at a point 2.5 m from the axis of rotation. What is the new angular speed when the man walks to a point 1 m from the center? Consider the merry-go-round is a solid 73 kg cylinder of radius of 2.5 m. Answer in units of rad/s.


(Part 2)


What is the change in kinetic energy due to this movement? Answer in units of J.

Respuesta :

Answer:

6.5 rad/s

11.695 kJ

Explanation:

Using the principle of conservation of momentum, we have

L1 = L2 and

I1w1 = I2w2, with I being the moment of inertia and w being the angular velocity.

w1 = 0.4 rev/s = 0.4 * 2π = 2.51 rad/s

Moment of Inertia, I = ½MR² + mr²

I1 = ½ * 73 * 2.5² +99 * 2.5²

I1 = 73 * 3.125 + 99 * 6.25

I1 = 228.125 + 618.75

I1 = 846.875 kgm²

The man has been said to have walked 1 m away from his position, so then

I2 = ½ * 73 * 2.5² + 99 * 1²

I2 = 228.125 + 99

I2 = 327.125 kgm²

From the formula stated above,

w2 = I1w1 / I2

w2 = (846.875 * 2.51) / 327.125

w2 = 2125.65725 / 327.125 rad/s

w2 = 6.5 rad/s

2

K = Iw²

K = 846.875 * 2.51²

K = 2125.65725 J

K'= Iw²

K' = 327.125 * 6.5²

K' = 13821.03125 J

Change in energy is given as, K' - K, so that

Energy, E = 13821.03125 - 2125.65725

Energy, E = 11.695 kJ