PART ONE
A merry-go-round rotates at the rate of
0.19 rev/s with an 97 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 0 m from the center?
Consider the merry-go-round is a solid 63 kg
cylinder of radius of 2.5 m.
Answer in units of rad/s

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

The problem can be solved by applying the law of conservation of angular momentum: in fact, the total angular momentum of the system must be conserved,

[tex]L_1 = L_2[/tex]

The initial angular momentum is given by:

[tex]L_1 = (I_d + I_m) \omega[/tex]

where

[tex]I_d = \frac{1}{2}MR^2[/tex] is the moment of inertia of the cylinder, with

M = 63 kg is its mass

R = 2.5 m is the radius

[tex]I_m = mr^2[/tex] is the moment of inertia of the man, with

m = 97 kg is the mass of the man

r = 2.5 m is the distance of the man fro mthe axis of rotation

[tex]\omega=0.19 rev/s \cdot 2 \pi = 1.19 rad/s[/tex] is the angular speed

The final angular momentum is given by

[tex]L_2 = I_d \omega'[/tex]

where [tex]\omega'[/tex] is the final angular speed, and where the angular momentum of the man is zero because he is at the axis of rotation.

Combining the two equations, we get:

[tex](\frac{1}{2}MR^2 + mr^2) \omega = \frac{1}{2}MR^2 \omega'\\\omega'=(1+2\frac{m}{M})\omega=(1+2\frac{97}{63})(1.19)=4.85 rad/s[/tex]

2)

The initial kinetic energy (which is rotational kinetic energy) of the system is given by

[tex]K_1 = \frac{1}{2}(I_d + I_m) \omega^2[/tex]

And substituting,

[tex]K_1 = \frac{1}{2}(\frac{1}{2}MR^2+mr^2)\omega^2=\frac{1}{2}(\frac{1}{2}(63)(2.5)^2+(97)(2.5)^2)(1.19)^2=568.6 J[/tex]

The final kinetic energy is given by

[tex]K_2 = \frac{1}{2}I_d \omega'^2[/tex]

And substituting,

[tex]K_2 = \frac{1}{2}(\frac{1}{2}MR^2)\omega'^2=\frac{1}{2}(\frac{1}{2}(63)(2.5)^2)(4.85)^2=2315 J[/tex]

So, the change in kinetic energy is

[tex]\Delta K = 2315 J - 568.6 J = 1746 J[/tex]

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

snehashish65 snehashish65 · Answer 2 · 2021-11-30T05:41:43+01:00

The new angular speed when the man walks to a point 0 m from the center is 4.85 rad/s.

The change in kinetic energy is 1746 J.

Given data:

The rate of revolution of merry go-round is, n = 0.19 rev/s.

The mass of man is, M = 97 kg.

The distance of man from axis of rotation is, d = 2.5 m.

The mass of merry-go round is, m = 63 kg.

The radius of cylinder is, r = 2.5 m.

(1)

The problem can be solved by applying the law of conservation of angular momentum: in fact, the total angular momentum of the system must be conserved,

[tex]L_{1} = L_{2}\\(\dfrac{1}{2}Md^{2}+mr^{2}) \times \omega=\dfrac{1}{2}Md^{2} \omega'[/tex]

Here, [tex]\omega'[/tex] is the new angular speed. and where the angular momentum of the man is zero because he is at the axis of rotation.

Solving as,

[tex](\dfrac{1}{2} \times 97 \times 2.5^{2}+(63 \times 2.5^{2}) \times (2 \pi \times 0.19)=\dfrac{1}{2} \times 97 \times 2.5^{2} \omega'\\\\\omega' = 4.85 \;\rm rad/s[/tex]

Thus, we can conclude that the new angular speed when the man walks to a point 0 m from the center is 4.85 rad/s.

(2)

The expression for the change in kinetic energy is,

Kinetic energy change = Final kinetic energy - Initial kinetic energy

[tex]\Delta KE =\dfrac{1}{2}(\dfrac{1}{2}Md^{2} \times \omega'^{2}) - \dfrac{1}{2}(\dfrac{1}{2}Md^{2}+mr^{2}) \omega^{2}\\\\\\\Delta KE =\dfrac{1}{2}(\dfrac{1}{2} \times 97 \times 2.5^{2} \times 4.85^{2}) - \dfrac{1}{2}(\dfrac{1}{2} \times 97 \times 2.5^{2}+(63 \times 2.5^{2})) (2 \pi \times 0.19)^{2}\\\\\\\Delta KE = 1746 \;\rm J[/tex]

Thus, we can conclude that the change in kinetic energy is 1746 J.

Learn more about the Angular momentum here:

https://brainly.com/question/22549612