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A person of mass 80 kg stands at the center of a rotating merry-go-round platform of radius 3.5 m and moment of inertia 950 kg*m^2 . The platform rotates without friction with angular velocity 0.85 rad/s . The person walks radially to the edge of the platform.
1. Calculate the angular velocity when the person reaches the edge.
w=______________ rad/s
2. Calculate the rotational kinetic energy of the system of platform plus person before the person's walk.
Ki=____________ J
3. Calculate the rotational kinetic energy of the system of platform plus person after the person's walk.
Kf=__________ J

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Answer with Explanation:

We are given that

Mass of person,M=80 kg

Radius,r=3.5 m

Moment of inertia of merry,I=[tex]950 kgm^2[/tex]

Angular velocity of platform=[tex]\omega=0.85rad/s[/tex]

1.Law of conservation of angular momentum

[tex]I\omega=I'\omega'[/tex]

Where[tex]I'=950+80(3.5)^2[/tex]

Substitute the values

[tex]950\times 0.85=(950+80(3.5)^2)\omega'[/tex]

[tex]\omega'=\frac{950\times 0.85}{950+80(3.5)^2}=0.418rad/s[/tex]

Angular velocity when the person reaches the edge =0.418 rad/s

2.Rotational kinetic energy of the system  before the person's walk

[tex]K_i=\frac{1}{2}I\omega^2=\frac{1}{2}(950)(0.85)^2=343.2 J[/tex]

3.Rotational kinetic energy of the system  after the person's walk

[tex]K_f=\frac{1}{2}(950+80(3.5)^2)\times (0.418)^2[/tex]

[tex]K_f=168.6 J[/tex]

(1) The angular velocity of the person at the edge is 0.42 rad/s.

(2) The initial rotational kinetic energy is 343J.

(3) the final rotational kinetic energy is 170J.

Rotational Motion:

(1) At the center of the merry-go-round, the moment of inertia of the person is zero, since the distance from the center is zero.

At the edge, the moment of inertia of the person is:

I = mr²

where m is the mass  = 80kg

r is the radius = 3.5m

I = 80×3.5²

I = 980 kgm²

The moment of inertia of the merry-go-round is 950 kgm².

According to the conservation of angular momentum:

Initial angular momentum = final angular momentum

Now, the angular momentum is given by:

L = Iω

where ω is the angular velocity

let the initial angular momentum be ω = 0.85 rad/s (given)

and the final angular momentum be ω'

So,

[tex]950\omega = (950+980)\omega'\\\\950\times0.85 = 1930\omega'[/tex]

ω' = 0.42 rad/s

(2) The initial rotational kinetic energy is given by:

[tex]KE_i=\frac{1}{2}I\omega^2\\\\KE_i = \frac{1}{2}\times950\times0.85^2[/tex]

KE[tex]_i[/tex] = 343J

(3) The final rotational kinetic energy is given by:

[tex]KE_f=\frac{1}{2}I'\omega'^2\\\\KE_f = \frac{1}{2}\times(950+980)\times0.42^2[/tex]

KE[tex]__f[/tex] = 170J

Learn more about rotational motion:

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