On the edge of a carousel shaped like a disk weighing 200 kg and

There are 5 people at the edge of the carousel, each of whom has a mass of 60 kg. The carousel has the form of a disk, weighing 200 kg and radius 2 m, rotating at a frequency of 1 rev/s. To find the rotation frequency and angular velocity of the carousel, it is necessary to move all people to the center at a distance equal to half the radius. In this case, people can be represented as point masses.

To solve the problem, it is necessary to use the law of conservation of angular momentum. The angular momentum of a closed system remains constant if it is not acted upon by external moments of forces. Moving people towards the center will change the moment of inertia of the system, but will not change the moment of momentum.

Initially, the angular momentum of the system is equal to the product of the moment of inertia and the angular velocity:

L = Iω

where L is the angular momentum, I is the moment of inertia, ω is the angular velocity.

The moment of inertia of a carousel with 5 people on the edge is equal to the sum of the moments of inertia of each person and the moment of inertia of the carousel without people:

I1 = 5mr^2/2 + mr^2 = 15mr^2/2

where m is the mass of one person, r is the radius of the carousel.

The moment of inertia of a carousel with people who are moved towards the center can be found in a similar way:

I2 = 5m(r/2)^2/2 + m(r/2)^2 = 5mr^2/8

Thus, the angular momentum of the system remains constant:

I1ω1 = I2ω2

where ω1 and ω2 are the angular speeds of the carousel before and after moving people.

Substituting the values ​​of the moments of inertia and angular velocity, we obtain:

15mr^2/2 * 2π = 5mr^2/8 * ω2

ω2 = 12π/5 ≈ 7.54 rad/s - the angular velocity of the carousel after people move to the center.

The rotation frequency of the carousel after moving people is:

f2 = ω2/2π = 12/5 ≈ 2.4 Hz.

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The problem is to find the rotational speed and angular velocity of a merry-go-round after five people, each weighing 60 kg, move to its center at a distance equal to half the radius.

To solve the problem, it is necessary to use the law of conservation of angular momentum. The angular momentum of a closed system remains constant if it is not acted upon by external moments of forces. Moving people towards the center will change the moment of inertia of the system, but will not change the moment of momentum.

Initially, the angular momentum of the system is equal to the product of the moment of inertia and the angular velocity: L = Iω

The moment of inertia of a carousel with 5 people on the edge is equal to the sum of the moments of inertia of each person and the moment of inertia of the carousel without people:

I1 = 5mr^2/2 + mr^2 = 15mr^2/2

where m is the mass of one person, r is the radius of the carousel.

The moment of inertia of a carousel with people who are moved towards the center can be found in a similar way:

I2 = 5m(r/2)^2/2 + m(r/2)^2 = 5mr^2/8

Thus, the angular momentum of the system remains constant:

I1ω1 = I2ω2

where ω1 and ω2 are the angular speeds of the carousel before and after moving people.

Substituting the values ​​of the moments of inertia and angular velocity, we obtain:

15mr^2/2 * 2π = 5mr^2/8 * ω2 ω2 = 12π/5 ≈ 7.54 rad/s - the angular velocity of the carousel after people move to the center.

The rotation frequency of the carousel after moving people is:

f2 = ω2/2π = 12/5 ≈ 2.4 Hz.

Thus, after moving people to the center, the rotation speed of the carousel will almost double, and the angular speed will increase by more than five times.

***

A disk-shaped carousel with a mass of 200 kg and a radius of 2 m is given, which rotates at a frequency of 1 rev/s. At the edge of the carousel there are five people weighing 60 kg each. To find the rotation frequency and angular velocity of the carousel if all people move to its center at a distance equal to half the radius, you need to use the laws of conservation of momentum and angular momentum.

First, let's find the moment of inertia of the carousel relative to its center, which is equal to:

$I = \frac{1}{2}mR^2 = \frac{1}{2} \cdot 200 \cdot 2^2 = 400$ кг·м²,

where m is the mass of the carousel, R is its radius.

Then we will find the moment of inertia of the carousel system and people relative to its center after all the people move towards it:

$I' = \sum_{i=1}^{5} m_i r_i^2 = m\left(\frac{R}{2}\right)^2 + m\left(\frac{R}{2}\right)^2 + mR^2 + m\left(\frac{R}{2}\right)^2 + m\left(\frac{R}{2}\right)^2 = 2.5mR^2$,

where m_i is the mass of the i-th person, r_i is the distance from the center of the carousel to the i-th person.

The law of conservation of angular momentum states that the angular momentum of a system remains unchanged in the absence of external torques:

$I\omega = I'\omega',

where ω is the angular speed of the carousel before people move, ω' is the angular speed of the carousel after people move.

Substituting the found values ​​of the moments of inertia, we obtain:

$\frac{1}{2} \cdot 200 \cdot 2^2 \cdot \omega = 2.5 \cdot 200 \cdot R^2 \cdot \omega'$

From here we find the angular velocity of the carousel after people move:

$\omega' = \frac{1}{5}\omega = \frac{1}{5}\cdot 2\pi = \frac{2\pi}{5}$ amount/с.

The rotational speed of the carousel is equal to the angular velocity divided by 2π:

$f = \frac{\omega'}{2\pi} = \frac{1}{5}$ rev/s.

So, the rotation frequency of the carousel after moving all the people to its center is 1/5 r/s, and the angular speed is 2π/5 rad/s.

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