Question

Шарик массой m = 60 г, привязанный к концу нити длиной Li = = 1,2 м, вращается с частотой ni = 2 об/с, опираясь на горизонтальную плос­кость. Нить укорачивается, приближая шарик к оси вращения до расстоя­ния L2 = 0,6 м. С какой частотой п2 будет вращаться шарик после этого?

Answer 1

это для кокого класса? очень сложное задание

Answer 2

Calculation of the Final Frequency

To determine the final frequency at which the ball will rotate after the string is shortened, we can use the principle of conservation of angular momentum. The angular momentum of an object is given by the product of its moment of inertia and its angular velocity.

The moment of inertia of a solid sphere rotating about an axis passing through its center is given by the formula:

I = (2/5) * m * r^2

Where: - I is the moment of inertia - m is the mass of the sphere - r is the radius of the sphere

In this case, the mass of the ball is given as 60 grams, which is equal to 0.06 kilograms. The initial length of the string is 1.2 meters, and the final length is 0.6 meters.

To find the initial radius, we can use the Pythagorean theorem:

Li^2 = r^2 + L1^2

Where: - Li is the initial length of the string - r is the initial radius of the ball - L1 is the distance between the center of the ball and the point where the string is attached

Simplifying the equation, we have:

r^2 = Li^2 - L1^2

Substituting the given values, we get:

r^2 = (1.2)^2 - (0.6)^2

Solving for r, we find:

r ≈ 1.039 meters

Now that we have the initial radius, we can calculate the initial moment of inertia:

Ii = (2/5) * m * r^2

Substituting the values, we get:

Ii ≈ (2/5) * 0.06 * (1.039)^2

Simplifying, we find:

Ii ≈ 0.051 kg·m^2

The initial angular velocity is given as 2 revolutions per second, which is equal to 2 * 2π radians per second.

To find the final angular velocity, we can use the principle of conservation of angular momentum:

Ii * ni = If * nf

Where: - Ii is the initial moment of inertia - ni is the initial angular velocity - If is the final moment of inertia - nf is the final angular velocity

We already know the values of Ii and ni. The final moment of inertia can be calculated using the same formula as before, but with the final radius:

If = (2/5) * m * r^2

Substituting the values, we get:

If ≈ (2/5) * 0.06 * (0.6)^2

Simplifying, we find:

If ≈ 0.0072 kg·m^2

Now we can solve for nf:

nf = (Ii * ni) / If

Substituting the values, we get:

nf ≈ (0.051 * 2 * 2π) / 0.0072

Simplifying, we find:

nf ≈ 88.57 rad/s

Therefore, the ball will rotate at a frequency of approximately 88.57 rad/s after the string is shortened.