A thin cylindrical ring starts from rest at a height h1 = 91 m. The ring has a radius R = 14 cm and a mass M = 9 kg.
If the ring rolls (without slipping) all the way to point 2, what is the ring's energy at point 2 in terms of h2 and v2?
Given h2 = 41 m, what is the velocity of the ring at point 2 in m/s?
What is the ring's rotational velocity in rad/s at point 2?
After passing point 2 the hill becomes frictionless and the ring's rotational velocity remains constant. What is the linear velocity of the ring at point 3 in m/s?

Respuesta :

To determine the energy, velocity, and rotational velocity of the ring at different points along the path:

1. **Energy at point 2 in terms of h2 and v2**:

- At point 1, the total initial energy of the ring is potential energy due to its height: \( PE = mgh_1 \).

- At point 2, the total energy of the ring is the sum of potential energy (due to height \( h_2 \)) and kinetic energy: \( PE + KE = mgh_2 + \frac{1}{2}mv_2^2 \).

2. **Calculating the velocity at point 2**:

- Using the principle of conservation of energy, the initial potential energy at point 1 is converted into the sum of potential and kinetic energy at point 2: \( mgh_1 = mgh_2 + \frac{1}{2}mv_2^2 \).

- Solve for velocity \( v_2 \) to find the velocity at point 2.

3. **Rotational velocity at point 2**:

- For a rolling object, the velocity is related to the angular velocity by \( v = Rω \), where \( R \) is the radius of the ring.

- The angular velocity at point 2 can be found using the relationship between linear velocity and angular velocity.

4. **Linear velocity at point 3**:

- As the hill is frictionless and the ring's rotational velocity remains constant, the linear velocity at point 3 is equal to the product of the radius and the angular velocity (\( v = Rω \)).

By following these steps and applying the relevant formulas, you can calculate the energy, velocity, and rotational velocity of the ring at points 2 and 3.

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