To solve this problem, we can make use of the principle of conservation of energy. The initial potential energy of the ball at point A is converted into kinetic energy as it rolls down the track. At any point on the track, the total mechanical energy (sum of potential energy and kinetic energy) remains constant.
Let's denote point A as the initial position where the ball is released and point B as the lowest point on the track. The height difference between points A and B is 5 meters, and the radius of the circular track is 1 meter.
- Velocity at point B: At point B, the ball is at its lowest position, so all of its initial potential energy is converted into kinetic energy. Using the conservation of energy, we can equate the initial potential energy to the kinetic energy at point B:
m * g * h = (1/2) * m * v_B^2
Here, m = mass of the ball = 1 kg g = acceleration due to gravity = 9.8 m/s^2 h = height difference between A and B = 5 m v_B = velocity at point B (to be determined)
Plugging in the values, we have:
1 * 9.8 * 5 = (1/2) * v_B^2 49 = (1/2) * v_B^2 98 = v_B^2 v_B = √98 v_B ≈ 9.9 m/s
Therefore, the magnitude of the velocity of the ball at point B is approximately 9.9 m/s.
- Velocity at point A: To find the velocity at point A, we can make use of the conservation of angular momentum. The angular momentum of the ball remains constant throughout its motion since there is no external torque acting on it. The angular momentum is given by:
L = I * ω
Here, L = angular momentum I = moment of inertia of the ball (rolling without slipping on a circular track) ω = angular velocity
For a solid sphere rolling without slipping, the moment of inertia is given by:
I = (2/5) * m * r^2
Here, m = mass of the ball = 1 kg r = radius of the ball = 1 m
Substituting the values, we get:
I = (2/5) * 1 * 1^2 I = (2/5) kg·m²
Since the ball rolls without slipping, the linear velocity (v) at point A is related to the angular velocity (ω) by the equation:
v = r * ω
Substituting the values, we get:
v_A = 1 * ω v_A = ω
Therefore, the magnitude of the velocity of the ball at point A is equal to the magnitude of the angular velocity.
To determine the angular velocity, we can equate the initial potential energy to the sum of the rotational kinetic energy and the linear kinetic energy at point A:
m * g * h = (1/2) * I * ω^2 + (1/2) * m * v_A^2
Plugging in the values, we have:
1 * 9.8 * 5 = (1/2) * (2/5) * ω^2 + (1/2) * 1 * v_A^2 49 = (1/5) * ω^2 + (1/2) * v_A^2
Since we know that v_A = ω, we can rewrite the equation as:
49 = (1/5) * ω^2 + (1/2) * ω^2 49 = (7/10) * ω^2
Simplifying the equation, we get:
ω^2 = (49 * 10) / 7 ω^2 = 70 ω = √70 ω ≈ 8.37 rad/s
Therefore, the magnitude of the velocity of the ball at point A is approximately 8.37 rad/s.
To summarize:
- The magnitude of the velocity of the ball at point B is approximately 9.9 m/s.
- The magnitude of the velocity of the ball at point A is approximately 8.37 rad/s.