How would you set this up?

A 250-N sphere 0.20 m in radius rolls without slipping 6.0 m down a ramp that is inclined at 28° with the horizontal. What is the angular speed of the sphere at the bottom of the slope if it starts from rest?

To find the angular speed of the sphere at the bottom of the slope, we can use the principle of conservation of energy. Here's how you can set up the problem:

1. Determine the initial potential energy (PE) and the final kinetic energy (KE) of the sphere at the bottom of the slope.

2. Calculate the change in potential energy as the sphere rolls down the ramp.

3. Use the relationship between rotational kinetic energy and angular speed to find the final rotational kinetic energy and angular speed.

4. Set the change in potential energy equal to the change in rotational kinetic energy and solve for the angular speed.

Let's walk through these steps in more detail:

1. Determine the initial potential energy and the final kinetic energy of the sphere at the bottom of the slope:

The initial potential energy (PE) of the sphere can be calculated using the formula PE = mgh, where m is the mass of the sphere, g is the acceleration due to gravity (9.8 m/s^2), and h is the vertical height.

In this case, we are given the weight of the sphere (250 N) but not its mass. To find the mass, we can use the formula W = mg, where W is the weight and g is the acceleration due to gravity. Rearranging the formula, we get m = W/g. So, m = 250 N / 9.8 m/s^2 = 25.51 kg.

The vertical height (h) can be calculated using the given angle of inclination (θ) and the horizontal distance traveled (d) using the formula h = d sin(θ). In this case, d = 6.0 m and θ = 28°. Plugging in the values, we get h = 6.0 m * sin(28°) = 2.75 m.

The initial potential energy (PE) = mgh = 25.51 kg * 9.8 m/s^2 * 2.75 m = 677 J.

The final kinetic energy (KE) of the sphere at the bottom of the slope can be calculated using the formula KE = (1/2) * I * ω^2, where I is the moment of inertia of the sphere and ω is the angular speed.

2. Calculate the change in potential energy:

The change in potential energy (ΔPE) can be calculated by subtracting the initial potential energy from the final potential energy. Since the sphere starts from rest, its final potential energy is zero. So, the change in potential energy ΔPE = 0 - 677 J = -677 J.

3. Use the relationship between rotational kinetic energy and angular speed:

The moment of inertia (I) of a solid sphere rotating about its diameter can be calculated using the formula I = (2/5) * m * r^2, where r is the radius of the sphere. In this case, the radius (r) is given as 0.20 m.

Plugging in the values, we get I = (2/5) * 25.51 kg * (0.20 m)^2 = 1.02 kg·m^2.

4. Set the change in potential energy equal to the change in rotational kinetic energy and solve for the angular speed:

Since the change in potential energy (ΔPE) is equal to the change in rotational kinetic energy (ΔKE), we can set -677 J = (1/2) * I * (ωf^2 - ωi^2), where ωf is the final angular speed and ωi is the initial angular speed (which is zero in this case).

Simplifying the equation, we get -677 J = (1/2) * 1.02 kg·m^2 * (ωf^2 - 0).

Solving for ωf, we have ωf^2 = (2 * -677 J) / (1.02 kg·m^2) = -1325.49 rad^2/s^2.

However, since angular speed cannot be negative, we take the square root of the absolute value of this result to get ωf = √(1325.49 rad^2/s^2) = 36.4 rad/s.

Therefore, the angular speed of the sphere at the bottom of the slope, starting from rest, is approximately 36.4 rad/s.