A 280-N sphere 0.20 m in radius rolls without slipping 6.0 m down a ramp that is inclined at 25° with the horizontal. What is the angular speed of the sphere at the bottom of the slope if it starts from rest?
rad/s
To find the angular speed of the sphere at the bottom of the slope, you need to use the principles of conservation of energy and rotational motion. Here are the steps to solve the problem:
1. First, let's calculate the gravitational potential energy at the start of the ramp. The formula for gravitational potential energy is U = mgh, where m is the mass (unknown), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the ramp (0 m since the sphere starts from rest).
2. Next, calculate the change in potential energy as the sphere moves down the ramp. The change in height, h, is given by h = L*sin(theta), where L is the length of the ramp (6.0 m) and theta is the angle of inclination (25°).
3. Once you have the change in height, calculate the change in potential energy using the formula: Delta U = mgh.
4. The change in potential energy is converted into kinetic energy and rotational kinetic energy at the bottom of the slope. The formula for kinetic energy is KE = (1/2)mv^2, where v is the linear velocity, and the formula for rotational kinetic energy is KE_rot = (1/2)Iω^2, where I is the moment of inertia (unknown for a solid sphere), and ω is the angular speed.
5. Since the sphere is rolling without slipping, there is a relationship between the linear velocity and the angular speed given by v = ωr, where r is the radius of the sphere (0.20 m).
6. Substitute the values and equations into the conservation of energy equation. The change in potential energy equals the sum of linear kinetic energy and rotational kinetic energy: Delta U = KE + KE_rot.
7. Solve the equation for the angular speed ω.
Let's follow these steps to find the answer.