Question: A uniform solid disk of dimeter 300mm and mass 2.0 kg rolls without slipping to the bottom of an inclined plane. If the velocity of the disk is 80 m/s at the bottom, what is the height of the inclined plane?
Answer: Approximately 48.47 meters.
Step-by-step explanation:
To solve this problem, we can use the principles of conservation of energy. At the top of the inclined plane, the initial potential energy of the disk is converted into both kinetic energy (rotational and translational) and potential energy at the bottom.
The total mechanical energy at the top is equal to the total mechanical energy at the bottom (assuming no energy losses due to friction or air resistance):
Initial Potential Energy + Initial Kinetic Energy = Final Potential Energy + Final Kinetic Energy
Initial Potential Energy:
PE_initial = m * g * h
Initial Kinetic Energy (at the top, the disk is initially not moving):
KE_initial = 0
Final Potential Energy (at the bottom, the potential energy is zero):
PE_final = 0
Final Kinetic Energy (at the bottom, the disk has both translational and rotational kinetic energy):
KE_final = 0.5 * I * ω^2 + 0.5 * m * v^2
Where:
- m is the mass of the disk (2.0 kg)
- g is the acceleration due to gravity (9.81 m/s²)
- h is the height of the inclined plane
- I is the moment of inertia of the disk (0.5 * m * r^2 for a solid disk)
- ω is the angular velocity of the disk (related to its linear velocity v by ω = v / r, where r is the radius of the disk)
- v is the velocity of the disk (80 m/s)
Given the diameter (300 mm), we can find the radius:
r = diameter / 2 = 0.15 m
Substitute these values into the equation for the final kinetic energy:
KE_final = 0.5 * (0.5 * m * r^2) * (v / r)^2 + 0.5 * m * v^2
Now, set the initial potential energy equal to the final kinetic energy and solve for the height h:
m * g * h = KE_final
m * g * h = 0.25 * m * v^2 + 0.5 * m * v^2
Simplify the equation:
g * h = 0.25 * v^2 + 0.5 * v^2
Solve for h:
h = (0.75 * v^2) / g
Now, substitute the given values:
h = (0.75 * (80^2)) / 9.81 ≈ 48.47 meters
So, the height of the inclined plane is approximately 48.47 meters.