A cylinder rolls down a hill. If the height of the hill is h, what is the speed of the center of mass of the cylinder at the bottom of the hill?
Practice Questions
1 question
Q1
A cylinder rolls down a hill. If the height of the hill is h, what is the speed of the center of mass of the cylinder at the bottom of the hill?
√(gh)
√(2gh)
√(3gh)
√(4gh)
Using conservation of energy, potential energy at the top (mgh) converts to kinetic energy (1/2 mv^2 + 1/2 Iω^2). For a solid cylinder, I = 1/2 mr^2, leading to v = √(2gh).
Questions & Step-by-step Solutions
1 item
Q
Q: A cylinder rolls down a hill. If the height of the hill is h, what is the speed of the center of mass of the cylinder at the bottom of the hill?
Solution: Using conservation of energy, potential energy at the top (mgh) converts to kinetic energy (1/2 mv^2 + 1/2 Iω^2). For a solid cylinder, I = 1/2 mr^2, leading to v = √(2gh).
Steps: 11
Step 1: Understand that the cylinder has potential energy at the top of the hill due to its height (h). This potential energy is calculated as mgh, where m is the mass of the cylinder and g is the acceleration due to gravity.
Step 2: When the cylinder rolls down the hill, this potential energy converts into kinetic energy at the bottom of the hill.
Step 3: The total kinetic energy at the bottom consists of two parts: translational kinetic energy (1/2 mv^2) and rotational kinetic energy (1/2 Iω^2), where v is the speed of the center of mass and I is the moment of inertia.
Step 4: For a solid cylinder, the moment of inertia I is given by the formula I = 1/2 mr^2, where r is the radius of the cylinder.
Step 5: The relationship between linear speed (v) and angular speed (ω) for rolling without slipping is ω = v/r.
Step 6: Substitute I and ω into the kinetic energy equation: Total kinetic energy = 1/2 mv^2 + 1/2 (1/2 mr^2)(v/r)^2.
Step 7: Simplify the equation to find the total kinetic energy in terms of v: Total kinetic energy = 1/2 mv^2 + 1/4 mv^2 = 3/4 mv^2.
Step 8: Set the potential energy equal to the total kinetic energy: mgh = 3/4 mv^2.
Step 9: Cancel the mass (m) from both sides of the equation: gh = 3/4 v^2.
Step 10: Solve for v by rearranging the equation: v^2 = (4/3)gh, then take the square root: v = √((4/3)gh).
Step 11: The final speed of the center of mass of the cylinder at the bottom of the hill is v = √(2gh).
