A cylinder rolls down a hill of height h. What is the speed of the center of mass when it reaches the bottom?
Practice Questions
1 question
Q1
A cylinder rolls down a hill of height h. What is the speed of the center of mass when it reaches the bottom?
√(2gh)
√(3gh)
√(4gh)
√(5gh)
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 and ω = v/R. Solving gives v = √(3gh).
Questions & Step-by-step Solutions
1 item
Q
Q: A cylinder rolls down a hill of height h. What is the speed of the center of mass when it reaches the bottom?
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 and ω = v/R. Solving gives v = √(3gh).
Steps: 11
Step 1: Identify the height of the hill as 'h' and the mass of the cylinder as 'm'.
Step 2: Understand that at the top of the hill, the cylinder has potential energy given by the formula PE = mgh.
Step 3: When the cylinder rolls down, this potential energy converts into kinetic energy.
Step 4: The total kinetic energy (KE) at the bottom consists of two parts: translational kinetic energy (1/2 mv^2) and rotational kinetic energy (1/2 Iω^2).
Step 5: For a solid cylinder, the moment of inertia (I) is (1/2)mR^2, where R is the radius of the cylinder.
Step 6: The relationship between linear speed (v) and angular speed (ω) for rolling without slipping is ω = v/R.
Step 7: Substitute I and ω into the kinetic energy formula: KE = 1/2 mv^2 + 1/2 (1/2)mR^2(v/R)^2.
Step 8: Simplify the equation to combine terms: KE = 1/2 mv^2 + 1/4 mv^2 = (3/4)mv^2.
Step 9: Set the potential energy equal to the total kinetic energy: mgh = (3/4)mv^2.
Step 10: Cancel 'm' from both sides and solve for v: gh = (3/4)v^2, leading to v^2 = (4/3)gh.
Step 11: Take the square root to find the speed: v = √(4/3)√(gh) = √(3gh).
