A solid sphere rolls down a frictionless incline. If it starts from rest, what is its final velocity at the bottom of the incline of height h?
Practice Questions
1 question
Q1
A solid sphere rolls down a frictionless incline. If it starts from rest, what is its final velocity at the bottom of the incline of height h?
√(gh)
√(5gh/7)
√(2gh)
√(3gh)
Using conservation of energy, the final velocity of the solid sphere at the bottom is v = √(5gh/7).
Questions & Step-by-step Solutions
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Q
Q: A solid sphere rolls down a frictionless incline. If it starts from rest, what is its final velocity at the bottom of the incline of height h?
Solution: Using conservation of energy, the final velocity of the solid sphere at the bottom is v = √(5gh/7).
Steps: 11
Step 1: Understand that the solid sphere starts from rest at a height h on a frictionless incline.
Step 2: Recognize that the total mechanical energy is conserved because there is no friction.
Step 3: Identify the initial energy of the sphere, which is all potential energy (PE) at height h: PE_initial = mgh, where m is mass and g is acceleration due to gravity.
Step 4: At the bottom of the incline, all potential energy converts to kinetic energy (KE). The kinetic energy of a rolling sphere is the sum of translational kinetic energy and rotational kinetic energy: KE = (1/2)mv^2 + (1/2)Iω^2.
Step 5: For a solid sphere, the moment of inertia I = (2/5)mr^2, and the relationship between linear velocity v and angular velocity ω is ω = v/r.
Step 6: Substitute I and ω into the kinetic energy equation: KE = (1/2)mv^2 + (1/2)(2/5)mr^2(v/r)^2.
Step 7: Simplify the kinetic energy equation: KE = (1/2)mv^2 + (1/5)mv^2 = (7/10)mv^2.
Step 8: Set the initial potential energy equal to the final kinetic energy: mgh = (7/10)mv^2.
Step 9: Cancel mass m from both sides (since m is not zero): gh = (7/10)v^2.
Step 10: Solve for v^2: v^2 = (10/7)gh.
Step 11: Take the square root to find v: v = √(10gh/7).
