A solid sphere of radius R rolls without slipping down an incline of height h. What is its speed at the bottom of the incline? (2021)
Practice Questions
1 question
Q1
A solid sphere of radius R rolls without slipping down an incline of height h. What is its speed at the bottom of the incline? (2021)
√(2gh)
√(3gh)
√(4gh)
√(5gh)
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. For a solid sphere, v = √(5gh/7). Thus, speed at the bottom is √(3gh).
Questions & Step-by-step Solutions
1 item
Q
Q: A solid sphere of radius R rolls without slipping down an incline of height h. What is its speed at the bottom of the incline? (2021)
Solution: Using conservation of energy, potential energy at the top = kinetic energy at the bottom. For a solid sphere, v = √(5gh/7). Thus, speed at the bottom is √(3gh).
Steps: 10
Step 1: Understand that the sphere starts at a height 'h' and has potential energy due to its height.
Step 2: Recognize that as the sphere rolls down, this potential energy converts into kinetic energy.
Step 3: Remember that the total kinetic energy of a rolling object includes both translational (movement) and rotational (spinning) energy.
Step 4: Write the equation for potential energy at the top: PE = mgh, where m is mass and g is acceleration due to gravity.
Step 5: Write the equation for kinetic energy at the bottom: KE = (1/2)mv^2 + (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity.
Step 6: For a solid sphere, the moment of inertia I = (2/5)mR^2 and the relationship between linear speed v and angular speed ω is ω = v/R.
Step 7: Substitute I and ω into the kinetic energy equation to get KE = (1/2)mv^2 + (1/2)(2/5)mR^2(v/R)^2.
Step 8: Simplify the kinetic energy equation to combine terms and express it in terms of v.
Step 9: Set the potential energy equal to the total kinetic energy: mgh = (1/2)mv^2 + (1/5)mv^2.
Step 10: Solve for v to find the speed at the bottom of the incline: v = √(5gh/7).
