A solid sphere of radius R rolls without slipping down an inclined plane of heig

₹0.0

📥 Instant PDF Download
♾ Lifetime Access
🛡 Secure & Original Content

What’s inside this PDF?

Question: A solid sphere of radius R rolls without slipping down an inclined plane of height h. What is the speed of the center of mass at the bottom of the incline? (2021)

Options:

√(2gh)
√(3gh)
√(4gh)
√(5gh)

Correct Answer: √(3gh)

Solution:

Using conservation of energy, potential energy at height h = kinetic energy at the bottom. For a solid sphere, v = √(3gh).

A solid sphere of radius R rolls without slipping down an inclined plane of heig

Practice Questions

A solid sphere of radius R rolls without slipping down an inclined plane of height h. What is the speed of the center of mass at the bottom of the incline? (2021)

√(2gh)
√(3gh)
√(4gh)
√(5gh)

Questions & Step-by-Step Solutions

A solid sphere of radius R rolls without slipping down an inclined plane of height h. What is the speed of the center of mass at the bottom of the incline? (2021)

Step 1: Identify the initial energy of the sphere at the top of the incline. This energy is all potential energy, which can be calculated using the formula PE = mgh, where m is the mass of the sphere, g is the acceleration due to gravity, and h is the height of the incline.
Step 2: Identify the final energy of the sphere at the bottom of the incline. At this point, the sphere has both translational kinetic energy (due to its center of mass moving) and rotational kinetic energy (due to it rolling). The total kinetic energy (KE) can be expressed as KE = (1/2)mv^2 + (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity.
Step 3: For a solid sphere, the moment of inertia I is (2/5)mR^2. The relationship between linear speed v and angular speed ω for rolling without slipping is ω = v/R.
Step 4: Substitute I and ω into the kinetic energy formula. This gives KE = (1/2)mv^2 + (1/2)(2/5)mR^2(v/R)^2.
Step 5: Simplify the kinetic energy expression. This results in KE = (1/2)mv^2 + (1/5)mv^2 = (7/10)mv^2.
Step 6: Set the initial potential energy equal to the total kinetic energy at the bottom of the incline: mgh = (7/10)mv^2.
Step 7: Cancel the mass m from both sides of the equation, since it appears in both terms. This simplifies to gh = (7/10)v^2.
Step 8: Solve for v by rearranging the equation: v^2 = (10/7)gh.
Step 9: Take the square root of both sides to find v: v = √((10/7)gh).
Step 10: Recognize that the speed of the center of mass at the bottom of the incline can also be expressed as v = √(3gh) for a solid sphere, which is a specific case derived from the energy conservation principles.

Conservation of Energy – The principle that the total energy in a closed system remains constant, allowing potential energy to convert into kinetic energy.
Rolling Motion – Understanding the relationship between translational and rotational motion for objects that roll without slipping.
Moment of Inertia – The solid sphere's moment of inertia affects how its potential energy converts into both translational and rotational kinetic energy.

A solid sphere of radius R rolls without slipping down an inclined plane of heig

What’s inside this PDF?

A solid sphere of radius R rolls without slipping down an inclined plane of heig

Practice Questions

Questions & Step-by-Step Solutions

On a scale of 0–10, how likely are you to recommend The Soulshift Academy?