If a solid cylinder rolls without slipping, what fraction of its total kinetic e
Practice Questions
Q1
If a solid cylinder rolls without slipping, what fraction of its total kinetic energy is translational?
1/3
1/2
2/3
1
Questions & Step-by-Step Solutions
If a solid cylinder rolls without slipping, what fraction of its total kinetic energy is translational?
Step 1: Understand that a solid cylinder has two types of kinetic energy: translational and rotational.
Step 2: The formula for total kinetic energy (KE_total) is KE_total = KE_translational + KE_rotational.
Step 3: The translational kinetic energy (KE_translational) is given by the formula (1/2)mv^2, where m is mass and v is velocity.
Step 4: The rotational kinetic energy (KE_rotational) for a solid cylinder is (1/2)(1/2)mR^2(ω^2), where R is the radius and ω is the angular velocity.
Step 5: Since the cylinder rolls without slipping, we can relate angular velocity (ω) to linear velocity (v) using the formula ω = v/R.
Step 6: Substitute ω = v/R into the rotational kinetic energy formula to get KE_rotational = (1/2)(1/2)mR^2(v/R)^2.
Step 7: Simplify the rotational kinetic energy formula to get KE_rotational = (1/4)mv^2.
Step 8: Now, substitute KE_translational and KE_rotational back into the total kinetic energy formula: KE_total = (1/2)mv^2 + (1/4)mv^2.
Step 10: To find the fraction of the total kinetic energy that is translational, divide KE_translational by KE_total: (KE_translational / KE_total) = ((1/2)mv^2) / ((3/4)mv^2).
