If a solid cylinder rolls without slipping, what is the ratio of its translation
Practice Questions
Q1
If a solid cylinder rolls without slipping, what is the ratio of its translational kinetic energy to its rotational kinetic energy?
1:1
2:1
1:2
3:1
Questions & Step-by-Step Solutions
If a solid cylinder rolls without slipping, what is the ratio of its translational kinetic energy to its rotational kinetic energy?
Step 1: Understand that a solid cylinder rolling without slipping has both translational and rotational motion.
Step 2: Identify the formula for translational kinetic energy (KE_trans), which is KE_trans = (1/2)mv², where m is mass and v is velocity.
Step 3: Identify the formula for rotational kinetic energy (KE_rot), which is KE_rot = (1/2)(Iω²), where I is the moment of inertia and ω is the angular velocity.
Step 4: For a solid cylinder, the moment of inertia I is (1/2)mr², where r is the radius of the cylinder.
Step 5: Relate the translational velocity v to the angular velocity ω using the formula v = rω.
Step 6: Substitute ω in the rotational kinetic energy formula: KE_rot = (1/2)(I)(v/r)².
Step 7: Replace I with (1/2)mr² in the KE_rot formula: KE_rot = (1/2)((1/2)mr²)(v/r)².
Step 8: Simplify the expression for KE_rot to find that KE_rot = (1/4)mv².
Step 9: Now, find the ratio of KE_trans to KE_rot: KE_trans:KE_rot = (1/2)mv² : (1/4)mv².
