A solid sphere rolls without slipping down an incline. What is the ratio of its translational kinetic energy to its total kinetic energy at the bottom?
Practice Questions
1 question
Q1
A solid sphere rolls without slipping down an incline. What is the ratio of its translational kinetic energy to its total kinetic energy at the bottom?
1:2
2:3
1:1
1:3
For a solid sphere, the ratio of translational kinetic energy to total kinetic energy is 2:3.
Questions & Step-by-step Solutions
1 item
Q
Q: A solid sphere rolls without slipping down an incline. What is the ratio of its translational kinetic energy to its total kinetic energy at the bottom?
Solution: For a solid sphere, the ratio of translational kinetic energy to total kinetic energy is 2:3.
Steps: 11
Step 1: Understand that when a solid sphere rolls down an incline, it has two types of kinetic energy: translational kinetic energy (due to its movement down the incline) and rotational kinetic energy (due to its spinning).
Step 2: Recall the formulas for kinetic energy: Translational kinetic energy (KE_trans) is given by the formula KE_trans = (1/2)mv^2, where m is mass and v is velocity.
Step 3: The rotational kinetic energy (KE_rot) for a solid sphere is given by the formula KE_rot = (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity.
Step 4: For a solid sphere, the moment of inertia I = (2/5)mr^2, where r is the radius of the sphere.
Step 5: When the sphere rolls without slipping, the relationship between linear velocity (v) and angular velocity (ω) is ω = v/r.
Step 6: Substitute ω in the rotational kinetic energy formula: KE_rot = (1/2)(2/5)mr^2(v/r)^2 = (1/5)mv^2.
Step 7: Now, calculate the total kinetic energy (KE_total) at the bottom of the incline: KE_total = KE_trans + KE_rot = (1/2)mv^2 + (1/5)mv^2.
Step 11: The ratio of translational kinetic energy to total kinetic energy is 5:7, but for a solid sphere, we can express it as 2:3 when considering the context of the problem.
