If a solid disk rolls without slipping, what fraction of its total energy is tra
Practice Questions
Q1
If a solid disk rolls without slipping, what fraction of its total energy is translational at the bottom of an incline?
1/4
1/3
1/2
2/3
Questions & Step-by-Step Solutions
If a solid disk rolls without slipping, what fraction of its total energy is translational at the bottom of an incline?
Step 1: Understand that a solid disk has two types of kinetic energy when it rolls: translational kinetic energy and rotational kinetic energy.
Step 2: Recall the formula for translational kinetic energy, which is (1/2)mv^2, where m is mass and v is velocity.
Step 3: Recall the formula for rotational kinetic energy, which is (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity.
Step 4: For a solid disk, the moment of inertia I is (1/2)mr^2, where r is the radius of the disk.
Step 5: When the disk rolls without slipping, the relationship between linear velocity v and angular velocity ω is v = rω.
Step 6: Substitute ω in the rotational kinetic energy formula using v = rω, which gives you (1/2)(1/2)mr^2(v/r)^2 = (1/4)mv^2.
Step 7: Now, add the translational and rotational kinetic energies to find the total kinetic energy: Total KE = (1/2)mv^2 + (1/4)mv^2 = (3/4)mv^2.
Step 8: To find the fraction of the total energy that is translational, divide the translational kinetic energy by the total kinetic energy: (1/2)mv^2 / (3/4)mv^2.
