A disk rolls down a slope of height h. What is the ratio of translational to rotational kinetic energy at the bottom?

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Q: A disk rolls down a slope of height h. What is the ratio of translational to rotational kinetic energy at the bottom?

Solution: At the bottom, the total kinetic energy is split equally between translational and rotational for a disk, hence the ratio is 1:1.

Steps: 9

Step 1: Understand that when a disk rolls down a slope, it has two types of kinetic energy: translational (movement) and rotational (spinning).
Step 2: Recognize that the total kinetic energy at the bottom of the slope is the sum of translational and rotational kinetic energy.
Step 3: For a solid disk, the formula for translational kinetic energy (KE_trans) is (1/2)mv^2, where m is mass and v is velocity.
Step 4: The formula for rotational kinetic energy (KE_rot) is (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity.
Step 5: For a solid disk, the moment of inertia I is (1/2)mr^2, and the relationship between linear velocity v and angular velocity ω is v = rω.
Step 6: Substitute the values into the rotational kinetic energy formula to express it in terms of v.
Step 7: At the bottom of the slope, the total kinetic energy is shared between translational and rotational energy.
Step 8: For a disk, it turns out that the translational and rotational kinetic energies are equal when it rolls without slipping.
Step 9: Therefore, the ratio of translational kinetic energy to rotational kinetic energy is 1:1.