Question

A solid disk and a thin-walled hoop each have a diameter of 8 cm. Both are released from rest at the same time at the top of a ramp that is 175 cm high and start rolling down the incline. What is the velocity of each at the bottom of the ramp? What percentage of the gravitational potential energy has been transformed into angular kinetic energy for each? Into what form of energy is the remaining percentage transformed?

Answer 1

Answer:

Explanation:using

Mgh= 1/2mv²+1/2Iw²

So for disk

mgh= 1/2mv²+1/2(1/2mR²)v²/R²

gh = 0.75v²

V= 4.8m/s

Then the translational KE

= (1/2mv²)/mgh x 100

= 66.7%

So rotational energy= 100- 67.7%

=33.3%

For hoop

mgh= 1/2mv²+1/2(1/2mR²)v²/R²

V= 4.14m/s

Then the translational KE

= (1/2mv²)/mgh x 100

= 50%

Then the remaining 50% is translational energy