228k views
5 votes
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?

2 Answers

3 votes

Final answer:

The velocity of the solid disk and thin-walled hoop at the bottom of the ramp can be calculated using the principle of conservation of energy. The percentage of energy transformed into angular kinetic energy and the remaining percentage transformed into translational kinetic energy can also be calculated.

Step-by-step explanation:

The velocity of the solid disk and the thin-walled hoop at the bottom of the ramp can be calculated using the principle of conservation of energy. At the top of the ramp, the gravitational potential energy is converted into both translational kinetic energy and rotational kinetic energy. The total kinetic energy at the bottom of the ramp is the sum of these two forms of energy.

For a solid disk rolling without slipping, the velocity at the bottom can be calculated using the equation:

v = [(2gh)/3]^(1/2)

For a thin-walled hoop rolling without slipping, the velocity at the bottom can be calculated using the equation:

v = [(2gh)]^(1/2)

The percentage of gravitational potential energy that is transformed into angular kinetic energy can be calculated using the formula:

Percentage = (Rotational KE / Gravitational PE) * 100%

The remaining percentage of energy is transformed into translational kinetic energy.

User Jatin Gupta
by
4.5k points
2 votes

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

User Pryma
by
4.8k points