# What is the rotational kinetic energy of the cylinder when it reaches the bottom of the ramp?

A 0.20-kg solid cylinder is released from rest at the top of a ramp 1.8 m long. The cylinder has a radius of 0.15 m, and the ramp is at an angle of 15 degrees with the horizontal.

### 2 Answers

- billrussell42Lv 72 months agoFavorite Answer
the height of the ramp is 1.8 sin 15 = 0.466 m

mgh = ½mV² + ½Iω² = 0.2•9.8•0.466 = 0.913 J

mV² + Iω² = 1.826 J

I = ½0.2•0.15² = 0.00225 kg•m²

ω = V/r = V/0.15

mV² + Iω² = 1.826 J

0.2V² + (0.00225)(V/0.15)² = 1.826 J

0.2V² + 0.1V² = 1.826 J

0.3V² = 1.826

V = 2.47 m/s

ω = 2.47/0.15 = 16.4 rad/s

rot KE = ½Iω² = ½(0.00225)(16.4)² = 0.303 J

I is moment of inertia in kg•m²

I = cMR²

M is mass (kg), R is radius (meters)

c = 1 for a ring or hollow cylinder

c = 2/5 solid sphere around a diameter

c = 7/5 solid sphere around a tangent

c = ⅔ hollow sphere around a diameter

c = ½ solid cylinder or disk around its center

cylinder or ball rolling down a slope

PE at top results in both linear KE and rotational KE

at the bottom.

mgh = ½mV² + ½Iω²

V = rω r is radius

V² = (2gh) / (1 + I/mr²)

for a solid cylinder that reduces to V² = (4/3)gh (1.33)

for a solid sphere, that is V² = (10/7)gh (1.43)

for a ring or hollow cylinder, that is V² = 2gh

for a hollow sphere, that is V² = (6/5)gh (1.2)

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- Andrew SmithLv 72 months ago
For a solid cylinder it requires half as much energy as the kinetic energy of a moving mass at the same speed.

ie mgh = 2*Er + Er -> Er = 1/3 mgh = 1/3 * 0.20 * 9.8 * 1.8*sin(15) = 0.30 J ( and of course Ek = 2*Er = 0.61J total = 0.91 J )

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