2 years ago

A solid uniform cylinder is rolling without slipping. What fraction of its kinetic energy is rotational?

Physics

1 answer:

tester [92]2 years ago

5 0

Answer:

Explanation:

Let m be the mass of cylinder and r be the radius. It is moving with velocity v and angular velocity is ω. Let I be the moment of inertia of the cylinder.

I = 0.5 mr²

Total kinetic energy, T = 0.5 mv² + 0.5 Iω²

T = 0.5 (mv² + 0.5 mr²ω²)

v = rω

So, T = 0.5 (mv² + 0.5 mv²) = 0.75 mv²

Rotational kinetic energy is

R = 0.5 Iω² = 0.5 x 0.5 mr²ω²

R = 0.25 mv²

So, R / T = 0.25 / 0.75 = 1/3

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A skydiver jumps from an airplane that is moving at 50 meters per second at a height of 1,000 meters what describes the skydiver

kkurt [141]

Answer:

11060M Joules, where M is the mass of the diver in kg

Explanation:

Mass of the skydiver missing, we're assuming it's M.

It's total energy is the sum of the contribution of his kinetic energy (K)- since he's moving at 50 m/s, and it's potential energy (U), since he's subject to earth gravity.

Energy is the sum of the two, so $E = K+U= \frac 12 M v^2 + Mgh = M (\frac 12 \cdot 50^2 + 9.81\cdot 1000) = M ( 1250 + 9810) = 11060\cdot M$