A 2.6 kg solid sphere (radius = 0.10 m ) is released from rest at the top of a r

Question

A 2.6 kg solid sphere (radius = 0.10 m ) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.85 m high and 5.5 m long.

Part A

When the sphere reaches the bottom of the ramp, what is its total kinetic energy?

Express your answer using two significant figures.

Part B

When the sphere reaches the bottom of the ramp, what is its rotational kinetic energy?

Express your answer using two significant figures.

Part C

When the sphere reaches the bottom of the ramp, what is its translational kinetic energy?

Express your answer using two significant figures.

Explanation / Answer

given,

mass = 2.6 kg

radius = 0.1 m

height = 0.85 m

length = 5.5 m

total kinetic energy = potential energy at top

total kinetic energy = mgh

total kinetic energy = 2.6 * 9.8 * 0.85

total kinetic energy = 21.658 J

rotational kinetic energy = 0.5 * I * w^2

translational kinetic energy = 0.5 * mv^2

21.658 = 0.5 * I * w^2 + 0.5 * mv^2

21.658 = 0.5 * (2/5) * m * R^2 * v^2 / R^2 + 0.5 * mv^2

21.658 = 0.5 * (2/5) * m * v^2 + 0.5 * mv^2

21.658 = 0.5 * (2/5) * 2.6 * v^2 + 0.5 * 2.6v^2

v = 3.4496 m/s

rotational kinetic energy = 0.5 * (2/5) * m * R^2 * v^2 / R^2

rotational kinetic energy = 0.5 * (2/5) * 2.6 * 3.4496^2

rotational kinetic energy = 6.1878 J

translational kinetic energy = 0.5 * 2.6 * 3.4496^2

translational kinetic energy = 15.469 J

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