1.6 kg solid sphere (radius = 0.20 m ) is released from rest at the top of a ram

Question

1.6 kg solid sphere (radius = 0.20 m ) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.90 m high and 5.9 m long.

Part A

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

Part B

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

Part C

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

1.6 kg solid sphere (radius = 0.20 m ) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is 0.90 m high and 5.9 m long.

Part A

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

Part B

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

Part C

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

Explanation / Answer

here,

height of the ramo , h = 0.9 m

mass of the sphere , m = 1.6 kg

radius of the sphere , r = 0.2 m

a)

using the conservation of energy

change in total kinetic energy = potential energy lost

change in total kinetic energy = m*g*h = 1.6 * 9.8 * 0.9 = 14.11 J

total kinetic energy at the bottom is 14.11 J

b)

total kinetic energy = rotational kinetic energy + translational kinetic energy

14.11 = 0.5 * I * w^2 + 0.5 * m*v^2

14..22 = 0.5 * 0.4 * m *r^2*( v/r)^2 + 0.5 * m *v^2

14.22 = 0.7 * 1.6 * v^2

v = 3.55 m/s

the rotational kinetic energy = 0.2 * 1.6 * 3.55^2 = 4.03 J

c)

translational kinetic energy =
total kinetic energy - rotational kinetic energy

translational kinetic energy = 14.11 - 4.03

translational kinetic energy = 10.08 J

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