A uniform solid sphere is released from rest at the top of an incline. The incli

Question

A uniform solid sphere is released from rest at the top of an incline. The incline is 3.60 m long and makes an angle of 18.4° relative to horizontal. The sphere has a mass of 0.420 kg, a radius of 3.90 cm, and rolls down the incline without slipping.

a. Taking the bottom of the ramp to be y = 0, what is the gravitational potential energy of the sphere at the top of the incline?

b. Assuming that no air-resistance acts on the sphere, what is the total kinetic energy of the sphere at the bottom of the incline?

c. What is the rotational inertia of the sphere?

d. What is the linear speed of the sphere at the bottom of the incline?

e. What is the rotational speed of the sphere at the bottom of the incline?

f. What is the linear kinetic energy of the sphere at the bottom of the incline?

g. What is the rotational kinetic energy of the sphere at the bottom of the incline?

h. What fraction of the sphere’s total kinetic energy is rotational kinetic energy at the bottom of the

incline?

Explanation / Answer

here,

L = 3.60 m

theta = 18.4 degree

r = 3.90 cm = 0.0390 m

m = 0.420 Kg

a)

gravitational potential energy = m * g * L * sin(theta)

gravitational potential energy = 0.420 * 9.8 * 3.6 * sin(18.4)

gravitational potential energy = 4.677 J

b)

at the bottom of the ramp

total kinetic energy of sphere = gravitationkal potential energy

total kinetic energy of sphere = 4.677 J

c)

rotational inertia of the sphere = 0.4 * m * r^2

rotational inertia of the sphere = 0.4 * 0.420 * 0.0390^2

rotational inertia of the sphere = 2.56 *10^-4 Kg.m^2

d)

let the linear speed of the sphere is v

Using conseravtion of energy

0.5 * m * v^2 + 0.5 * I * (v/r)^2 = 4.677

0.5 * 0.420 * v^2 + 0.5 * 2.56 *10^-4 * (v/.0390)^2 = 4.677

solving for v

v = 3.99 m/s

the linear speed of the sphere at the bottom is 3.99 m/s

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