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A uniform sphere of m = 0.85 kg and radius r = 4.2 cm.

a) Calculate the moment of inertia of the sphere.
b) Determine the gravitational potential energy.
c) Find the angular momentum about a given axis.
d) Identify the surface area of the sphere.

1 Answer

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Final answer:

a) The moment of inertia of the sphere is 0.002162 kg*m^2. b) The gravitational potential energy is 0. c) The angular momentum cannot be calculated without additional information. d) The surface area of the sphere is 0.022 m^2.

Step-by-step explanation:

a) To calculate the moment of inertia of the sphere, we can use the formula for the moment of inertia of a solid sphere:

I = (2/5) * m * r^2

where I is the moment of inertia, m is the mass of the sphere, and r is the radius of the sphere.

Plugging in the values: I = (2/5) * 0.85 kg * (0.042 m)^2 = 0.002162 kg*m^2

b) The gravitational potential energy of the sphere can be calculated using the formula:

PE = m * g * h

where PE is the gravitational potential energy, m is the mass of the sphere, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the sphere.

As the sphere is not lifted or moved vertically, the height h is 0. Therefore, the gravitational potential energy is 0.

c) The angular momentum about a given axis can be calculated using the formula:

L = I * ω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

However, the angular momentum about a given axis is not provided in the question, so an answer cannot be calculated without additional information.

d) The surface area of the sphere can be calculated using the formula:

A = 4 * π * r^2

where A is the surface area of the sphere and r is the radius of the sphere.

Plugging in the values: A = 4 * π * (0.042 m)^2 = 0.022 m^2

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