Question

A uniform bar has two small balls glued to its ends. The bar is 2.00 m long and with mass 4.00 kg, while the balls each have mass 0.500 kg and can be treated as point masses. A) Find the moment of inertia of this combination about an axis perpendicular to the bar through its center. B) Find the moment of inertia of this combination about an axis perpendicular to the bar through one of the balls. C) Find the moment of inertia of this combination about an axis parallel to the bar through both balls. D) Find the moment of inertia of this combination about an axis parallel to the bar and 0.500 m from it.

Answer 1

A) The moment of inertia about the center axis is 2mR². B) The moment of inertia about an axis through one of the balls is 4mR². C) The moment of inertia about an axis parallel to the bar through both balls is (1/12)mL² + 2mR². D) The moment of inertia about an axis parallel to the bar and 0.500 m from it is (1/12)mL² + 2mR² + mL².

Step-by-step explanation:

A) To find the moment of inertia of the combination about an axis perpendicular to the bar through its center, we can treat the two balls as point masses. Using the parallel-axis theorem, the moment of inertia about the center axis is given by: I₁ = 2mR², where m is the mass of each ball and R is the distance from the center to each ball.

B) To find the moment of inertia about an axis through one of the balls, we can consider the bar as a point mass at the center and the two balls as point masses located at the ends. The moment of inertia about the axis passing through one of the balls is given by: I₂ = 4mR².

C) To find the moment of inertia about an axis parallel to the bar through both balls, we can use the parallel-axis theorem again. The moment of inertia is the sum of the moment of inertia of the bar and the moment of inertia of the two balls about the center axis: I₃ = (1/12)mL² + 2mR², where L is the length of the bar.

D) To find the moment of inertia about an axis parallel to the bar and 0.500 m from it, we can use the parallel-axis theorem once more. The moment of inertia is the sum of the moment of inertia of the bar and the moment of inertia of the two balls about the center axis, plus the moment of inertia of the bar about the new axis: I₄ = (1/12)mL² + 2mR² + mL².