Question

A. What is the moment of inertia of a solid iron sphere 20 inches in diameter about an axis passing through the sphere half way from its center to the surface (you will have to look up its density)? If the sphere rolls down an incline plane that makes an angle 20 degrees with the horizontal, what will be its acceleration?

B. Derive the formula for acceleration of a circular disk that rolls down an inclined plane that makes an angle θ with the horizontal. (draw a diagram and say in words what you are doing)

Answer 1

A) Let , M be the mass of the sphere and R be its radius.

R = 20 inches = 0.508 m

Density of iron = 7000 kg/m^{​​​​​​3}

Moment of inertia of a solid sphere about an axis passing through its centre is

I₁ = (2/5)MR²

Moment of inertia of a solid sphere about an axis at a distance R/2 from the centre of the sphere can be calculated using parallel axis theorem

I₂ = I₁ + M(R/2)²

I₂ = (2/5)MR² + MR²/4

I₂ = (13/20)MR²

As, mass = volume × density

M = (4/3)R³ ×

Using R = 0.508 m and = 7000 kg/m³

M = 3845.5 kg

I₂ = (13/20)×3845.5×(0.508)²

I₂ = 645.05 kgm²

B) For a circular disk