A bowling ball is far from uniform. Lightweight bowling balls are made of a relatively low-density core surrounded by a thin shell with much higher density. A 7.0 lb (3.2 kg) bowling ball has a diameter of 0.216 m; 0.196 m of this is a 1.6 kgcore, surrounded by a 1.6 kg shell. This composition gives the ball a higher moment of inertia than it would have if it were made of a uniform material. Given the importance of the angular motion of the ball as it moves down the alley, this has real consequences for the game.

(a)Model a real bowling ball as a 0.196-m-diameter core with mass 1.6 kg plus a thin 1.6 kg shell with diameter 0.206 m (the average of the inner and outer diameters). What is the total moment of inertia?

Express your answer with the appropriate units.

(b)Find the moment of inertia of a uniform 3.2 kg ball with diameter 0.216 m.

Express your answer with the appropriate units.

The moment of inertia is a scalar by which an additive magnitude, we can add the moments of inertia of each part of the system, taking into account the axis of rotation.

I = I core + I shell

The moment of inertia of a solid sphere is

I sphere = 2/5 MR²

The moment of inertia of a thin spherical shell is

I shell = 2/3 M R²

a) Let's apply to our system, first to the core of weight 1.6 kg and diameter 0.196m, the radius is half the diameter

R = d / 2

R= 0.196 m / 2 = 0.098 m

I core = 2/5 1.6 0.098²

I core = 6.147 10-3 kg m²

Let's calculate the moment of inertia of the shell of mass 1.6 kg with a diameter of 0.206 m

R = 0.206 / 2

R = 0.103 m

I shell = 2/3 1.6 0.103²

I shell = 1,132 10-2 kg m²

The moment of inertia of the ball is the sum of these moments of inertia,

I = I core + I shell

I = 6,147 10⁻³ + 1,132 10⁻² = 6,147 10⁻³ + 11.32 10⁻³

I = 17.47 10⁻³ kg m²

I = 1,747 10-² kg m²

b) Now the ball is report with mass 3.2kg and diameter 0.216 m

R = 0.216 / 2

R = 0.108 m

It is a uniform sphere

I = 2/5 M R²

I = 2/5 3.2 0.108²

I = 1.49 10⁻² kg m²