A hollow sphere is attached to the end of a uniform rod. The sphere has a radius of 0.60 m and a mass of 0.44 kg. The rod has a length of 1.28 m and a mass of 0.48 kg. The rod is placed on a fulcrum (pivot) at X = 0.46 m from the left end of the rod. Calculate the moment of inertia (click for graphical table) of the contraption around the fulcrum.

Answer:Moment of inertia is the property of a body due to which it resists angular acceleration. It is the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation.

Explanation: I = m*r*r

Given mass of sphere = 0.4kg

Radius of sphere = 0.6m

Moment of Inertia of sphere = (2/3)*0.4* 0.6

Inertia of sphere = 0.096 kgm2

Given mass of rod = 0.48kg

Length of rod = 1.28m

Inertia of rod = (m*L*L)/3

I = (0.48 * 1.28 * 1.28)/3

I = 0.218 kgm2

Total Moment of inertia I = 0.096 + 0.218

I = 0.314 kgm2

AFOKE88 AFOKE88 · Answer 2 · 2021-10-24T16:05:48+02:00

The moment of inertia of the contraption around the fulcrum is;

J = 0.5755 kg•m²

We are given;

Mass of sphere; m_s = 0.44 kg

Mass of rod; m_r = 0.48 kg

Radius of sphere; r = 0.6 m

Length of rod; L = 1.28

Distance from left end of rod to the pivot; X = 0.46 m

The polar moment of inertia J of the contraption around the fulcrum is given by the formula;

J = Jℓ + Jr

Where;

Jℓ is polar moment of inertia around length of rod

Jr is polar moment of inertia around the sphere of radius

Let's calculate Jℓ

Jℓ = [((0.46²/3) × 0.48 × 0.46)/1.28] + (0.44 × (0.46 + 0.6)²)

Jℓ = 0.506551 kg•m²

Now, let's Calculate Jr

Jr = ((1.28 - 0.46)²/3) × ((1.28 - 0.46)/1.28) × 0.48

Jr = 0.068921 kg•m²

Thus;

J = 0.506551 + 0.068921

J = 0.5755 kg•m²

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