Question

Calculate the moment of inertia of each of the following uniform objects about the axes indicated. Consult Table Moments of Inertia of Various Bodies in the Textbook as needed.(A)A thin 3.70-kg rod of length 80.0cm, about an axis perpendicular to it and passing through one end.(B)A thin 3.70-kg rod of length 80.0cm, about an axis perpendicular to it and passing through its center.(C)A 5.00-kg sphere 25.0cm in diameter, about an axis through its center, if the sphere is solid.(D)A 5.00-kg sphere 25.0cm in diameter, about an axis through its center, if the sphere is a thin-walled hollow shell.(E)An 6.00-kg cylinder, of length 15.0cm and diameter 24.0cm, about the central axis of the cylinder, if the cylinder is thin-walled and hollow.(F)An 6.00-kg cylinder, of length 15.0cm and diameter 24.0cm, about the central axis of the cylinder, if the cylinder is solid.

Answer 1

To find the moment of inertia of each object, we use different formulas for different shapes. The moments of inertia about the given axes are calculated using the formulas and the given values. The moments of inertia for the objects are: (A) 0.79 kg·m^2, (B) 0.10 kg·m^2, (C) 0.031 kg·m^2, (D) 0.042 kg·m^2, (E) 0.043 kg·m^2, and (F) 0.07 kg·m^2.

Step-by-step explanation:

To calculate the moment of inertia of each object, we can use the formulas for moment of inertia for different shapes. Here are the calculations for each object:

(A) Thin rod about an axis perpendicular to it and passing through one end:

The formula for the moment of inertia of a thin rod about an axis through one end is I = (1/3) * m * L^2, where m is the mass of the rod and L is its length. Plugging in the given values, we get:

I = (1/3) * 3.70 kg * (0.80 m)^2 = 0.79 kg·m^2

(B) Thin rod about an axis perpendicular to it and passing through its center:

The formula for the moment of inertia of a thin rod about an axis through its center is I = (1/12) * m * L^2. Plugging in the given values, we get:

I = (1/12) * 3.70 kg * (0.80 m)^2 = 0.10 kg·m^2

(C) Solid sphere about an axis through its center:

The formula for the moment of inertia of a solid sphere about an axis through its center is I = (2/5) * m * R^2, where m is the mass of the sphere and R is its radius. Plugging in the given values, we get:

I = (2/5) * 5.00 kg * (0.125 m)^2 = 0.031 kg·m^2

(D) Hollow shell sphere about an axis through its center:

The formula for the moment of inertia of a thin-walled hollow shell sphere about an axis through its center is I = (2/3) * m * R^2. Plugging in the given values, we get:

I = (2/3) * 5.00 kg * (0.125 m)^2 = 0.042 kg·m^2

(E) Hollow cylinder about its central axis:

The formula for the moment of inertia of a thin-walled hollow cylinder about its central axis is I = (1/2) * m * R^2, where m is the mass of the cylinder and R is its radius. Plugging in the given values, we get:

I = (1/2) * 6.00 kg * (0.12 m)^2 = 0.043 kg·m^2

(F) Solid cylinder about its central axis:

The formula for the moment of inertia of a solid cylinder about its central axis is I = (1/12) * m * (3R^2 + h^2), where m is the mass of the cylinder, R is its radius, and h is its height. Plugging in the given values, we get:

I = (1/12) * 6.00 kg * (3 * 0.12 m)^2 + (0.15 m)^2 = 0.07 kg·m^2

Answer 2

The moment of inertia for each object is calculated using specific formulas. The moment of inertia of the thin rod about an axis passing through one end is 0.79 kg.m², and about an axis passing through its center is 0.10 kg.m². The moment of inertia of the solid sphere is 0.02 kg.m², of the thin-walled hollow sphere is 0.10 kg.m², of the thin-walled hollow cylinder is 0.04 kg.m², and of the solid cylinder is also 0.04 kg.m².

Step-by-step explanation:

To calculate the moment of inertia of each object, we can use the formulas provided in Table Moments of Inertia of Various Bodies in the Textbook. Let's calculate the moment of inertia for each object:

(A) For a thin rod, the moment of inertia about an axis perpendicular to it and passing through one end is given by:

I = (1/3) * m * L², where m is the mass of the rod and L is its length. Plugging in the values, we get:

I = (1/3) * 3.70 kg * (0.80 m)² = 0.79 kg.m²

(B) For the same thin rod, but with the axis passing through its center, the moment of inertia is given by:

I = (1/12) * m * L². Plugging in the values, we get:

I = (1/12) * 3.70 kg * (0.80 m)² = 0.10 kg.m²

(C) For a solid sphere, the moment of inertia about an axis through its center is given by:

I = (2/5) * m * R², where m is the mass of the sphere and R is its radius. Plugging in the values, we get:

I = (2/5) * 5.00 kg * (0.125 m)² = 0.02 kg.m²

(D) For a thin-walled hollow sphere, the moment of inertia about an axis through its center is given by:

I = (2/3) * m * R². Plugging in the values, we get:

I = (2/3) * 5.00 kg * (0.125 m)² = 0.10 kg.m²

(E) For a thin-walled hollow cylinder, the moment of inertia about its central axis is given by:

I = (1/2) * m * R², where m is the mass of the cylinder and R is its radius. Plugging in the values, we get:

I = (1/2) * 6.00 kg * (0.12 m)^2 = 0.04 kg.m²

(F) For a solid cylinder, the moment of inertia about its central axis is given by:

I = (1/2) * m * R². Plugging in the values, we get:

I = (1/2) * 6.00 kg * (0.12 m)^2 = 0.04 kg.m²