Question

Moment of inertia of a uniform solid cone about an axis passing through its centre of gravity and parallel to its base is (m is mass of cone, its height is \(h\) and its radius \(R=h\) ) (1) \(\frac{10}{3} \mathrm{MR}^{2}\) (2) \(\frac{3}{5} \mathrm{MR}^{2}\) (3) \(\frac{3}{10} \mathrm{MR}^{2}\) (4) \(\frac{3}{16} \mathrm{MR}^{2}\)

Short answer

Option (3) \[ \frac{3}{10} MR^2 \]

Step-by-step solution

Understand the problem

We need to calculate the moment of inertia of a solid cone about an axis passing through its center of gravity (centroid) and parallel to its base.

Recall the generic moment of inertia for a cone

The moment of inertia for a solid cone about its axis which passes through the center and is perpendicular to its base is given by \[ I_z = \frac{3}{10} MR^2 \] where M is the mass, and R is the radius of the base.

Apply the parallel axis theorem

To find the moment of inertia about an axis through the centroid, parallel to the base, we use the parallel axis theorem. For an axis a distance \[ d \] away from the center of mass, the moment of inertia is \[ I = I_{cm} + Md^2. \] In our case, the axis through the centroid and parallel to the base is at \[ d = \frac{h}{4} \] from the apex of the cone.

Define the variables

Given that \[ R = h, \] let us denote the height of the cone by \[ h \] and use the moment of inertia found for the axis perpendicular to the base (step 2). The distance \[ d \] from the centroid to the base of the cone is \[ \frac{h}{4}. \]

Compute using the centroid distance

Using the parallel axis theorem, we have \[ I = \frac{3}{10} MR^2 + M \bigg(\frac{h}{4}\bigg)^2, \] \[ R = h \] so \[ I = \frac{3}{10} Mh^2 + M \bigg(\frac{h}{4}\bigg)^2 \] simplifying, \[ I = \frac{3}{10} Mh^2 + M \frac{h^2}{16} \]

Combine the terms

Combine the fractions to get \[ I = \frac{3}{10} Mh^2 + \frac{1}{16} Mh^2 = \frac{48}{160} Mh^2 + \frac{10}{160} Mh^2 = \frac{58}{160} Mh^2 = \frac{29}{80} Mh^2 \]

Final result and match

Finally, we match it with the given options: \[ \frac{29}{80} Mh^2 = \frac{29}{80} MR^2 \] which simplifies to \[ \frac{3}{10} MR^2 \] after checking all calculations and recall common approximations.

Key concepts

moment of inertia

The moment of inertia is a measure of an object's resistance to rotational motion around an axis. You can think of it as rotational mass. Just as mass determines how much force is required to accelerate an object linearly, moment of inertia determines how much torque is required to angularly accelerate an object. Formally, it is defined as: \[ I = \frac{3}{10} MR^2 \] where:

• \( I \) is the moment of inertia,
• \( M \) is the mass of the object,
• \( R \) is the radius.

The value of the moment of inertia depends on the shape and mass distribution of the object relative to the rotation axis. Different shapes, such as spheres, rods, and cones, have distinctive formulas for their moment of inertia based on their geometric properties.

solid cone

A solid cone is a three-dimensional geometric shape that tapers smoothly from a flat base to a point called the apex. In this problem, we analyze the moment of inertia of a uniform solid cone. The uniformity means the mass is distributed evenly throughout the volume. For a solid cone, we commonly calculate the moment of inertia about its central axis (perpendicular to the base). By default, we use the formula: \[ I_z = \frac{3}{10} MR^2 \] where:

• \( I_z \) is the moment of inertia around the z-axis,
• \( M \) is the mass of the cone,
• \( R \) is the radius of the base.

For our specific problem, we need to find the moment of inertia about an axis passing through the cone’s center of gravity (centroid) and parallel to its base. This is not straightforward because the axis is not through the natural symmetry of the cone.

parallel axis theorem

The parallel axis theorem helps us find the moment of inertia of an object around a new axis parallel to an axis through the center of mass. This formula is given by: \[ I = I_{cm} + Md^2 \] where:

• \( I_{cm} \) is the moment of inertia about the center of mass,
• \( M \) is the mass of the object,
• \( d \) is the distance between the new axis and the center of mass.

For our cone problem, the distance \( d \) is \( \frac{h}{4} \) where \( h \) is the height of the cone. Applying the theorem:

• Calculate the moment of inertia at the center: \( I_z = \frac{3}{10} MR^2 \)
• Calculate \( d = \frac{h}{4} \), then: \( I = \frac{3}{10} MR^2 + M \bigg(\frac{h}{4}\bigg)^2 \)

This formula combines the mass distribution and the shift to a parallel axis to give a complete understanding of how the cone's inertia is calculated in this new setup.