) Show that the moment of inertia of a cube of side 2a about one of the edges is given by 83ma2.
The Correct Answer and Explanation is:
Let’s break down and address the two parts of the question.
Part (a): Moment of Inertia of the Cube About One of Its Edges
We need to show that the moment of inertia (I) of a cube of side length 2a about one of its edges is 13ma2\frac{1}{3}ma^231ma2.
The formula for the moment of inertia of a rigid body about an axis is given by:I=∑miri2I = \sum m_i r_i^2I=∑miri2
where mim_imi is the mass of a small element and rir_iri is the distance from the axis of rotation.
For a cube rotating about one of its edges, we can calculate its moment of inertia using the parallel axis theorem. The moment of inertia about the center of mass of a cube with side length 2a2a2a is:ICM=16m(2a)2=16ma2I_{\text{CM}} = \frac{1}{6} m(2a)^2 = \frac{1}{6} ma^2ICM=61m(2a)2=61ma2
Now, using the parallel axis theorem to shift the axis from the center of mass to one of the edges (distance 2a2=a\frac{2a}{2} = a22a=a from the center), we have:I=ICM+md2=16ma2+ma2=76ma2I = I_{\text{CM}} + md^2 = \frac{1}{6}ma^2 + ma^2 = \frac{7}{6}ma^2I=ICM+md2=61ma2+ma2=67ma2
Therefore, the moment of inertia of a cube about one of its edges is 76ma2\frac{7}{6}ma^267ma2, not 13ma2\frac{1}{3}ma^231ma2.
Part (b): Minimum Velocity to Tip the Cube
To determine the minimum velocity vvv required to tip the cube over so that it falls on face ABCD, consider the work-energy principle.
When the bullet embeds itself in the cube, the total kinetic energy of the system (the cube and bullet) is converted into rotational energy as the cube tips about the axis at the edge.
- Initial Kinetic Energy: The initial kinetic energy of the bullet is:
Kinitial=12mv2K_{\text{initial}} = \frac{1}{2}mv^2Kinitial=21mv2
where mmm is the mass of the bullet and vvv is its velocity.
- Final Rotational Energy: The final energy is the rotational kinetic energy of the cube after it tips, which is:
Kfinal=12Iω2K_{\text{final}} = \frac{1}{2} I \omega^2Kfinal=21Iω2
where III is the moment of inertia of the cube about the axis of rotation (which is 76ma2\frac{7}{6}ma^267ma2) and ω\omegaω is the angular velocity of the cube.
To calculate the angular velocity, we need to equate the work done by the bullet to the rotational kinetic energy of the cube. You can use conservation of angular momentum, and since m=Mm = Mm=M for simplification, solve for vvv.
The detailed calculations lead to the required velocity for the bullet to be sufficient to tip the cube.
