Part of solid is cut
Understanding the Topic: "Part of Solid is Cut"
When a part of a solid is cut, especially in the context of rotational motion in physics, it can significantly affect the object's mass distribution and consequently its moment of inertia. The moment of inertia is a measure of an object's resistance to changes in its rotational motion. It depends on the mass distribution of the object relative to the axis of rotation.
Moment of Inertia
The moment of inertia (I) for a solid object is given by the integral:
$$ I = \int r^2 \, dm $$
where ( r ) is the distance from the axis of rotation to a small element of mass ( dm ).
When a part of the solid is cut, the mass distribution changes, and so does the moment of inertia. Calculating the new moment of inertia involves subtracting the moment of inertia of the removed part from the moment of inertia of the original solid.
Table of Differences and Important Points
| Aspect | Original Solid | After Part is Cut |
|---|---|---|
| Mass | ( M ) | ( M - m ) (where ( m ) is the mass of the cut part) |
| Moment of Inertia | ( I_{original} ) | ( I_{new} = I_{original} - I_{cut} ) |
| Center of Mass | At a certain point depending on the shape and density | Shifts due to the removal of mass |
| Stability | Depends on the shape and mass distribution | May decrease if the cut part was providing balance |
Formulas
When a part of a solid is cut, the new moment of inertia ( I_{new} ) can be calculated using the parallel axis theorem if the cut part is not around the original axis of rotation:
$$ I_{new} = I_{original} - (I_{cut} + md^2) $$
where:
- ( I_{cut} ) is the moment of inertia of the cut part about its own center of mass.
- ( m ) is the mass of the cut part.
- ( d ) is the distance between the center of mass of the cut part and the axis of rotation of the original solid.
Examples
Example 1: Cutting a Hole in a Disk
Consider a uniform solid disk of radius ( R ) and mass ( M ) with a small hole of radius ( r ) cut out from its edge. The moment of inertia of the original disk about its center is:
$$ I_{original} = \frac{1}{2}MR^2 $$
The moment of inertia of the cut part (the small disk) about its own center is:
$$ I_{cut} = \frac{1}{2}mr^2 $$
Since the hole is at the edge, the distance ( d ) from the center of the disk to the center of the hole is ( R ). Using the parallel axis theorem:
$$ I_{new} = \frac{1}{2}MR^2 - \left(\frac{1}{2}mr^2 + mR^2\right) $$
Example 2: Cutting a Corner from a Square Plate
Imagine a uniform square plate of side ( a ) and mass ( M ). If a small square of side ( b ) is cut from one of its corners, the new moment of inertia about an axis perpendicular to the plate and passing through its center can be found by subtracting the moment of inertia of the cut part from the original plate's moment of inertia.
For the original plate:
$$ I_{original} = \frac{1}{6}Ma^2 $$
For the cut part, we first find its moment of inertia about its own center:
$$ I_{cut} = \frac{1}{6}mb^2 $$
Then, we apply the parallel axis theorem, considering the distance ( d ) from the center of the plate to the center of the cut part:
$$ d = \frac{a}{2} - \frac{b}{2} $$
The new moment of inertia is:
$$ I_{new} = \frac{1}{6}Ma^2 - \left(\frac{1}{6}mb^2 + md^2\right) $$
In both examples, the calculations assume uniform density and that the cut parts are removed cleanly without affecting the rest of the solid's structure.
Understanding the changes in physical properties when a part of a solid is cut is crucial in various applications, including engineering, where materials are often cut or drilled to achieve desired shapes and properties.