Problem 94 - University Physics with Modern Physics

Question

A pendulum is made of a uniform solid sphere with mass \(M\) and radius \(R\) suspended from the end of a light rod. The distance from the pivot at the upper end of the rod to the center of the sphere is \(L\) . The pendulum's moment of inertia I for rotation about the pivot is usually approximated as \(M L^{2} .\) (a) Use the parallel-axis theorem to show that if \(R\) is 5\(\%\) of \(L\) and the mass of the rod is ignored, \(I_{p}\) is only 0.1\(\%\) greater than \(M L^{2} .\) (b) If the mass of the rod is 1\(\%\) of \(M\) and \(R\) is much less than \(L,\) what is the ratio of \(I_{\text { rod }}\) for an axis at the pivot to \(M L^{2} ?\)

Step-by-Step Solution

Answer

(a) The moment of inertia \(I_p\) is 0.1% greater than \(ML^2\). (b) The ratio of \(I_{rod}\) to \(ML^2\) is 0.0033.

Step 1: Calculate moment of inertia of the sphere about its center

The moment of inertia of a uniform solid sphere about its center is given by \( I_c = \frac{2}{5} M R^2 \).

Step 2: Apply the parallel-axis theorem for the sphere

According to the parallel-axis theorem, the moment of inertia of the sphere about the pivot point \(I_p\) is given by \( I_p = I_c + M d^2 \), where \(d = L\) is the distance from the center of the sphere to the pivot. Substitute \( I_c = \frac{2}{5} M R^2 \) to get \( I_p = \frac{2}{5} M R^2 + M L^2 \).

Step 3: Express R in terms of L

Given that \( R = 0.05L \), substitute \( R \) into \( I_p \) to rewrite it as \( I_p = \frac{2}{5} M (0.05L)^2 + M L^2 \).

Step 4: Simplify the expression for I_p

Simplify the expression: \( I_p = \frac{2}{5} M (0.0025L^2) + M L^2 = 0.001 M L^2 + M L^2 \). Thus, \( I_p = 1.001 M L^2 \), showing that \( I_p \) is 0.1% greater than \( M L^2 \).

Step 5: Calculate the moment of inertia of the rod

The moment of inertia of the rod about its end is \( I_{rod} = \frac{1}{3}mL^2 \), where \( m = 0.01M \) (1% of the sphere's mass). Thus, \( I_{rod} = \frac{1}{3} (0.01M)L^2 = 0.0033ML^2 \).

Step 6: Find the ratio of I_rod to ML^2

The ratio of the rod's moment of inertia to the sphere's approximation is given by \( \frac{I_{rod}}{ML^2} = \frac{0.0033ML^2}{ML^2} = 0.0033 \).

Key Concepts

Parallel-Axis Theorem

The Parallel-Axis Theorem is an important principle in physics that allows us to calculate the moment of inertia of a body around an axis that is parallel to an axis through its center of mass. This theorem is crucial when the axis of rotation is not at the center of mass.

The theorem's formula is:

• \[ I_p = I_c + Md^2 \]

Where:

• \( I_p \) is the moment of inertia about the parallel axis.
• \( I_c \) is the moment of inertia about the center of mass.
• \( M \) is the mass of the object.
• \( d \) is the distance between the two axes.

In our exercise, we use the parallel-axis theorem to find the moment of inertia of the pendulum, considering the distance from the sphere's center to the pivot to understand how close the real moment of inertia is to the approximation \( ML^2 \). It shows the moment of inertia is just 0.1% more when \( R \) is 5% of \( L \).

Uniform Solid Sphere

A uniform solid sphere has a constant density throughout, which means its mass is evenly distributed. This makes calculating properties like moment of inertia a bit simpler.

The moment of inertia of a uniform solid sphere around an axis through its center is given by:

• \[ I_c = \frac{2}{5} MR^2 \]

where \( M \) is the mass and \( R \) is the radius of the sphere. This formula provides a fundamental understanding when applying the parallel-axis theorem for further calculations. In the exercise problem, we used this concept to determine the sphere's resistance to rotational motion about its center, later applying the parallel-axis theorem to find its inertia about the pivot point.

Pendulum Dynamics

Pendulum dynamics involves studying the motion of a pendulum, which is a weight suspended from a pivot so it can swing freely. Understanding pendulum dynamics is crucial for applications in clocks, measuring instruments, and seismographs.

The motion is generally influenced by:

• The mass of the pendulum.
• The length of the arm from the pivot to the center of mass of the pendulum.
• The gravitational force acting on it.

For a pendulum involving a solid sphere, the moment of inertia, distance from the pivot, and small angle approximations (because large angles result in non-harmonic motion) are important factors. This exercise demonstrates how additional components, like rod mass and sphere radius, affect the moment of inertia and, subsequently, the pendulum's swinging behavior.

Rotational Motion

Rotational motion refers to the motion of a body about an axis. It is different from linear motion, which occurs in a straight line.

For rotational motion:

• The rotational equivalent of mass is the moment of inertia, which determines the resistance of an object to changes in its rotational state.
• The more distributed the mass is relative to the axis of rotation, the larger the moment of inertia.

More specifically, when considering the uniform solid sphere in the pendulum exercise:

• The distribution of mass (characteristic radius compared to pendulum length) helps calculate the effective moment of inertia using the parallel-axis theorem.
• This allows for better understanding of how the sphere interacts with the rotational dynamics of a pendulum.

The interplay of these elements is key to understanding why certain assumptions or approximations (like ignoring rod mass or treating radius as a small fraction of length) hold true in engineering and physics applications.