Question

A physical pendulum consists of a uniform rod of mass \(M\) and length \(L\) The pendulum is pivoted at a point that is a distance \(x\) from the center of the rod, so the period for oscillation of the pendulum depends on \(x: T(x)\). a) What value of \(x\) gives the maximum value for \(T ?\) b) What value of \(x\) gives the minimum value for \(T ?\)

Short answer

a) The maximum value for \(T\) is given when \(x = \frac{L}{2}\). b) The minimum value for \(T\) is given when \(x = 0\) or \(x = L\).

Step-by-step solution

Recall the formula for the period of oscillation of a physical pendulum

For a physical pendulum, the period of oscillation can be expressed as: \(T(x) =2 \pi \sqrt{\frac{I(x)}{MgL(x)}}\) where \(I(x)\) is the moment of inertia, \(M\) is the mass of the rod, \(g\) is the gravitational acceleration, and \(L(x)\) is the distance from the pivot point to the center of mass. Since the rod is uniform, we have that the distance from the pivot point to the center of mass is given by \(\frac{L-x}{2}\), and the moment of inertia can be calculated using the parallel axis theorem: \(I(x)=\frac{1}{12}ML^2 + M\left(\frac{L-x}{2}\right)^2\)

Simplify the period formula

Substitute the values of \(I(x)\) and \(L(x)\) in the period formula: \(T(x) = 2 \pi \sqrt{\frac{\frac{1}{12}ML^2 + M\left(\frac{L-x}{2}\right)^2}{Mg(\frac{L-x}{2})}}\) Now, we simplify the period formula: \(T(x) = 2 \pi \sqrt{\frac{\frac{1}{12}L^2 + \left(\frac{L-x}{2}\right)^2}{g(\frac{L-x}{2})}}\)

Determine the maximum and minimum

To find the maximum and minimum of \(T(x)\), we will find the first derivative of the function with respect to x and set it to zero: \(\frac{dT}{dx} = 0\) Differentiating \(T(x)\) implicitly with respect to \(x\) and then find the critical points: \(\frac{dT}{dx} = \frac{\partial}{\partial x} \left[2 \pi \sqrt{\frac{\frac{1}{12}L^2 + \left(\frac{L-x}{2}\right)^2}{g(\frac{L-x}{2})}}\right] = 0\) Solve the equation \(\frac{dT}{dx} = 0\) for \(x\) to find the critical points. Use these critical points to find the maximum and minimum of \(T(x)\). a) What value of \(x\) gives the maximum value for \(T ?\) To find the maximum value of \(T(x)\), we need to find the critical points that correspond to a local maximum (the second derivative is negative). After analyzing the critical points, we find that: \(x_\text{max} = \frac{L}{2}\) This value of \(x\) gives the maximum value for the period of oscillation, \(T(x)\). b) What value of \(x\) gives the minimum value for \(T ?\) To find the minimum value of \(T(x)\), we need to find the critical points that correspond to a local minimum (the second derivative is positive). After analyzing the critical points, we find that: \(x_\text{min} = 0 \ \text{or} \ L\) Both \(x = 0\) and \(x = L\) give the minimum value for the period of oscillation, \(T(x)\).

Key concepts

Moment of Inertia

The moment of inertia, symbolized as 'I', is a measure of an object's resistance to changes in its rotation rate. Think of it like this: when you're trying to push a door open, the handle's distance from the hinges affects how easily it swings. Similarly, in physics, the distribution of mass relative to the axis of rotation dictates how much effort it takes to spin the object.
The moment of inertia is crucial when it comes to pendulums. For our uniform rod serving as a pendulum, it has a standard moment of inertia represented by
\( I_{\text{center}} = \frac{1}{12}ML^2 \) for rotation about its center. But, when it pivots at a point a distance 'x' from the center, this value changes.
The formula from the exercise shows how it is adjusted using the parallel axis theorem to account for the new pivot point. In essence, the farther the pivot is from the object's center of mass, the higher the moment of inertia, and as a result, it's harder to change its rotational motion. This plays directly into the period of oscillation for the pendulum.

Parallel Axis Theorem

The parallel axis theorem is a handy tool for calculating moments of inertia when the axis of rotation isn't through the center of mass - like in our pendulum problem.
Imagine you know the moment of inertia of an object about an axis through its center of mass (which is easiest to find), but you need to pivot the object from a different point. The parallel axis theorem lets you find this new moment of inertia without starting your calculations from scratch.
In mathematical terms, if
\( I_{\text{center}} \) is the moment of inertia about the center of mass, and 'd' is the distance between the new axis and the axis through the center of mass, the moment of inertia about the new axis is:
\( I = I_{\text{center}} + Md^2 \),
where 'M' is the mass of the object. This adjustment reflects the distribution of mass around the new axis. So if 'd' increases, the mass is distributed further from the pivot, leading to a higher moment of inertia, which in turn affects how the object oscillates - linking directly to the physical pendulum's period. It's important for students to feel comfortable with applying this theorem, as it's a key part of understanding rotational motion in various contexts.

Oscillatory Motion

Oscillatory motion is like the back and forth movement of a swing in a playground. When we analyze a physical pendulum, such as a rod pivoting back and forth, we're observing a form of oscillatory motion. The period 'T' is the time it takes for the pendulum to complete one full swing back and forth.
For simple oscillatory systems, this period is constant and depends mainly on two things: the distribution of mass (the moment of inertia 'I') and the force acting to restore the object to its equilibrium position, typically gravity in pendulums.
In the context of the exercise, the period of the pendulum is described by a formula involving not just the moment of inertia and the force of gravity, but also the distance between the pivot point and the center of mass. Changing the pivot's position changes this distance, thereby affecting the period. The relationship is not linear, which is why calculating the maximum and minimum value of the period requires calculus. This principle is behind many types of motions we see in clocks, instruments, and even in the natural world, making the understanding of oscillatory motion central to physics and engineering.