A physical pendulum in the form of a thin rod of length 1.04 m is pivoted at point P, a distance x from the pendulum's center of mass. Find the value of x for which the period of the pendulum will be a minimum.
To find the value of x for which the period of the pendulum is a minimum, we need to use the concept of the center of mass and the parallel axis theorem.
The period of a physical pendulum is given by the formula:
T = 2π √(I/mgd)
Where:
- T is the period of the pendulum
- I is the moment of inertia of the pendulum about the pivot point
- m is the mass of the pendulum
- g is the acceleration due to gravity
- d is the distance between the pivot point and the center of mass of the pendulum
For a thin rod of length L rotating about one end, the moment of inertia is given by:
I = (1/3) mL^2
Now, let's consider the situation where the pendulum is pivoted at a distance x from the center of mass.
The distance between the pivot point and the center of mass will now be (L/2 - x).
To find the minimum period, we want to minimize the expression T = 2π √(I/mgd).
To minimize T, we can minimize the expression inside the square root [√(I/mgd)].
To do that, we can take the derivative of this expression with respect to x and set it equal to zero:
d/dx (√(I/mgd)) = 0
Now, let's solve this equation to find the value of x for which the period of the pendulum will be a minimum.
1. Square both sides of the equation to eliminate the square root:
(I/mgd) = (I/mgd)
2. Substitute the moment of inertia of the pendulum about the pivot point:
(1/3) mL^2 / (m * g * d) = (1/3) L^2 / (L/2 - x) * g
3. Simplify the equation:
1 / (2 - (2x/L)) = 1 / (2 - (2x/L))
4. Now, let's find the value of x that satisfies this equation:
2 - (2x/L) = 2 - (2x/L)
Simplifying further, we get:
2x/L = 2x/L
5. Solving for x, we find:
x = L/2
Therefore, the value of x for which the period of the pendulum will be a minimum is equal to half the length of the pendulum rod.