Solved Examples on Physical Pendulum

Example 1: A rectangular plate with dimensions of 0.5 meters in length and 0.2 meters in breadth makes up a physical pendulum. The 2-kilogram plate pivots at one end and has a mass. What is the pendulum’s period if the pivot point and centre of mass are separated by 0.1 meters?

Solution:

First, we need to find the moment of inertia of the rectangular plate, which is 

I = (1/12) × m × (l² + w²) 

  = (1/12) × 2 × (0.5² + 0.2²) 

  = 0.067 kg m²

Substituting the values into the formula, we get

T = 2π√(I/mgd) 

   = 2π√(0.067/29.810.1) 

   = 0.97 seconds

Therefore, the period of the pendulum is 0.97 seconds.

Example 2: A rectangular rod having a length of 0.4 meters and a radius of 0.1 meters makes up a physical pendulum. The 3-kilogram rod pivots at one end and has a mass. What is the pendulum’s period if the pivot point and centre of mass are separated by 0.2 meters?

Solution:

First, we need to find the moment of inertia of the cylindrical rod, which is 

I = (1/12) × m × (3r² + h²) 

  = (1/12) × 3 × (3 × 0.1² + 0.4²) 

  = 0.094 kg m²

Substituting the values into the formula, we get:

T = 2π√(I/mgd) 

   = 2π√(0.094/39.810.2) 

   = 0.305 sec

Example 3: A one-meter-long, uniformly thin rod weighing 0.5 kg makes up a physical pendulum. The rod has a pivot point at one end, and there are 0.3 meters between the pivot point and the centre of mass. How long has the pendulum been swinging?

Solution:

The moment of inertia of a thin, uniform rod rotating about one end is given by 

I = (1/3) × m × L²,    (where L is the length of the rod)

I = (1/3) × 0.5 × 1² 

  = 0.167 kg m²

T = 2π√(I/mgd) 

   = 2π√(0.167/0.59.810.3) 

   = 1.09 seconds

Therefore, the period of the pendulum is 1.09 seconds.

Example 4: A round disc with a radius of 0.2 meters and a mass of 1 kilogram makes up a physical pendulum. The disc pivots in its centre, and there are 0.2 meters between the pivot and the centre of mass. How long has the pendulum been swinging?

Solution:

The moment of inertia of a circular disk rotating about its centre is given by 

I = (1/2) × m × r²      (where r is the radius of the disk)

I = (1/2) × 1 × 0.2L2 

  = 0.02 kg m²

T = 2π√(I/mgd) 

   = 2π√(0.02/19.810.2) 

   = 0.63 seconds

Therefore, the period of the pendulum is 0.63 seconds.

Example 5: A physical pendulum consists of a rectangular plate with a length of 0.3 meters and a width of 0.1 meters. The plate has a mass of 2.5 kg and is pivoted at one end. If the distance between the pivot point and the centre of mass is 0.05 meters, what is the period of the pendulum?

Solution:

The moment of inertia of a rectangular plate rotating about one end is given by 

I = (1/3) × m × L²       (where L is the length of the plate)

I = (1/3) × 2.5 × 0.3²

  = 0.225 kg m²

T = 2π√(I/mgd) 

   = 2π√(0.225/2.59.810.05) 

   = 0.85 seconds

Therefore, the period of the pendulum is 0.85 seconds.

A rigid body that is capable of rotating around an axis makes up a physical pendulum, a particular kind of pendulum. The physical pendulum can be shaped into a straight rod, a rectangular plate, or a circular disc, in contrast to a basic pendulum, which consists of a tiny mass hanging by a string. The moment of inertia, the separation between the pivot point and the center of mass, and the gravitational pull all affect how a physical pendulum swings.