How to Solve a Physics Problem Step by Step (With Example)

By Pankaj Kumar|Last Updated: July 12, 2026|

Key Takeaways

A physical pendulum differs from a simple pendulum because mass is distributed, not concentrated.
Start from the displaced position and apply the restoring torque to derive the period.
For a uniform bar pivoted at one end, the period formula is T = 2π√(2l/3g).
Small-angle approximation (sin θ ≈ θ) is essential for arriving at SHM criteria.
Plugging T = 1.2 s into the formula gives a bar length of approximately 0.537 m.

This article will see how to solve a physics problem from scratch, step by step. To demonstrate this, we will solve a physical pendulum problem with a detailed explanation as given in a textbook. If you find these kinds of problems challenging, working with an online physics tutor can help you build the logical reasoning needed to tackle them independently.

Question: A Uniform Steel Bar Swings from a Pivot at One End with a Period of 1.2 Seconds. How Long Is the Bar?

This problem deals with PHYSICAL PENDULUM and calculating its period. We will explain the solution to this problem right from scratch. We will not use any secondary formulas, and all the steps will be based only on the most fundamental equations.

Also, we will explain how to approach and solve a problem, where to start from, and what strategy to adopt to solve it most efficiently and logically.

It is explained in such a way that anyone knowing even fundamental physics can understand it easily.

We will try not to solve the problem step by step but also provide helpful insight into what goes on while solving a problem. It will help us solve other problems in your physics studies as well. If you have ever wondered why physics homework takes so long, understanding this kind of structured approach is a good place to start.

Approaching the Problem

The bar given is uniform. By this, we mean the dimensions are uniform, and the density is the same everywhere in the bar.

This situation is entirely different from a simple pendulum where the mass “m” is concentrated in the bob at a distance “l” from the pivot point. Here the whole mass “m” is distributed uniformly throughout the length of the bar pendulum. Students preparing for AP Physics will encounter this distinction frequently in exam problems.

Free Body Diagram of the Physical Pendulum

Free body diagram of a physical pendulum showing a uniform bar pivoted at one end

Building Our Strategy

We know from our experience that when we displace a uniform rod pivoted at one end slightly by a small angle “θ,” it oscillates about the pivot. So, our analysis should start from the point where the uniform bar gets displaced from its equilibrium position or the vertical position by a small angle “θ.”

The bar will try to come back to its original position. When it comes back to its equilibrium position (vertical), the bar gets kinetic energy due to loss in gravitational potential energy. This momentum does not let the rod settle and instead takes it away again from the vertical position.

This cycle continues till it loses all its energy and becomes upright again. Understanding this energy exchange is also central to AP Physics 1 mechanics topics that appear on the exam.

The Physics and Math with Explanations

We will assume the bar’s mass as “m” and its length as “l.”

Since the bar is uniform, its center of mass must be at its geometrical center; for a linear rod, the center of mass is at a distance of “l/2” from any end.

The component of the weight W=mg which is normal to the rod, is “mg*sinθ.”

It acts at “l/2” from the pivot point, so the moment of this restoring torque about the pivot point is:

T= (l/2) *mg*sinθ and for small angles sinθ= θ this becomes T= (l/2) mgθ

Moment of inertia about the pivot is:

I=1/3*ml^2

The resulting angular acceleration α is:

α=T/I= (l/2)*mg*θ/(1/3*ml^2)

Which is α= 3gθ/2l

The SHM criteria is α= w^2*θ

It satisfies the SHM criteria with angular velocity w= sqrt(3g/2l)

Now the period is T=2*pi/w

So, T=2pi*sqrt(2l/3g)

It completes our derivation. Students taking AP Physics C will recognise this rotational dynamics derivation as a core skill tested in the Mechanics free-response section.

Solving for the Bar Length

In the given problem, the period is 1.2 seconds. Plug this in, and we will get:

1.2=2pi*sqrt(2l/3*9.81)

Solving this, we will get l= 0.537 m

Finally, it completes our explanation.

If you follow the logic and strategy that we used to solve this problem, you will be able to solve physics problems better in the future. The same structured approach applies across many topics — including those covered in chemistry tutoring, where deriving results from first principles is equally important.

If you need a private online physics tutor who can logically explain physics problems in easy-to-understand steps, contact us on WhatsApp at +91 8971 383660 or by email at meb@myengineeringbuddy.com. Before committing, you can also review what a physics tutor typically costs to set your expectations.

If you are studying for a high-stakes exam, the strategies in 7 smart ways to use predicted papers for A-Level Physics are worth reading alongside this worked example. You can also explore AP Physics 2 tutoring if your coursework extends into electricity, magnetism, and thermodynamics.