# A Physical Pendulum Problem

About 2 weeks ago, Former MIT Professor Walter H. G. Lewin posted a Physics problem on his YouTube channel as a part of his series of problems for the viewers of his channel. This question was divided into two parts – the first one, which was quite simple and a trickier sequel to it. The videos have been embedded and the solution is derived along the problem.

## The First Part

### Question 1.

What is the time period of a swinging pendulum made out of a uniform rod which is oscillated about one of its ends ?

#### Possible Solution

We have an uniform rod of mass M, length L and the acceleration of gravity is taken as g. The pendulum is oscillated about one of its end. The center of mass and center of gravity of an uniform rod lies at the center of the rod i.e. at a distance L/2 from one end of the rod.

The given problem can be visualized according to the figure drawn below :We know that according to Newton’s Laws of Motion, an object remains at rest or continues to move in the same trajectory, unless acted upon by an external force. A rotational analogue for the same would be that an object at rest, remains at rest or continues to rotate with constant angular velocity, unless acted on  by an external torque.

By the laws of rotational dynamics,  torque is given by the dot product of Force and distance to the line of force (force lever) i.e. τ = Force . Lever arm = $\inline&space;M&space;g&space;\frac{L}{2}&space;sin(b)&space;=&space;M&space;g&space;\frac{L}{2}&space;b${As sin(b)=b for small angles (ignoring higher terms in the expansion of the sine function and doing small angle approximation).}

Moment of inertia of an uniform rod for an axis passing through one of its ends is given by, $\inline&space;I&space;=&space;\frac{1}{3}ML^{2}$ .

For rotational equilibrium,  τ = Iα.  Hence,   ( $\inline&space;\frac{1}{3}$ ML2 ) α = ( Mg $\inline&space;\frac{L}{2}$ b ) or,  α = $\inline&space;\frac{3g}{2L}$ x b

We know that α = ω² r. So, from this relation   $\inline&space;\omega&space;^{2}&space;=&space;\frac{3g}{2L}$

Now,  T = $\inline&space;\frac{2\pi}{\omega&space;}$From here we get our solution to the problem which is  $\inline&space;T&space;=&space;2\pi\sqrt{}&space;\frac{2L}{3g}$.

Alternatively, a very easy and sneaky-peaky answer to this problem is given by the expression given below, where, I is the inertial factor and k is the spring factor of the solid.

$\inline&space;T&space;=&space;2\pi\sqrt{\frac{I}{a}}&space;=2\pi\sqrt\frac{\frac{1}{3}ML^2}{Mg\frac{L}{2}}&space;=2\pi\sqrt{\frac{2L}{3g}}$

## The Second Part

### Question 2.

What shall be the other point on the rod from where it should be supported so that time period of this physical pendulum is same as the time period when it was supported from one of its ends ?

#### Possible Solution

Now we should know that Torque = Sum of anti CW moments and CW moments, following a sign convention that anti CW moments are to be taken as negative and CW moments are to be taken as positive. Using this definition, we solve the problem.

Our results are in agreement with the solutions Prof. Walter Lewin gave to this problem. If you want to check out his channel and refer to his solution, please follow the link given: Lectures by Walter Lewin. They will make you ♥ Physics.