Q.The bob of a pendulum is released from a horizontal position. If the length of the pendulum is $1.5\ \text{m}$, what is the speed with which the bob arrives at the lowermost point, given that it dissipated $5\%$ of its initial energy against air resistance?

This problem applies the Conservation of Energy principle, considering the work done by a non-conservative force (air resistance).

1. The bob starts from rest at a horizontal position, so its initial height is $L$. Its initial mechanical energy is purely potential: $E_i = mgL$.
2. At the lowermost point, the height is zero, and the energy is purely kinetic: $E_f = \frac{1}{2}mv^2$.
3. $5\%$ of the initial energy is dissipated, meaning the final mechanical energy is $95\%$ of the initial energy.
4. Equating the final energy to the initial energy minus the dissipated energy: $\frac{1}{2}mv^2 = E_i - 0.05 E_i = 0.95 E_i = 0.95 mgL$. …

When a pendulum bob is released from a horizontal position, it swings downwards due to gravity. As it falls, its gravitational potential energy is converted into kinetic energy. This is a classic application of the work-energy theorem or, more simply, the principle of conservation of mechanical energy. However, the problem states that $5\%$ of the initial energy is dissipated against air resistance. This means that not all of the initial potential energy will be converted into kinetic energy; $95\%$ of it will.

Our approach will be to:

1. Determine the initial potential energy of the bob.
2. Calculate the amount of energy remaining after dissipation.
3. Equate this remaining energy to the kinetic energy at the lowermost point.
4. Solve for the speed.

Let's break this down step-by-step.

1. Identify the initial and final states and define a reference level for potential energy.

   • Initial State: The bob is released from a horizontal position. This means its initial height is equal to the length of the pendulum, $L$, relative to the lowermost point of its swing. Since it's released, its initial speed is $v_i = 0$.
   • Final State: The bob arrives at the lowermost point. At this point, its height is $h_f = 0$ (if we set the lowermost point as our reference level for potential energy). Its speed will be maximum, let's call it $v_f$.
   Tip

   Choosing the lowermost point as the reference level ($h=0$) simplifies calculations because the final potential energy becomes zero.

2. Calculate the initial mechanical energy.

   The initial mechanical energy ($E_i$) is the sum of its initial potential energy ($PE_i$) and initial kinetic energy ($KE_i$).

   The length of the pendulum is $L = 1.5\ \text{m}$.

   The initial height of the bob is $h_i = L = 1.5\ \text{m}$.

   We use the formula for gravitational potential energy:

   $PE = mgh$

   And for kinetic energy:

   $KE = \frac{1}{2}mv^2$

   So, the initial potential energy is $PE_i = mgL$.

   Since the bob is released from rest, its initial kinetic energy is $KE_i = \frac{1}{2}m(0)^2 = 0$.

   Therefore, the total initial mechanical energy is:

3. Account for energy dissipated against air resistance. The problem states that $5\%$ of the initial energy is dissipated. This means $5\%$ of $E_i$ is lost. Energy dissipated $= 0.05 \times E_i = 0.05 mgL$. …

Concept: Energy Conservation with a Dissipative Loss

Step 1: Initial mechanical energy (released from horizontal, at rest)

Step 2: Account for the 5% loss to air resistance

Step 3: Equate the remaining energy to kinetic energy at the bottom

$\tfrac12 mv^2 = 0.95\,mgL \implies v = \sqrt{1.9\,gL}$ …