Question

The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 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 ?

Answer 1

Length of the pendulum, \(\text I\) = \(1.5 \,m\) 
Mass of the bob = \(m\) 
Energy dissipated = \(5\)% 
According to the law of conservation of energy, the total energy of the system remains constant. 
At the horizontal position: 
Potential energy of the bob, \(E_p\) = \(mgl\) 
Kinetic energy of the bob, \(E_k\) = \(0\) 
Total energy = \(mgl\)        … (i) 
At the lowermost point (mean position): Potential energy of the bob, \(E_p\) = \(0\) 
Kinetic energy of the bob, \(E_k\)= \(\frac{1}{2}mv^2\)
Total energy \(E_x\)= \(\frac{1}{2}mv^2\) … (ii) 
As the bob moves from the horizontal position to the lowermost point, \(5\)% of its energy gets dissipated. The total energy at the lowermost point is equal to \(95\)% of the total energy at the horizontal point, i.e.,

\(\frac{1}{2}\) \(mv^2\) = \(\frac{95}{100}\times\,mgl\)

\(\therefore\)  \(v\) = \(\sqrt{\frac{2\times 95\times 1.5\times 9.8}{100}}\)= 

\(5.28\;m/s\)