Question

If a simple pendulum of length $L$ has maximum angular displacement $\alpha$ then the maximum kinetic energy of bob of mass $M$ is

• A. $\frac{1}{2}\frac{ML}{g}$
• B. $\frac{Mg}{2L}$
• C. $MgL(1-\cos \alpha )$
• D. $\frac{MgL\sin \alpha }{2}$

Answer 1

The correct option is C $MgL(1-\cos \alpha )$

Here, mass of the bob =M=M
Length of the pendulum =L=L
From figure
h=L−Lcosα=L(1−cosα)h=L−Lcos⁡α=L(1−cos⁡α)
Potential energy at A=Mgh=MgL(1−cosα)A=Mgh=MgL(1−cos⁡α)
Potential energy at B=0B=0
Kinetic energy at A=12mv2=0(∵Asv=0)A=12mv2=0(∵Asv=0)
Mechanical energy at A=0+MgL(1−cosα)A=0+MgL(1−cos⁡α)
Let KBKB is the kinetic energy at B=KB+0B=KB+0
According to conservation of mechanical energy AA and BB, we get
0+MgL(1−cosα)=KB+00+MgL(1−cos⁡α)=KB+0
or KB=MgL(1−cosα)KB=MgL(1−cos⁡α)
Kinetic energy is maximum at point BB
Therefore maximum kinetic energy of bob mass MM is MgL(1−cosα)