Question

The length of a simple pendulum is 0.81 mand the mass of the particle (the "bob") at the end of the cable is0.23 kg. The pendulum is pulled away fromits equilibrium position by an angle of 8.10° and released from rest. Assume that frictioncan be neglected and that the resulting oscillatory motion issimple harmonic motion.
(a) What is the angular frequency of themotion?
1(No Response) rad/s

(b) Using the position of the bob at its lowest point as thereference level, determine the total mechanical energy of thependulum as it swings back and forth.
2(No Response) J

(c) What is the bob's speed as it passes through the lowest pointof the swing?
3(No Response) m/s

Answer 1

Answer:

$\displaystyle w=3.478\ rad/sec$

$M=0.0182\ J$

$v=0.398\ m/s$

Explanation:

Simple Pendulum

It's a simple device constructed with a mass (bob) tied to the end of an inextensible rope of length L and let swing back and forth at small angles. The movement is referred to as Simple Harmonic Motion (SHM).

(a) The angular frequency of the motion is computed as

$\displaystyle w=\sqrt{\frac{g}{L}}$

We have the length of the pendulum is L=0.81 meters, then we have

$\displaystyle w=\sqrt{\frac{9.8}{0.81}}$

$\displaystyle w=3.478\ rad/sec$

(b) The total mechanical energy is computed as the sum of the kinetic energy K and the potential energy U. At its highest point, the kinetic energy is zero, so the mechanical energy is pure potential energy, which is computed as

$U=mgh$

where h is measured to the reference level (the lowest point). Please check the figure below, to see the desired height is denoted as Y. We know that

$H+Y=L$

And

$H=L\ cos\alpha$

Solving for Y

$Y=L(1-cos\alpha )$

$Since\ \alpha=8.1^o, L=0.81\ m$

$Y=0.0081\ m$

The potential energy is

$U=mgh=0.23\ kg(9.8\ m/s^2)(0.0081\ m)$

$U=0.0182\ J$

The mechanical energy is, then

$M=K+U=0+U=U$

$M=0.0182\ J$

(c) The maximum speed is achieved when it passes through the lowest point (the reference for h=0), so the mechanical energy becomes all kinetic energy (K). We know

$\displaystyle K=\frac{mv^2}{2}$

Equating to the mechanical energy of the system (M)

$\displaystyle \frac{mv^2}{2}=0.0182$

Solving for v

$\displaystyle v=\sqrt{\frac{(2)(0.0182)}{0.23}}$

$v=0.398\ m/s$