Question

Multiple-Concept Example 11 provides some pertinent back- ground for this problem. A pendulum is constructed from a thin, rigid, and uniform rod with a small sphere attached to the end opposite the pivot. This arrangement is a good approximation to a simple pendulum (period ???? 0.66 s), because the mass of the sphere (lead) is much greater than the mass of the rod (aluminum). When the sphere is removed, the pendulum is no longer a simple pendulum, but is then a physical pendulum. What is the period of the physical pendulum?

Answer 1

Answer:

$T = 0.54 s$

Explanation:

When physical pendulum is a system of rod and sphere then we will have

$\tau = I\alpha$

so we will have

$(MgL + \frac{mgL}{2})\theta = (\frac{mL^2}{3} + ML^2)\alpha$

so we will have

$\alpha = \frac{(M + \frac{m}{2})g}{(\frac{m}{3} + M)L}\theta$

since mass of ball is much higher than the mass of rod

so we have

$T = 0.66 s = 2\pi\sqrt{\frac{L}{g}}$

now when we use the rod as physical pendulum then we have

M = 0

$\alpha = \frac{g/2}{L/3}\theta$

so we have

$T = 2\pi\sqrt{\frac{2L}{3g}}$

so its time period is given as

$T = (\sqrt{\frac{2}{3}})0.66$

$T = 0.54 s$