A pendulum consists of a solid sphere of radius r = 14.0 cm tied to a light string so that the center of mass of the sphere is a distance L = 21.8 cm from the pivot point.

(a) What is the period (in s) of this physical pendulum?
(b) What error (in s) is made in calculating the period by treating the sphere as a point mass?

a. 0.9370 s =Ta

b. 0.8510 s = Tb

c . Error = 0.086 s = Ta -Tb

Explanation:

a) This is a simple harmonic motion problem = T =2π√l/g

T = Period in s

I = length or distance l = 21.8 cm = 21.8/100 = 0.218  m note 100 cm = 1m

g = acceleration due to gravity = 9.81 m/s^ 2

π = pie = 22/7

substitute

T = 2* 22/7 *√0.218/9.81 = 0.937018962 s

                                        approximate = 0.9370 s = Ta

b) As a point mass implies the entire object as it is concentrated in space neglecting its dimension in space

for  a pendulum 2πf = [(MgL)/l]

l = moment of inertia of a solid sphere = 2/5MR^2

M = mass

R = radius  = 14.0 cm =  14.0 / 100 = 0.14 m

f = frequency = 1/T

g = acceleration due to gravity = 9.81m/s^2

L = I = length or distance l = 21.8 cm = 21.8/100 = 0.218  m note 100 cm = 1m

Thus 2π1/T =  [(MgL)/l ] ^1/2

make subject formula and substitute l = moment of inertia of a solid sphere = 2/5MR^2 in the equation

square both sides

4*π^2/ T^2 =  [(MgL)/l ]

T^2 = 4*π^2 *l / [(MgL]

      = 4*π^2* 2/5MR^2 /[(MgL]

      = 8* π^2*R^2/gL

     = 8* 22/7^2 *0.14^2/ 9.81* 0.218

    = 0.7242188742

T   =√0.7242188742 = 0.8510105018 s

       approximate = 0.8510 s  = Tb

Error = Ta - Tb =  0.9370 s - 0.8510 s = 0.086 s

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr