A golden sphere (density of gold is 19.3 ×103 kg/m3) has aradius of 12 cm. The s

Question

A golden sphere (density of gold is 19.3 ×103 kg/m3) has aradius of 12 cm. The sphere is attached to a stiff but nearlymassles rod so that the center of the sphere is 1 m away from theother end of the rod. This other end

1is suspended on a nail. The whole setup acts as a simple pendulum.a) What is the period of oscillations? b) What is the speed of thesphere 10 seconds after its release from rest 10o awayfrom the vertical direction? c) The setup is fully submerged inwater. Assume that the drag force is negligible. What is the periodof oscillations now?

Explanation / Answer

a) What is the period ofoscillations? This is a physical pendulum and its period T is T = 2( I/ (mg LCM) ) Using parallel axis theorem I = mLCM2 + (2/5)mr2   where
m - mass of the pendulum (ball)
Lcm distance from pivot to ceneter of mass (CM)
r - radius of the sphere
I=  m( LCM2 + (2/5) r2   )   = (4/3)r3 (  LCM2 + (2/5) r2   )
m= 19.3 ×103 (4/3) (0.12)3
m= 140 kg
I= 140 ((1.00 + 0.12)2 + (2/5) (0.12)2)
I= 176.0 kg m2
Now since
T = 2( I/ (mg LCM) )
T=2(176 /( 140 x 9.8x (1.12))
T=2 ( 0.339)
T=2.13s
b) What is the speed of the sphere 10 seconds after itsrelease from rest 10o away from the verticaldirection?
Since we can express the oscilations as
(t) = o cos( t )
given (0) =10 deg = (1/18)radians
o = (0) that gives us a maximum or the aplitude =2f = 2(1/T) =2(1/2 ( 0.339)) = 2.95 rad/s wehave to compute the position of the pendulum 10 s afterrelease
(t)= o cos( t ) (Note: angle in radians) (10)= (1/18) cos( 2.95 x10)=- 0.0188  rad =- 3.38 deg We have tocompute kinetic energy Ke Ke=Pemax - Pe Pemax= mghmax= mgLCM(1-cos(10)) = Pemax=140 x 9.8 x 1.12 (1 -cos(10) = 1537 ((1 - cos(10)= 23.6 J (Note: angles in degrees) Pe= mgh=mgL CM(1-cos(10- 3.38 )) = 1537(1-cos(10- 3.38 ))= 10.2 J BRB
  

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