(2.1) A pendulum consists of a rod AB of length 2a and mass M, rotating about a

Question

(2.1) A pendulum consists of a rod AB of length 2a and mass M, rotating about a horizontal axis through its one end B. Let 6 denote the counterclockwise angle that the rod forms with the vertical line drawn down from point B. (17) (a) Find the equation of rotation of the pendulum (b) What is the total mechanical energy of the pendulum, if the rod is initially held horizontal and then released? Assume that the zero energy level of the gravita- tional energy goes through point B. (c) Give all the initial conditions (specify the initial angle of rotation and angular velocity) which will ensure that the kinetic energy of the pendulum will always stay the same.

Explanation / Answer

2.1 given pendulum

rod AB

length l = 2a

mass = M

rotating about horizontal axis about one of its ends

I = M(2a)^2/3 = 4Ma^2/3

a. for theta tilt

from energy balance

0.5Iw^2 + mga(1 - cos(theta)) = Eo

hence

differentiating

Iw*w' + mga(0 + sin(theta)*w) = 0

Iw' = -mga*sin(theta)

w' = alpha

for small angle

sin(theta) = theta

theta' = w

hence

I*alpha = -mga*theta

this is energy balance for gerrneal pendulum

the time perios is

T = 2*pi/w

w = sqrt(-theta/alpha) = sqrt(I/mga)

w = sqrt(4a/3g)

b. total mechanical energy = Eo = 0.5Iw^2 + mga(1 - cos(theta))

c. let intieial angle at t= 0 be thteao

initial angular velocoty = wo

then

Eo = 0.5Iwo^2 + mga(1 - cos(tehtao))

but the KE will keep on changing

hence if we keep intiial Eo = 0, thenintiial KE = 0
hence for initial angular speed wo = 0 rad/s and initial angular displacement of 0 rad, the KE remainss the same

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