PHYSICS - physical pendulum - Need your help

[nyyfan0729]

A physical pendulum consists of a uniform rod of length L = 1.5 meters. The mass of the rod is 5kg and the rod is pivoted about its end. The rod is pulled out to an angle of θ0=0.2rad and is then given a push so that its instantaneous angular velocity is -0.5 rad/sec.
a) What is the period of the oscillator?
b) Determine the function for θ(t), the angular displacement from vertical as a function of time.
c) What is the angular kinetic energy of the rod after 1.0 sec?
d) What is the maximum torque on the rod during the course of its oscillation?

 

[vaishakh]

Will you show your works so that it might be easy to help. The basic thing to start with is initial kinetic energy and then use energy conservatin to relate it to find other things. Specific helps will be given only after you show that you tried it.

 

[kyle8921]

a) The period of the oscillator can be calculated using the formula T = 2π√(I/mgd), where I is the moment of inertia of the rod, m is the mass, g is the acceleration due to gravity, and d is the distance from the pivot point to the center of mass. In this case, the moment of inertia can be calculated as I = (1/3)mL^2 = (1/3)(5kg)(1.5m)^2 = 3.75 kgm^2. Plugging in the values, we get T = 2π√(3.75kgm^2/5kg(9.8m/s^2)(1.5m)) = 1.35 seconds.

b) To determine the function for θ(t), we can use the equation θ(t) = θ0cos(ωt) + (v0/ω)sin(ωt), where θ0 is the initial angle, ω is the angular frequency (given by ω = 2π/T), and v0 is the initial angular velocity. In this case, θ0 = 0.2 rad, ω = 2π/1.35 = 4.67 rad/s, and v0 = -0.5 rad/s. Plugging in the values, we get θ(t) = 0.2cos(4.67t) - (0.5/4.67)sin(4.67t).

c) The angular kinetic energy of the rod can be calculated using the formula KE = (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity. After 1 second, the angular velocity will be ω = ω0 + αt = -0.5 + (0/1) = -0.5 rad/s. Plugging in the values, we get KE = (1/2)(3.75kgm^2)(-0.5rad/s)^2 = 0.47 J.

d) The maximum torque on the rod will occur when the rod is at its maximum displacement from the vertical, which is at θ = θ0. The maximum torque can be calculated using the formula τmax = mgdθ0sinθ0, where m is the mass, g is the acceleration due