A physical pendulum consists of a disc of radius R and mass m fixed at the end of a rod of mass m and length l .
(a) Find the period of the pendulum for small angles of oscillation. Express your answer in terms of m, R, l and acceleration due to gravity g as needed (enter m for m, R for R, l for l, g for g and pi for π).
Tfixed=
(b) For small angles of oscillation, what is the new period of oscillation if the disk is mounted to the rod by a frictionless bearing so that it is perfectly free to spin? Express your answer in terms of m, R, l and acceleration due to gravity g as needed (enter m for m, R for R, l for l, g for g and pi for π).
Tfree=
hey have u got this@Anonymous
try googling MIT OCW with the problem text .
q7, the answer is
a)2*pi*sqrt(((1/3*m+m)*l^2+1/2*m*R^2)/((1/2*m+m)*g*l))
b)2*pi*sqrt(((1/3*m+m)*l^2)/((1/2*m+m)*g*l)) 4a: (m_2*g)/(m_2+I_c/R^2)
b: (I_c*m_2*g)/(m_2*R^2+I_c)
c: sqrt((2*h*m_2*g*R^2)/(m_2*R^2+I_c))
To find the period of the physical pendulum for small angles of oscillation, we need to consider the rotational motion of both the disc and the rod.
(a)
The moment of inertia for the disc about its axis of rotation is given by:
I_disc = (1/2) * m * R^2
The moment of inertia for a uniform rod rotating about its end is given by:
I_rod = (1/3) * m * l^2
To find the equivalent moment of inertia for the combined system (disc and rod), we use the parallel axis theorem. The axis of rotation for the combined system is at the fixed end of the rod, so the distance between the center of mass of the rod and the axis is l/2:
I_combined = I_disc + m * (l/2)^2 + I_rod
Next, we can use the equation for the period of a physical pendulum in terms of moment of inertia and gravitational acceleration:
T_fixed = 2 * pi * sqrt(I_combined / (m * g))
Substituting the values for I_combined, m, and g, we get:
T_fixed = 2 * pi * sqrt((1/2) * m * R^2 + m * (l/2)^2 + (1/3) * m * l^2) / (m * g)
Simplifying, we have:
T_fixed = 2 * pi * sqrt((1/2) * R^2 + (1/4) * l^2 + (1/3) * l^2) / sqrt(g)
(b)
If the disk is mounted to the rod by a frictionless bearing, it is free to spin independently of the rod's motion. In this case, the rotational motion of the disk is decoupled from the oscillations of the rod. The period of oscillation for the disk can be found using the moment of inertia for the disk alone.
The moment of inertia for a single disk rotating about its axis through its center is given by:
I_disk_only = (1/2) * m * R^2
Using the equation for the period of a physical pendulum as before, we have:
T_free = 2 * pi * sqrt(I_disk_only / (m * g))
Substituting the value for I_disk_only and simplifying, we get:
T_free = 2 * pi * sqrt((1/2) * R^2) / sqrt(g)
Simplifying further, we have:
T_free = pi * R / sqrt(g)
So, the period of oscillation for the physical pendulum when the disk is perfectly free to spin is pi times R divided by the square root of g.