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=
3 answers
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))
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))
1 answer