Question

A uniform disk of radius r and mass md rolls without slipping on a cylindrical surface and is attached to a uniform slender bar AB of mass mb. The bar is attached to a spring of constant K and can rotate freely in the vertical plane about point A as shown in the figure . If the bar AB is displaced by small angle 0 and released, determine The energy of the system in terms of theta and theta '. The equation of motion in terms of theta and theta '' using Lagrange's equations. The natural frequency of vibration of the system.

Answer 1

Let theta=Th and D(theta)/Dt= DTh and D(D(theta)/Dt)/Dt =DDTh

a) The total energy = energy of spring + rotational energy of slender rod and disc + kinetic energy of disc

= 1/2*k*(L/2*Th)² + 1/2*(1/12*m_b*L²)*(DTh²) + 1/2*(.5*m_d*r²)*(L/2r*DTh)² + 1/2*m_d*(L/2*DTh)²

As total energy remains constant, differentiating the above equation we get,

[ k*Th*(L/2)² + (1/12*m_b*L²)*DDTh + (.5*m_d*r²)*( L/2r + 1 )²*DDTh + m_d*(L/2)²*DDTh ]*DTh = 0

or k/4*Th + 1/12*m_b*DDTh + 1/2*(1/2+r/L)²*m_d*DDTh + 1/4*m_d*DDTh=0

or 3k*Th + ( m_b + 6*(1/2+r/L)²*m_d + 3*m_d )*DDTh =0 ....b)

c) Natural frequency = [ 3k/( m_b + 6*(1/2+r/L)²*m_d) + 3*m_d ]^.5

Answer 2

Answer 3