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 and released, determine
The energy of the system in terms of theta and theta . The equation of motion in terms of theta and theta . The equation of motion in terms of theta and theta Lagrange\'s equations. The natural frequency of vibration of the system.
Solution
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 *L 2 )*(DTh 2 ) + 1/2*(.5*m d *r 2 )*(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 *L 2 )*DDTh + (.5*m d *r 2 )*( 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
.
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A uniform disk of radius r and mass md rolls without slipping on a cyl.docx
1. 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 and released, determine
The energy of the system in terms of theta and theta . The equation of motion in terms of theta
and theta . The equation of motion in terms of theta and theta Lagrange's equations. The natural
frequency of vibration of the system.
Solution
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 *L 2
)*(DTh 2
) + 1/2*(.5*m d *r 2
)*(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 *L 2
)*DDTh + (.5*m d *r 2
)*( 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
![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 and released, determine
The energy of the system in terms of theta and theta . The equation of motion in terms of theta
and theta . The equation of motion in terms of theta and theta Lagrange's equations. The natural
frequency of vibration of the system.
Solution
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 *L 2
)*(DTh 2
) + 1/2*(.5*m d *r 2
)*(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 *L 2
)*DDTh + (.5*m d *r 2
)*( 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](https://image.slidesharecdn.com/auniformdiskofradiusrandmassmdrollswithoutslippingonacyl-230203012626-9dcbd24f/85/A-uniform-disk-of-radius-r-and-mass-md-rolls-without-slipping-on-a-cyl-docx-1-320.jpg)
