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Method 1: Using force analysis [Vo Thanh Minh Tue]
The force exerted by the pivot on the rod is composed of two components Nx amd Ny Moment of inertia about O: I = ml2 + m(3l)2 = 10ml2 Net torque at the initial condition:
Iθ&& = mgl + 3mg (0) = mgl
θ&& =
g 10l
The distance from the center of mass to the origin:
R=
1 2 10 l + (3l ) 2 = l 2 2
Now let consider the motion of the center of mass. The motion of the center of mass can be divided into two components in tangential and radical directions Hence:
r ˆ R && − θ& 2 R aCM = θˆ Rθ&& + 2θ&R& + R
[
] [
]
Since the R is a constant and object starts from rest θ& = 0
r aCM = θˆ Rθ&& = θˆ
[ ]
g 2 10
g 3 3g = 2 10 10 20 g 1 g sin φ = = 2 10 10 20
a x = aCM cos φ = a y = aCM
The net-force on the center of mass:
3g 3 F x = 2ma x = N x = 2m = mg (to the right) 20 10 g mg Fy = 2ma y = 2mg − N y = 2m = 20 10
1 19 N y = mg 2 − = mg 10 10 (in the positive y direction) Hence total force exerted by the pivot on the rod
r 3 19 N = mgˆi + mgˆj 10 10 mg 2 370 N= 3 + 19 2 = mg 10 10 Therefore total force exerted by the rod on the pivot
r − 3 ˆ − 19 ˆ Nr = mgi + mgj 10 10 Method 2: Using energy approach [Hoang Kim Dinh]: Let the coordinate of the first ball be (x1, y1) and the coordinate of the second ball be (x2, y2) and the origin at the pivot. Total kinetic energy:
KE =
m 2 m m m m 2 x&1 + y&12 + x& 22 + y& 22 = l 2θ& 2 + (3l ) θ& 2 = 10l 2θ& 2 = 5ml 2θ& 2 2 2 2 2 2
[
]
[
]
Total potential energy: Suppose that the system is displaced by an angle θ
U = mg (3l − l sin θ ) + mg (3l − 3l cos θ ) Since total energy is conserved:
E = KE + U = constant Differentiating with respect to time:
d 5ml 2θ& 2 + mg (3l − l sin θ ) + mg (3l − 3l cos θ ) = 0 dt d 5ml 2θ& 2 + mg (3l − l sin θ ) + mg (3l − 3l cos θ ) = 0 dt 10lθ&& + g (3 sin θ − cos θ ) θ& = 0 10lθ&& + g (3 sin θ − cos θ ) = 0
(
)
(
)
[
]
Initial condition: θ=0
θ&& =
g 10l
Since the system’s center of mass is rotating around the pivot with fixed radius R:
[ ]
a = Rθ&& =
g 2 10
Similarly solving for the force by the system on the pivot:
r − 3 ˆ − 19 ˆ Nr = mgi + mgj 10 10 [All things come to an end. See you next year!]
The force exerted by the pivot on the rod is composed of two components Nx amd Ny Moment of inertia about O: I = ml2 + m(3l)2 = 10ml2 Net torque at the initial condition:
Iθ&& = mgl + 3mg (0) = mgl
θ&& =
g 10l
The distance from the center of mass to the origin:
R=
1 2 10 l + (3l ) 2 = l 2 2
Now let consider the motion of the center of mass. The motion of the center of mass can be divided into two components in tangential and radical directions Hence:
r ˆ R && − θ& 2 R aCM = θˆ Rθ&& + 2θ&R& + R
[
] [
]
Since the R is a constant and object starts from rest θ& = 0
r aCM = θˆ Rθ&& = θˆ
[ ]
g 2 10
g 3 3g = 2 10 10 20 g 1 g sin φ = = 2 10 10 20
a x = aCM cos φ = a y = aCM
The net-force on the center of mass:
3g 3 F x = 2ma x = N x = 2m = mg (to the right) 20 10 g mg Fy = 2ma y = 2mg − N y = 2m = 20 10
1 19 N y = mg 2 − = mg 10 10 (in the positive y direction) Hence total force exerted by the pivot on the rod
r 3 19 N = mgˆi + mgˆj 10 10 mg 2 370 N= 3 + 19 2 = mg 10 10 Therefore total force exerted by the rod on the pivot
r − 3 ˆ − 19 ˆ Nr = mgi + mgj 10 10 Method 2: Using energy approach [Hoang Kim Dinh]: Let the coordinate of the first ball be (x1, y1) and the coordinate of the second ball be (x2, y2) and the origin at the pivot. Total kinetic energy:
KE =
m 2 m m m m 2 x&1 + y&12 + x& 22 + y& 22 = l 2θ& 2 + (3l ) θ& 2 = 10l 2θ& 2 = 5ml 2θ& 2 2 2 2 2 2
[
]
[
]
Total potential energy: Suppose that the system is displaced by an angle θ
U = mg (3l − l sin θ ) + mg (3l − 3l cos θ ) Since total energy is conserved:
E = KE + U = constant Differentiating with respect to time:
d 5ml 2θ& 2 + mg (3l − l sin θ ) + mg (3l − 3l cos θ ) = 0 dt d 5ml 2θ& 2 + mg (3l − l sin θ ) + mg (3l − 3l cos θ ) = 0 dt 10lθ&& + g (3 sin θ − cos θ ) θ& = 0 10lθ&& + g (3 sin θ − cos θ ) = 0
(
)
(
)
[
]
Initial condition: θ=0
θ&& =
g 10l
Since the system’s center of mass is rotating around the pivot with fixed radius R:
[ ]
a = Rθ&& =
g 2 10
Similarly solving for the force by the system on the pivot:
r − 3 ˆ − 19 ˆ Nr = mgi + mgj 10 10 [All things come to an end. See you next year!]
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