Q: A ( 1kg ) ball of clay moving with a velocity ( vbi ) collides and sticks to the end of a ( 120cm ) rod of uniform mass ( 2kg ). Assuming that the ball and rod are at rest upon a frictionless surface:
( a ) Where is the new center of mass located after the clay sticks to the rod?
( b ) What linear velocity is acquired by the rod?
( c ) What is the rotational inertia about the new center of mass ( c.o.m. )?
( d ) What is the angular velocity about the new ( c.o.m. )?
A: The center of mass is a hypothetical location where the mass of a system ( mr ) is concentrated with respect to a chosen point of reference. If we use the uppermost end of the rod as a point of reference, the before-collision ( c.o.m. ) is as follows:
c.o.m. = ∑mrx/M
The distance ( x ) from the rod’s end to the ( c.o.m. ) is ( 0.06m ). Since the ball has yet to bind to the rod, the rod’s mass ( mr ) comprises the total mass ( Mt ) of the system:
mr = Mt
c.o.m. = ( m/Mt )( x )
c.o.m. = ( 2 kg/ 2 kg )( 0.6m )
c.o.m. = 0.6m
When the ball of clay sticks to the rod, it becomes part of the system. Since new mass is added to the end of the rod, the system’s ( c.o.m. ) will shift upward along the rod. The new center of mass position is determined using the mass ( mb ) of the ball and the mass of the rod ( mr ):
M = mb + mr
c.o.m. = [ ( mb )( x1 ) + ( mr )( xr com ) ]/( mb + mr )
c.o.m. = [ ( 1kg )( 0 ) + ( 2kg )( 0.6m ) ]/( 3kg )
c.o.m. = [ ( 2kg )( 0.6m ) ]/( 3kg )
c.o.m. = ( 1.2 kg*m )/( 3kg )
c.o.m. = 0.2m
The system will revolve with an angular velocity ( ⍵ ) around the new center-of-mass location after the collision occurs. Additionally, the center of mass acquires a linear velocity ( vcm ) that propels the system forward:
As is always the case, the linear momentum of the system is preserved. Initially, the ball of clay was the only moving part of the system, so ( pi = pb ). The final momentum of the system ( pf ) is a product of the linear velocity acquired by the center of mass ( vcm ) and the mass ( M ) of the system. Since ( mb ), ( vbi ), and ( M ) are known values, we are now able to solve for ( vcm ):
pi = pf
mbvbi = ( M )( vcm )
vcm = ( mb/M )( vbi )
vcm = ( 1kg/3kg )( vbi )
vcm = ⅓ vbi
A system’s moment of inertia ( I = mr2 ) is a measure of its ability to resist a change in angular motion. We must use appropriate values of ( r ) to obtain correct values. Since the ball essentially revolves around the ( c.o.m. ), the moment of inertia of the ball is calculated as follows:
Itotal = Irod + Iball
Iball = mballr2
Iball = ( 1kg )( 0.4m )2
Iball = ( 1kg )( 0.16m2 )
Iball = 0.16 kg*m2
Itotal = Irod + 0.16 kg*m2
The rod’s ( c.o.m. ) shifts upward ( 0.2m ) during the collision. As a consequence, we must use the parallel axis theorem to determine the new ( Irod ) value. The standalone value for the rod’s moment of inertia will increase in proportion to how far the center of mass shifted from its initial position:
Irod = Irod standalone + mrodr2
Irod standalone = 1/12 mr2
Irod standalone = ( 1/12 )( 2kg )( 1.2m )2
Irod standalone = ( 1/12 )( 2kg )( 1.4m2 )
Irod standalone = 0.24 kg*m2
And likewise:
mrodr2 = ( 2kg )( 0.2m )2
mrodr2 = ( 2kg )( 0.04m2 )
mrodr2 = 0.08 kg*m2
Solving for ( Itotal ), we obtain the following result:
Irod = 0.24 kg*m2 + 0.08 kg*m2
Irod = 0.32 kg*m2
Itotal = 0.32 kg*m2 + 0.16 kg*m2
Itotal = 0.48 kg*m2
We have now arrived at the most conceptually challenging part of solving the problem. The ball has angular momentum ( Lb ) before colliding with the rod. The angular momentum of the ball is defined by the momentum it will deliver to the system relative to the newly formed ( c.o.m. ) that is ( 0.4m ) away:
L = mvr
Lb = ( mb )( vbi )( r )
Lb = ( 1kg )( vbi )( 0.4m )
Like its linear counterpart, angular momentum is also conserved; however, the final value of angular momentum must be written in terms of the moment of inertia of the system:
Li = Lf
Li = Lb
Lf = ( m )( v )( r )
v = r⍵
Lf = ( m )( r⍵ )( r )
Lf = ( mr2 )( ⍵ )
Lf = I⍵
We are now able to determine the angular velocity ( ⍵ ) of the system after the collision occurs:
( 1kg )( vbi )( 0.4m ) = I⍵
( 1kg )( vbi )( 0.4m ) = ( 0.48 kg*m2 )( ⍵ )
⍵ = [ ( 0.4 kg*m )/( 0.48 kg*m2 ) ]( vbi )
⍵ = ( 0.08m-1 )( vbi )
⍵ ≅ 5/6 vbi
RENAISSANCE PHYSICS 