Question

4. A ball of clay, of mass m, traveling at speed vo, collides and sticks to a stationary stick. The ball approaches the stick in a direction perpendicular to the length, striking the stick at a distance D from the CM of the stick. The stick + ball system then moves, rotating about a new center of mass, CM’.
1. Where is the new CM' relative to the CM 'of the stick?
2.What is the final speed of the new CM'?
3. What is the final angular velocity of the sustem about the new CM'?

Answer 1

Answer:

Explanation:

1) The center of mass of the stick ball system is given below :

CM' = [m*D + M*0]/[m+M]

= m*D/M above CM where M is the mass of stick

2) When momentum is conserved, mv0 = [M+m]v

final velocity v = mv0/[m+M]

3) When angular momentum about new CM' is conserved,

mv0*(D-m*D/M) = I*w , where  w is final angular speed and I is moment of inertia of system about CM',

w = [mv0*(D-m*D/M)]/I

= [mv0*(D-m*D/M)]/[ ML^2/12 + M*(m*D/M)^2 + m*(D-m*D/M)^2]

Answer 2

Answer:

1) X' = mD/(M + m)

2) $V = \frac{mv_{c}}{m + M}$

3) $w = mv_{c}[\frac{MD}{I(m+M)} ]$

Explanation:

1)  mass of stick  = M

mass of clay = m

Location of original center of mass of stick, $X_{s}$ =  0 (at the origin)

Location of center of mass of clay, $X_{c}$ = D

X' = location of center of mass of the stick + ball system

The equation below applies for center of mass

X'(M+m) = $MX_{s} + mX_{c}$

But $X_{s} = 0$ and $X_{c} = D$

X'(M+m) = M (0) + m (D)

X' = mD/(M + m)

2)  Let $v_{c}$ = speed of the clay ball before collision

$v_{s} =$ speed of the stationary stick before collision = 0 m/s

V = speed of the stick-ball system after collision

Applying the principle of momentum conservation

$mv_{c} + Mv_{s} = (m+M)V$

$mv_{c} + M(0)= (m+M)V\\mv_{c} = (m+M)V\\V = \frac{mv_{c}}{m + M}$

3)

I = moment of inertia of the stick about the center of mass of the system  

Using conservation of angular momentum

$mv_{c} X_{c} -mv_{c}X' = Iw\\mv_{c} D -mv_{c}X' = Iw$

But X' = mD/(M + m)

$mv_{c} D -mv_{c}\frac{mD}{m+M} = Iw$

$mv_{c} [D -\frac{mD}{m+M} ] = Iw\\mv_{c} [\frac{mD + MD-mD}{m+M} ] = Iw\\mv_{c}[\frac{MD}{I(m+M)} ] = w$