contestada

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'?

Respuesta :

Answer:

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

2) [tex]V = \frac{mv_{c}}{m + M}[/tex]

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

Explanation:

1)  mass of stick  = M

mass of clay = m

Location of original center of mass of stick, [tex]X_{s}[/tex] =  0 (at the origin)

Location of center of mass of clay, [tex]X_{c}[/tex] = D

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

The equation below applies for center of mass

X'(M+m) = [tex]MX_{s} + mX_{c}[/tex]

But [tex]X_{s} = 0[/tex] and [tex]X_{c} = D[/tex]

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

X' = mD/(M + m)

2)  Let [tex]v_{c}[/tex] = speed of the clay ball before collision

[tex]v_{s} =[/tex] 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

[tex]mv_{c} + Mv_{s} = (m+M)V[/tex]

[tex]mv_{c} + M(0)= (m+M)V\\mv_{c} = (m+M)V\\V = \frac{mv_{c}}{m + M}[/tex]

3)

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

Using conservation of angular momentum

[tex]mv_{c} X_{c} -mv_{c}X' = Iw\\mv_{c} D -mv_{c}X' = Iw[/tex]

But X' = mD/(M + m)

[tex]mv_{c} D -mv_{c}\frac{mD}{m+M} = Iw[/tex]

[tex]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[/tex]

Ver imagen kollybaba55
onyebuchinnaji

(a) The new center mass relative to the center mass of the stick is [tex]\frac{m_1 D}{m_1 + m_2}[/tex].

(b) The final speed of the of the new center mass is [tex]\frac{m_1 v_0}{m_1 + m_2}[/tex].

(c) The final angular momentum of the system is [tex]\frac{m_1D^2 \omega_1}{X_c_m^2(m_1 + m_2)}[/tex].

The given parameters;

  • mass of the clay = m₁
  • initial velocity of the clay, u₁ = v₀
  • let the mass of the stick = m₂

The new center mass relative to the center mass of the stick is calculated as;

[tex]X_{cm} = \frac{m_1 X_1 \ + \ m_2X_2}{m_1 + m_2} \\\\X_{cm} = \frac{m_1(D) \ + \ m_2(0)}{m_1 + m_2} \\\\X_{cm} = \frac{m_1 D}{m_1 + m_2}[/tex]

The final speed of the of the new center mass is calculated as;

[tex]V_{cm} = \frac{m_1(v_0)}{m_1 + m_2} \\\\V_{cm} = \frac{m_1 v_0}{m_1 + m_2}[/tex]

The final angular momentum of the system is calculated as;

[tex]\ I_i \omega_i = I_f \omega _f\\\\m_1D^2\omega_1 + 0 = \omega _f[X_c_m^2(m_1 + m_2)]\\\\\omega_f = \frac{m_1D^2 \omega_1}{X_c_m^2(m_1 + m_2)}[/tex]

Learn more here:https://brainly.com/question/19320213

Otras preguntas