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Center of Mass Linear Momentum
Chapter 9
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Center of Mass the point in which the mass of an object can be considered to be concentrated the objects motion acts as if it’s mass is concentrated at that point the translational acceleration calculated by Newton’s second law Fnet = ma determines the motion of the object at it’s center of mass all gravitational forces exerted on the object is applied at that point
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Calculating Center of Mass
The center of mass of an object can be found in all dimensions. (m1x1 + m2x mnxn) → ^ ^ ^ ______________________ vectorcom = rcom = xcomi + ycomj + zcomk xcom = (m1 + m mn) → → ___ 1 n Σ miri (m1y1 + m2y mnyn) rcom = ______________________ M ycom = i = 1 (m1 + m mn) (m1z1 + m2z mnzn) ______________________ zcom = (m1 + m mn)
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Newton’s Second Law for a System of Particles
→ → Fnet = Macom → Fnet : net force of all external forces (internal forces not included) M = total mass of the system (closed system) → acom = acceration of the center of mass of the system → → Fnet,x = Macom,x → → Fnet,y = Macom,y → → Fnet,z = Macom,z
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Conversion of Center of Mass Equation
→ rcom : gives position → → → → → dri → ___ Mrcom = m1r1 + m2r mnrn vi = dt → → → → → [Mrcom]` = Mvcom = m1v1 + m2v2 + … + mnvn → dvi → ___ ai = → → → → → dt [Mvcom]` = Macom = m1a1 + m2a2 + … + mnan → → → → → → m1a1 + m2a2 + … + mnan = F1 + F2 + … + Fn → → Fi = miai
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Linear Momentum linear momentum of a particle = total mass of the system * velocity of its center of mass → → p = mv → the momentum of an particle acts in the same direction as the velocity of that particle because mass is a positive scaler quantity → → → P = p1 + p2 + … + pn → → → = m1v1 + m2v2 + … + mnvn → → P = Mvcom time rate of change of the momentum of a particle = net force acting on that particle and in the same direction → → dP ___ Fnet = dt → → → dp d → dvcom ___ ___ _____ Fnet = = (Mvcom) = M → = Macom dt dt dt therefore net external force changes the momentum of an object, thus momentum can only change due to an external force
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Impulse ∫ ∫ ∫ → → → → impulse is defined as: dp = F(t) dt → → →
tf tf → ∫ → ∫ impulse is defined as: dp = F(t) dt ti ti → → → pf – pi = ∆p tf → → ∫ F(t)dt J = ti → → thus ∆p = J linear momentum – impulse theorem → → J = Favg∆t → → series of collisions : J = -n∆p n = # of projectiles → → → → → __ J __ -n -n __ ___ ∆m ∆p m∆v – ∆v Favg = = = = ∆t ∆t ∆t ∆t
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Conservation of Linear Momentum
→ no external forces : Fnet = 0 momentum stays constant in a closed system → → Pi = Pf law of conservation of linear momentum if a component of net external force in a closed system = 0 along an axis, then the component of the linear momentum of the system along that system can not change
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Collisions elastic collision: momentum AND kinetic energy conserved →
P1,i + P2,i = P1,f + P2,f conservation of momentum → → → → m1v1,i + m2v2,i = m1v1,f + m2v2,f K1,i + K2,i = K1,f + K2,f conservation of kinetic energy → → → → 1/2m1v1,i2 + 1/2m2v2,12 = 1/2m1v1,f2 + 1/2m2v2,f2 inelastic collision: momentum is conserved, kinetic energy IS NOT conserved → → → → P1,i + P2,i = P1,f + P2,f conservation of momentum → → → → m1v1,i + m2v2,i = m1v1,f + m2v2,f completely inelastic: bodies stick together
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Systems With Varying Mass: A Rocket
→ → velocity of rocket relative to frame velocity of rocket relative to products velocity of products relative to frame Pi = Pf + = → → → m v vR-F = vR-P + vP-F → vrel M + dM -dM v + dv U → → → → → → → → Mv = -dMU + (M +dM)(v + dv) (v + dv) = vrel + U → → → → → → P of gas P of rocket afterwards U = v + dv - vrel → → -dMvrel = Mdv → → → dM ___ → dv __ dv = – vrel ____ -dM Vrel = M dt M dt vf ∫ Mf → → ∫ dM ___ → → dv = – vrel Rvrel = Ma first rocket equation M vi Mi | | → → → Mi __ vf - vi = vrelln second rocket equation Mf
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What is Newton’s Second Law for a system of particles?
→ → Fnet = Macom
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How does Mrcom become Mvcom?
→ → → drcom → ____ Mvcom = M dt
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Why is the direction of a particle’s momentum the same as the direction of its velocity?
mass is a scaler quantity in the equation: → → p = mv
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If there is no external force (Fnet = 0), P is constant in a _________ system.
closed
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The equation for the linear momentum of a system is?
→ → p = mv
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Impulse is defined as? → tf → ∫ J = F(t)dt ti
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What is the impulse – momentum theorem?
→ → → → J = ∆P = Pf - Pi
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If there is no net external force momentum ( does, does not ) change.
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In an elastic collision both _______________ and _______________ are conserved.
kinetic energy momentum
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In an inelastic collision _____________ is conserved and _____________ is not conserved.
momentum kinetic energy
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In a completely inelastic collision what occurs?
the bodies stick together
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What is the Law of Conservation of Momentum?
Pi = Pf
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vR-P vP-F vR-F = _______ + ________ where P = products R = rocket F = frame of reference
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Vrel = ________ vR-G
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The vcom ( is, isn’t ) changed by a collision.
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In the absence of external forces a rocket accelerates at an instonaneous rate given by: (first rocket equation) Rvrel = Ma
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The point in which the mass of an object could be concentrated is?
center of mass
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What is the equation for the center of mass of an object on the x axis?
(m1x1 + m2x mnxn) ____________________ xcom = (m1 + m mn)
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The thrust of an object is given by:
Rvrel
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All gravitation forces are exterted on what point?
center of mass
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The path in which a baseball bat’s center of mass is thrown into the air is in the shape of a ______________. parabola
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Find the center of mass of these 3 objects:
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Find the impulse from 1 second to 5 seconds if Fnet(t) = t.
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In a completely elastic collision:
v m 2m Find the final velocity of the second particle. → → V2 = 2v
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In a completely inelastic collision:
v m 2m Find the final velocity. vf = 2/3v
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If a boat is moving at 10 mph against a river current of 2 mph, how fast is the boat going past the shoreline? 6 mph
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If a 2000 kg (including fuel) rocket burns 10 kg of fuel per second going at m/s past earth ejecting fuel at 500 m/s past earth. What is the rocket’s acceleration? 4.75 m/s 2
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A 2 kg mass hits a 4 kg mass at 5 m/s in an elastic collision
A 2 kg mass hits a 4 kg mass at 5 m/s in an elastic collision. Find the final velocity of the 2 kg mass. – 1.67 m/s
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In a completely in elastic collision:
2v v 2m 4m What is the final velocity? 0 m/s
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Find the center of mass:
2 2m 2 -2 4m 3m - 2 (0.1,- 0.8)
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