Dynamics of Systems of Particles. Hasbun, Ch 11 Thornton & Marion, Ch 9

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1 Dynamics of Systems of Particles Hasbun, Ch 11 Thornton & Marion, Ch 9

2 Center of Mass Discrete System Center of Mass R CM 1 m i r i M i v CM = RCM 1 m i v i M a CM = v CM = RCM 1 M i m i a i i Center of Gravity R cg i m ig i r i i m ig i v cg i a cg i m i g i v i / i m i g i a i / i m i g i m i g i

3 Example The positions of 3 particles of masses m 1 = 1 kg, m 2 = 2 kg, m 3 = 3 kg are given as r 1 = (3 m + 2 m/s 2 t 2 ) i + (4 m) j, r 2 = (-2 m + 1 m s/ t) i + (2 m/s t) j, and r 3 = (1 m) i (3 m/s 2 t 2 ) j. (a) Obtain the center of mass position, velocity, and acceleration of the 3-particle system as a function of time. (b) Evaluate the expressions at t=1sec. Confirm your results using partic1.m

4 Center of Mass Continuous System Center of Mass R CM 1 M v CM 1 M a CM 1 M i i i r dm v dm a dm Center of Gravity R cg r g(r) dm/ v cg v g(r) dm/ a cg a g(r) dm/ g(r) dm g(r) dm g(r) dm

5 Example Find the center of mass for a solid hemisphere of radius a.

6 Multi-particle Systems: Linear Momentum The COM moves as if it were a single particle of mass M, acted on by the total external force, and independent of the internal forces. P = i m i ri = M R P = M R = F

7 Multiparticle Systems: Linear Momentum The linear momentum of the system is the same as if a single particle of mass M were located at the center of mass and moving with the velocity and acceleration of the center of mass (COM). In the absence of external forces, the total linear momentum of the system is constant and equal to the linear momentum of the COM.

8 Multiparticle Systems: Angular Momentum Angular momentum of a single particle: L i = m i r i r i = r i p i Total angular momentum is obtained by summing over particles: L = i m i ( R cm + r i) ( Rcm + r i) L = R cm P + i r i p i

9 Multiparticle Systems: Angular Momentum The total angular momentum about some point is the sum of the angular momentum of the COM about that point and the angular momentum of the system about the COM. L = R cm P + i r i p i

10 Multiparticle Systems: Torque If the net external torque about some axis is zero, then the total angular momentum about that axis is constant. τ = d L dt = R cm P + i τ i The total internal angular momentum must vanish if the internal forces are central (Newton s third law).

11 Multiparticle Systems: Energy The total kinetic energy of the system is equal to the sum of the KE of the COM and the KE of individual particles relative to the COM. T = 1 2 Mv2 cm The potential energy of the ith particle is the sum of the potential energy associated w/the external force and the potential energy due to internal interactions with the rest of the particles. U = i V i + i i k>i m i V ik r 2 i

12 Multiparticle Systems: Energy The total energy is constant for a conservative system, where E is given by

13 Example The positions of 3 particles of masses m 1 = 1 kg, m 2 = 2 kg, m 3 = 3 kg are given as r 1 = (3 m + 2 m/s 2 t 2 ) i + (4 m) j, r 2 = (-2 m + 1 m s/ t) i + (2 m/s t) j, and r 3 = (1 m) I (3 m/s 2 t 2 ) j. Find the system s (a) linear momentum, (b) angular momentum, (c) kinetic energy, (d) net force, and (e) net torque at time t=1s. Check your answers using the matlab program partic2.m

14 Example 11.4 A projectile is fired at an angle with an initial velocity of v 0. At the highest point, the projectile explodes into two fragments, m 1 and m 2. Each fragment continues in the original horizontal direction. If the energy lost in the explosion is E 0, by what distance are the fragments separated when they land?

15 Variable Mass Rocket Motion Newton s 2 nd Law: F = dp dt = d (mv) =mdv dt dt + v dm dt Consider a rocket: mass m under external force F u = speed of ejected gas relative to rocket v = time-dependent speed of rocket

16 Variable Mass Rocket Motion Let rocket s initial momentum be p i = mv As soon as rocket begins to burn fuel, its mass decreases. m m + dm Velocity also changes v v + dv The ejected gas is moving with speed v + u relative to the ground

17 Variable Mass Rocket Motion Some time later, the total momentum of the rocket-gas system is p =(m + dm)(v + dv) dm(v + u) p f mv + mdv udm The net force is the change in momentum over change in time: F = p f p i t = (mv + mdv udm) mv dt

18 Variable Mass Rocket Motion Rearranging gives F = p f p i t = (mv + mdv udm) mv dt m dv dt = udm dt + F This is known as the rocket equation.

19 Example 11.5 Assuming a constant gas exhaust speed, obtain the velocity as a function of time of a rocket in free space if its initial speed is v 0.

20 Rocket Take-off Simulation Rocket equation m dv dt = udm dt + F For a rocket launched from earth: m dv dt = udm dt GMm r 2 Assume that the fuel is burnt at a constant rate, = -dm/dt ˆr

21 Rocket Take-off Simulation We then get 3 time-dependent coupled differential equations: dv x dt = αθ tu x m(t) GM x (x 2 + y 2 ) 3 2 dv y dt = αθ tu y m(t) dm dt = αθ t GM y (x 2 + y 2 ) 3 2 Θ t = 1 0 <t<tfmax 0 t t fmax

22 Rocket Take-Off Simulation Check out rocket.m What parameters do you need to change to make the rocket escape from Earth? Need to run this on desktop matlab. Download the following from the Chapter11 folder on web-based matlab: rocket.m rocket_der.m stepf.m

Nov. 27, 2017 Momentum & Kinetic Energy in Collisions elastic collision inelastic collision. completely inelastic collision

Nov. 27, 2017 Momentum & Kinetic Energy in Collisions In our initial discussion of collisions, we looked at one object at a time, however we'll now look at the system of objects, with the assumption that