lecturenotes-Feb17

lecturenotes-Feb17 - 9.2 Quick Review: the Center of Mass x...

Info iconThis preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
x com = 1 M m i x i i = 1 N y com = 1 M m i y i i = 1 N z com = 1 M m i z i i = 1 N M = m i i = 1 N x com = 1 M xdm y com = 1 M ydm z com = 1 M zdm ρ = dm dV = M V Here “mass density” replaces mass x com = 1 V xdV 9.2 Quick Review: the Center of Mass (1) Center of mass of a symmetric object always lies on an axis of symmetry. (2) Center of mass of an object does NOT need to be on the object.
Background image of page 2
O m 1 m 3 m 2 F 1 F 2 F 3 x y z com net Ma F = net, com, net, com, net, com, x x y y z z F Ma F Ma F Ma = = = Quick Review: 9.3 Newton’s 2 nd Law for a System of Particles net dp F dt = F net = d P dt = d p 1 dt + ... + d p n dt - Linear Momentum Δ P = 0 - Conservation of Linear Momentum
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
x cm = x CM boat ( ) ( ) m b + Dm d m b + m d = x CM ( boat ) − Δ x B ( ) m b + D − Δ x dog / boat ( ) + Δ x B ( ) m d m b + m d Step I: choose a system F net x = 0 Step II: Find net force COM remains unchanged Step III: choose coordinating system Step IV: boat ( m b ) + dog ( m d ) Problem # 16, Ch. 9: A 4.5 kg dog stands on an 19 kg ±atboat at distance D= 6.1 m for the shore. It walks 2.4 m along the boat toward shore and then stops. Assuming no friction between the boat and the water, ²nd how far the dog is from the shore. x 0 Δ x B = Δ x dog / boat ( ) m d m b + m d ( )
Background image of page 4
m v p Linear momentum of a particle of mass and velocity The SI unit for linear momentum is the kg.m is defined a /s s . . p mv p m v = p mv = The time rate of change of the linear momentum of a particle is equal to the magnitude of net force acting on t Below we will prove the fol he particle and has the dir lowing statem ection of the ent: for ( ) net net In equation form: . We will prove this equation using Newton's second law: This equation is stating that the linear momentum of a particle can be c c a . h e n dp F dt dp d dv p mv mv m ma F dt dt dt = = = = = = ged only by an external force. If the net external force is zero, the linear momentum cannot change: net dp F dt = 9.4 Linear Momentum of a Particle
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Linear Momentum of a Particle p = m v d p dt = m d v dt = m a = F net , ext Linear Momentum It is a vector, having the same direction as velocity; Any change in p is due to external force F 0; Conservation of linear momentum when F = 0. more vectors !
Background image of page 6
An astronaut drops a golf ball that is initially at rest from a cliff on the surface of the moon. The ball falls freely under the inFuence of gravity. Which one of the following statements is true concerning the ball as it falls? Neglect any frictional effects. a) The ball will gain an equal amount of kinetic energy during each second. b) The ball will gain an equal amount of momentum during each second. c) The ball will gain an equal amount of momentum during each meter through which it falls.
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 8
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 37

lecturenotes-Feb17 - 9.2 Quick Review: the Center of Mass x...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online