ω = 0 a = 0 = α P = constant L = constant dt = 0 = d Equilibrium when: τ i = 0 τ net τ i Static Equilibrium when: F z = 0 F net = F i = ma = d P

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Bartholomew Sullivan
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2 Equilibrium when: F net = F i τ net = τ i a = 0 = α dp = 0 = d L = ma = d P = 0 = I α = d L = 0 P = constant L = constant F x = 0 τ i = 0 F y = 0 F z = 0 Static Equilibrium when: P = 0 L = 0 v com = 0 ω = 0

3 1) Draw a free-body diagram showing all forces acting on body and the points at which these forces act. 2) Draw a convenient coordinate system and resolve forces into components. 3) Using letters to represent unknowns, write down equations for: F x =0, F y =0, and F z =0 4) For τ = 0 equation, choose any axis perpendicular to the xy plane. But choose judiciously! Pay careful attention to determining lever arm and sign! [for xy-plane, ccw is positive & cw is negative] 5) Solve equations for unknowns.

4 Chapter 13: Gravita/on Isaac Newton (1687) : What keeps the moon in a nearly circular orbit about the earth? If falling objects accelerate, they must experience a force. This Force = Gravity No contact! Every body attracts every other body

5 Gravita/on For 2 particles, the magnitude of the attractive force between them is F = G m 1m 2 r 2 m 1 and m 2 are masses and r is distance between them and... F 12 = F 21 Newton s third law G = N m 2 /kg 2 = m 3 /kg s 2 Gravitational Constant ( g, 9.8 m/s 2 )

Newton s Law of Gravita/on F = G m 1 m 2 r 2 This is always attractive. G = 6.

6 Newton s Law of Gravita/on F = G m 1 m 2 r 2 This is always attractive. G = N m 2 kg 2 F = m 1 g = G m 1 m 2 r 2 = m 1 G m EARTH 2 r EARTH Where does g come from? What is force of gravity at surface of earth? N m 2 = m kg 2 = m m s 2 = m 1 g kg ( m) 2

7 Gravita/on 1) Objects are independent to each other 2) Gravitational Force is a VECTOR - unit vector notation 3) Principle of superposition F 12 = G m 1 m 2 r 12 2 F 21 = G m 1 m 2 F 21 r Force on m 1 due to m 2 r ˆ 4) A uniform spherical shell of matter attracts an object on the outside as if all the shell s mass were concentrated at its center (note: this defines the position) height = R E + h ˆ r = F 12 F 1,net = F 12 + F 13 + F F r ˆ 21 Force on m 2 due to m 1 r ˆ 21 = 1n = n i=1 F 1i 12 = r 21 r 21 r 12 r 12 = ˆ r 12 F 21 = F 12 VECTOR ADDITION!!

8 Checkpoint What are the gravitational forces on the particle of mass m 1 due to the other particles of mass m. What is the direction of the net gravitational force on the particle of mass m 1 due to the other particles.

9 Checkpoint The figure shows four arrangements of three particles of equal mass. Rank the arrangements according to the magnitude of the net gravitational force on the particle labeled m, greatest first.

10 Question Question 13-1 The moon does not fall to Earth because: 1. The net force on it is zero. 2. It is beyond the main pull of Earth s gravity 3. It is being pulled by the Sun and the planets as well as by Earth. 4. All of the above 5. None of the above

11 Problem 13 6 Three 5 kg spheres are located in the xy-plane. What is the magnitude and direction of the net gravitational force on the sphere at the origin due to the other two spheres? 3 1 2

12 Gravita/on and the earth F g = G M E 2 R E m apple = m apple g Net force points towards center of earth g differs around the earth (equator & north pole m/s 2 ) 1) Earth is not a perfect sphere - height (R E is not constant): - On Mount Everest (8.8 km) g=9.77 m/s 2 (0.2% smaller) - At Equator earth bulges by 21 km 2) Earth is not uniform density: gravity irregularities ( )g gravimeters can measure down to 10-9 g 2) Earth is rotating: centripetal force makes apparent weight change At poles: At equator: W mg = 0 W = mg W mg = m v2 R E W = m g v2 R E Weight is less (0.3%)

13 Gravity and Spheres A uniform spherical shell of matter attracts a particle that is outside the shell as if all the shell s mass were concentrated at its center. A uniform shell of matter exerts no net gravitational force on a particle located inside it. net vector force is zero inside

m 2 m 1 m 2 r m 1 F 1 Gravitation Inside the Earth Newton proved that the net gravitational force on a particle by a shell depends on the position of the particle with respect to the shell.

14 m 2 m 1 m 2 r m 1 F 1 Gravitation Inside the Earth Newton proved that the net gravitational force on a particle by a shell depends on the position of the particle with respect to the shell. If the particle is inside the shell, the net force is zero. m1m 2 If the particle is outside the shell, the force is given by: F1 = G. r 2 Consider a mass m inside the Earth at a distance r from the center of the Earth. If we divide the Earth into a series of concentric shells, only the shells with ins radius less than r exert a force on m. The net force on m is: F =. 2 Here M ins GmM r is the mass of the part of the Earth inside a sphere of radius r : 3 4π r 4π Gmρ M ins = ρvins = ρ F = r F is linear with r. 3 3 (13-7)

15 Problem13 13 With what gravitational force does the hollowed-out sphere attract a small sphere of mass m?

Gravita/onal Poten/al Energy From Section 8.3 ΔU = W done by force Conservative force-path independent ΔU g = W done = x f x i F g dx At Earth s surface, F g ~const.

16 Gravita/onal Poten/al Energy From Section 8.3 ΔU = W done by force Conservative force-path independent ΔU g = W done = x f x i F g dx At Earth s surface, F g ~const. ΔU g = W done = m g y f ( ) dy = mgδy y i W = r f G mm r f dr = GmM 1 dr r 2 r i r 2 r i If we define U = 0 at, then the work done by taking mass m from R to U U (r) = W = GmM 0 1 R U (r) = GmM r F(r) = du(r) dr d dr GmM = GmM r r 2 Note: 1) As before, Grav. Pot. Energy decreases as separation decreases (more negative) 2) Path independent 3) MUST HAVE AT LEAST TWO PARTICLES TO POTENTIAL ENERGY (& force) 4) Knowing potential, you can get force.

17 Gravita/onal Poten/al Energy ΔU g = W done = x f x i F g dx Conservative force-path independent If we define U = 0 at, then the work done by taking mass m from R to U(r) = G m 1m 2 r 12 F 12 = G m 1 m 2 r 12 2 ˆ r 12 Note: Potential energy increases as the separation gets larger: If r then GmM and U gets less negative (larger) r

18 Example Three spheres with mass m A, m B, and m C. You move sphere B from left to right. How much work do you do on sphere B? How much work is done by the gravitational force? Gravita(onal Poten(al Energy

19 x com = 1 M N i=1 m ix i x com = 1 M xdm x com = 1 V xdv y com = 1 M N i=1 m iy i y com = 1 M ydm z com = 1 M N i=1 m iz i z com = 1 M zdm M = N i=1 m i ρ = dm dv = M V Here mass density replaces 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.

20 F 1 m 2 F 2 x m 1 O z F 3 m 3 y Ma com = F net F F F net, x net, y net, z = = = Ma Ma Ma com, x com, y com, z p = mv - Linear Momentum F net = dp F net = d P = d p d p n Δ P = 0 - Conservation of Linear Momentum

t i Impulse Change in Momentum is equal to Impulse acting on it F L R (t)

21 P R,after Collision What changes momentum of each? P R,before p f d p R = F L R (t) dp R = p i Definition p f J t f t i t f t i Impulse Change in Momentum is equal to Impulse acting on it F L R (t) F net (t) p i = Δp = J Impulse Vector! Must satisfy for each direction! Impulse-momentum theorem

22 Collisions: Elastic vs. Inelastic Elastic collision : TOTAL KE is conserved (~ Conservative forces ) AND if system is closed and isolated, the total linear momentum P cannot change (whether the collision is elastic or inelastic!). ΔKE = K f K i = ( K 1f K nf ) ( K 1i K ni ) = 0 Δ P = P f Pi = P1f P nf P1i+...+ P ni = 0 Inelastic collision : KE is not conserved (~ thermal energy ) However, if system is closed and isolated, the total linear momentum P cannot change (whether the collision is elastic or inelastic!). ΔKE = K f K i = ( K 1f K nf ) ( K 1i K ni ) 0 Δ P = P f Pi = P1f P nf P1i+...+ P ni = 0

23 Transla(onal vs. Rota(onal Mo(on Translational Motion x v a θ ω α Rotational Motion v = v 0 + at ω = ω 0 + αt (eq. 1) x = x o + v 0 t + at 2 2 θ = θ + ω 0 t + αt 2 2 (eq. 2) v 2 v 2 0 = 2a( x x ) o ω 2 ω 2 0 = 2α ( θ θ ) 0 (eq. 3)

24 Rela(ng Transla(onal Rota(onal Variables Rota(onal posi(on and distance moved s = θ r (only radian units) Rota(onal and transla(onal speed v = d r = ds = d θ r v = ω r Rota(onal and transla(onal accelera(on v = ω r ccw is positive cw is negative a t = dv dr = d ω r a t = α r Rela(ng period and rota(onal speed [ distance = rate time] T = 2πr v = 2π units of (me (s) ω ωt = 2π

Rela(ng Transla(onal Rota(onal Accelera(on Rota(onal and transla(onal accelera(on a) from v = ω r dv = d ω r a t = d ω r = α r a t = α r tangen/al accelera/on b) from before we know there

25 Rela(ng Transla(onal Rota(onal Accelera(on Rota(onal and transla(onal accelera(on a) from v = ω r dv = d ω r a t = d ω r = α r a t = α r tangen/al accelera/on b) from before we know there s also a radial component a r = v2 r = ω 2 r a r = ω ω r ( ) radial accelera/on c) must combine two dis(nct rota(onal accelera(ons 2 a tot = a 2 r + a 2 t a = ω 2 r 2 tot = a t + a r + αr 2

26 Kine(c energy of Rota(on KE system = rota(onal iner(a (moment of iner(a) about some axis of rota(on 1 m 2 2 iv i ( ) 2 1 = m 2 i ωr i = 1 2 m 2 [ ( i r i )]ω 2 I where v i = ωr i trans rot All par(cles have same ω Energy of rota(onal mo(on KE rot = 1 2 Iω 2 [ KE trans = 1 2 ] mv2

27 Rota(onal Iner(as If h is a perpendicular distance between a given axis and the axis through the center of mass (these two axes being parallel). Then the rotational inertia I about the given axis is I = I COM + Mh 2

28 Rota/onal Work and Energy We can compare linear variables with rotational variables x v a Δt F m θ ω α Δt τ I s = rθ v T = rω a T = rα The same can be done for work and energy: For translational systems W = F x KE = 1 2 mv 2 For rotational systems W = τ θ KE = 1 2 Iω 2

29 Newton s 2 nd law for rota(on F net = { F i } = m a τ net = τ i = r F = r m a = m r (r α )= Iα Work and Rotational Kinetic Energy NET Work done ON system W = ΔKE rot ( ) = 1 I ω f ω i W = ΔKE trans = 1 m v 2 2 v 2 f i ( ) Rotational work, fixed axis rotation (if torque is const) W = θ 2 θ 1 τ net dθ ( ) = τ θ f θ i x 2 W = Fdx 1 D motion x 1 Power, fixed axis rotation P = dw = d ( τθ ) = τω P = dw = d F ( x ) = F v

30 Rolling (ch. 11) = Rota(on (ch. 10) + Transla(on (ch. 1 9) Wheel moving forward with constant speed v com s = θr displacement: translation rotation v com = ds = d θr a com = dv com ( ) = d ( ωr ) = ωr = αr Only if NO SLIDING [smooth rolling] 1 2 I comω Mv com 2 = KE tot Note: rotation about COM and translation of COM combine for total KE

31 Linear and angular rela(ons Force Linear momentum (one) Linear momentum (system) Linear momentum (system) Newton s second law (system) Conserva(on Law (closed,isolated) F v = p m p = P Mv com = P dp = F net Ma com = F net 0 = Δ P τ = r F l = r p L = l L = Iω τ net = d L τ net = I α ΔL = 0 Torque Angular momentum (one) Angular momentum (system) Angular momentum (system, fixed axis) Newton s second law (system) Conserva(on Law (closed,isolated) τ net = d L τ net has no meaning unless the net torque, and the τ L net total rota/onal momentum, are defined with L respect to the same origin

32 Equilibrium Equilibrium when: F net = F i τ net = τ i a = 0 = α dp = 0 = d L = ma = d P = 0 = I α = d L = 0 P = constant L = constant F x = 0 τ i = 0 F y = 0 F z = 0 Static Equilibrium when: P = 0 L = 0 v com = 0 ω = 0

33 Procedure for Solving Equilibrium Problems 1) Draw a free-body diagram showing all forces acting on body and the points at which these forces act. 2) Draw a convenient coordinate system and resolve forces into components. 3) Using letters to represent unknowns, write down equations for: F x =0, F y =0, and F z =0 4) For τ = 0 equation, choose any axis perpendicular to the xy plane. But choose judiciously! Pay careful attention to determining lever arm and sign! [for xy-plane, ccw is positive & cw is negative] 5) Solve equations for unknowns.

ω = ω 0 θ = θ + ω 0 t αt ( ) Rota%onal Kinema%cs: ( ONLY IF α = constant) v = ω r ω ω r s = θ r v = d θ dt r = ω r + a r = a a tot + a t = a r

ω = ω 0 θ = θ + ω 0 t αt ( ) Rota%onal Kinema%cs: ( ONLY IF α = constant) v = ω r ω ω r s = θ r v = d θ dt r = ω r + a r = a a tot + a t = a r θ (t) ( θ 1 ) Δ θ = θ 2 s = θ r ω (t) = d θ (t) dt v = d θ dt r = ω r v = ω r α (t) = d ω (t) dt = d 2 θ (t) dt 2 a tot 2 = a r 2 + a t 2 = ω 2 r 2 + αr 2 a tot = a t + a r = a r ω ω r a t = α r ( ) Rota%onal