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Lesson 7 Physics 168 Monday, October 24, 16 Luis Anchordoqui 1 Eruption of a large volcano on Jupiter’s moon When volcano erupts speed of effluence exceeds escape speed of Io and so a stream of particles is projected into space Material in stream can collide with and stick to surface of asteroid passing through stream We now consider effect of impact of this material on motion of asteroid Monday, October 24, 16 2 Continuously varying mass Overview Consider continuous stream of matter moving at velocity ~u which impacts object of mass M that is moving with velocity This impacting particles stick to object increasing its mass by M during time v changes by During time t velocity ~ ~v t ~v Applying impulse momentum theorem to this system F~net, ext t = C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 P~ = P~f ~i = [(M + P M )(~v + ~v )] [M~v + M ~u] 2 3 Continuously varying mass (cont’d) F~net, ext Rearranging terms Overview t = M ~v + M (~v Dividing by F~net, ext Taking limit t = M ~v + t M (~v t ~u) + M ~v t ~u) + t ! 0 that also means M ! 0 and d~v dM ~ Fnet, ext = M + (~v ~u) dt dt M t ~v ~v ! 0 Rearranging terms we obtain Newton’s second law for a system that has a continuously changing mass dM d~v ~ Fnet, ext + ~vrel = M dt dt ~vrel = ~u ~v C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 2 4 Rocket Propulsion Overview Momentum conservation works for a rocket as long as we consider rocket and its fuel to be one system and account for mass loss of rocket C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 2 5 Rocket Propulsion Overview Rocket propulsion is a striking example of conservation of momentum in action Use Newton’s law in form Fext = dP/dt Consider a rocket moving with speed v relative to earth ☛ If fuel is burned at constant rocket’s mass at time t is ☛ m(t) = m0 Momentum of system at time At a later time t+ R = |dm/dt| t is ☛ Rt Pi = mv t rocket has expelled gas of mass R If gas is exhausted at speed uex relative to rocket velocity of gas relative to Earth is Rocket then has a mass m C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 R v t uex t and is moving at speed v + v 2 6 Rocket Propulsion Momentum of system at t + t is Overview Pf = (m R t)(v + v) + R = mv + m v vR t ⇡ mv + m v uex R we dropped term R R t(v t uex ) v + vR t uex R t t t v which is product of two very small quantities Change in momentum is P = Pf Pi = m v uex R t and P = m t As t approaches zero C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 v t uex R v/ t approaches derivate dv/dt ☛ acceleration 2 7 Rocket Propulsion For a rocket moving upward near surface of earth Setting dP/dt = Fext = mg Fext = mg gives us rocket equation dv m = Ruex + Fext = Ruex dt mg rocket equation or dv Ruex = dt m Quantity Ruex Ruex g = m0 Rt g ( ✴) is force exerted on rocket by exhausting fuel This is called thrust Fth = Ruex Monday, October 24, 16 dm = uex dt 8 Rocket Propulsion (✴) is solved by integrating both sides with respect to time For a rocket starting at rest at t = 0 result is ⇣m Rt ⌘ 0 v = uex ln gt m0 as can be verified by taking time derivative of v Payload of a rocket is final mass mf after all fuel has been burned Rtb or Burn time tb is given by mf = m0 tb = m0 mf R A rocket starting at rest with mass m0 and payload of attains a final speed vf = mf uex ln m0 gtb mf final speed of rocket assuming acceleration of gravity to be constant Monday, October 24, 16 9 Saturn V: America’s Moon Rocket Overview The Saturn V rocket used in Apollo moon-landing program had: 6 initial mass m0 = 2.85 ⇥ 10 kg 73% of which was fuel a burn rate ↵ = 13.84 ⇥ 103 kg/s and a thrust Fth = 34 ⇥ 106 N Find (a) the exhaust speed relative to the rocket; (b) the burn time; (c) the acceleration at liftoff (d) the acceleration at just before burnout; (e) the final speed of the rocket C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 2 10 (a) Fth Saturn V: America’s Moon Rocket dm = uOverview ex ) uex = 2.46 km/s dt mfuel = ↵ tb (b) mb = 0.27m0 = 7.70 ⇥ 105 kg mfuel m0 mb tb = = = 150 s ↵ ↵ (c) dvy uex dm = dt m0 dt dvy uex dm (d) = dt mb dt (e) vy = uex ln C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 ✓ g = 2.14 m/s g = 34.3 m/s m0 m0 ↵t ◆ 2 2 gt = 1.75 km/s 2 11 Rotational Dynamics Monday, October 24, 16 12 Angular Quantities Overview In purely rotational motion all points on object move in circles around axis of rotation ☛ O Radius of circle is r All points on straight line drawn through axis move through same angle in same time Angle ✓ in radians is defined ☛ C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 l ✓ = r arc length 2 13 titulo A laser jet is directedOverview at Moon, 380.000 km from Earth Beam diverges at an angle ✓ = 1.4 ⇥ 10 5 What diameter spot will it make on Moon? C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 2 14 titulo A laser jet is directedOverview at Moon, 380.000 km from Earth Beam diverges at an angle ✓ = 1.4 ⇥ 10 5 What diameter spot will it make on Moon? diameter ✓= ) diameter = ✓rEM = 5.3 ⇥ 103 m rEM C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 2 15 Angular Quantities (cont’d) Angular displacement Overview ✓ = ✓2 ✓1 Average angular velocity is defined as total angular displacement divided by time ! ¯ = ✓ t Instantaneous angular velocity: ! = lim t!0 C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 ✓ t 2 16 Angular Quantities (cont’d) Overview Angular acceleration is rate at which angular velocity changes with time ↵ ¯= !2 !1 t ! t = Instantaneous acceleration ↵ = lim a t!0 C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 ! t 2 17 Angular Quantities Overview Every point on a rotating body has an angular velocity ! and a linear velocity v They are related ☛ C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 v = r! 2 18 Angular Quantities (cont’d) Overview Therefore objects farther from axis of rotation will move faster C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 2 19 Star tracks in a time exposure of night sky Monday, October 24, 16 20 Angular Quantities (cont’d) Overview If angular velocity of a rotating object changes, it has a tangential acceleration atan = r ↵ Even if angular velocity is constant each point on object has centripetal acceleration aR C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 v2 (r!)2 = = = !2 r r r 2 21 Angular Quantities Overview Here is correspondence between linear and rotational quantities LINEAR TYPE ROTATIONAL x DISPLACEMENT ✓ x = r✓ v VELOCITY ! v = r! atan ACCELERATION ↵ atan = r ↵ C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 RELATION 2 22 Angular Quantities (cont’d) Overview ✓ Frequency is number of complete revolutions per second ! f = 2⇡ ✓ Frequencies are measured in hertz 1 Hz = 1 s 1 ✓ Period is time one revolution takes 1 T = f C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 2 23 Rotational Kinetic Energy Overview Kinetic energy of rigid object rotating about fixed axis is sum of kinetic energy of individual particles that collectively make object 1 K = mi vi2 2 vi = ri ! gives rotational kinetic energy Kinetic energy of the i-th particle ☛ Summing over all particles using K= X1 i I= X 2 mi vi2 1 2X 1 2 1X 2 2 2 mi ri ! = ! mi ri = I! = 2 i 2 2 i mi ri2 ☛ moment of inertia for axis of rotation i Object that has both translational and rotational motion also has both translational and rotational kinetic energy 1 1 2 KE = M vCM + ICM ! 2 2 2 C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 2 24 Quantity I = X Moment of Inertia is called rotational inertia of an object Overview mi ri2 Distribution of mass matters here these two objects have same mass but one on left has a greater rotational inertia as so much of its mass is far from axis of rotation Integral form ☛ C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 I = Z r2 dm 2 25 Estimating moment of inertia Overview Estimate moment of inertia of a thin uniform rod of length L and mass M about an axis perpendicular to rod and through one end Execute this estimation by modeling rod as three point masses each point mass representing 1/3 of rod C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 2 26 Estimating moment of inertia Overview I= X mi ri2 = C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 ✓ 1 M 3 ◆✓ 1 L 6 ◆2 + ✓ 1 M 3 ◆✓ 3 L 6 ◆2 + ✓ 1 M 3 ◆✓ 5 L 6 ◆2 = 35 M L2 108 2 27 Moment of inertia of a thin uniform road Overview Find moment of inertia of a thin uniform rod of length L and mass M about an axis perpendicular to rod and through one end C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 2 28 Moment of inertia of a thin uniform road Overview I= Z r2 dm = Z x2 dm M dm = dx = dx L I= C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 Z 2 x dm = Z L 0 2M 1 x dx = M L2 L 3 2 29 Steiner’s Theorem Overview Parallel-axis theorem relates moment of inertia about an axis through CM to moment of inertia about a second parallel axis Let I be moment of inertia Let Icm be moment of inertia about a parallel-axis through center of mass Let M be total mass of object and h distance between two axes Parallel axis theorem states that ☛ I = Icm + M h2 C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 2 30 Steiner’s Theorem (cont’d) Overview Consider object rotating about fixed axis that does not pass through CM Kinetic energy of such a system is 1 2 K = I! 2 Moment of inertia about fixed axis Kinetic energy of a system can be written as sum of its translational kinetic energy and kinetic energy relative to its CM For an object that is rotating kinetic energy relative to its CM 1 Icm ! 2 2 Moment of inertia about axis through cm Total kinetic energy of object is ☛ C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 1 1 2 K = M vcm + Icm ! 2 2 2 2 31 Steiner’s Theorem (cont’d) Overview cm moves along a circular path of radius Substituting h ! vcm = h! 1 2 1 1 2 2 I! = M h ! + Icm ! 2 2 2 2 Multiplying through this equation by 2 2/! 2 leads to I = M h + Icm C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 2 32 Application of Steiner’s theorem A thin uniform rod of mass M and length L on x axis has one end at origin Overview Using parallel-axis theorem, find moment of inertia about parallel to y C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 y 0 axis, which is axis, and through center of rod 2 33 Application of Steiner’s theorem Overview Iy = ICM + M h2 ) ICM = Iy C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 M h2 = 1 M L2 3 1 1 M L2 = M L2 4 12 2 34 Torque Overview To make an object start rotating a force is needed Position and direction of force matter as well Perpendicular distance from axis of rotation to line along which force acts is called lever arm C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 2 35 Torque (Cont’d) Overview Torque is defined as ☛ C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 ⌧ = r? F 2 36 Torque (Cont’d) Lever arm for Overview FA is distance from knob to hinge Lever arm for FD is zero Lever arm for FC is as shown C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 2 37 Rotational Dynamics ☛ Torque and Rotational Inertia Knowing that Overview F = ma ☛ ⌧ = mr2 ↵ This is for a single point mass What about an extended object? F~ r m As angular acceleration is same for whole object we can write C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 X ⌧i, net = ( X mi ri2 ) ↵ 2 38 titulo Overview Rotational inertia of an object depends not only on its mass distribution but also location of axis of rotation – compare (f) and (g), for example - C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 2 39 titulo Overview To get a flat, uniform cylindrical satellite spinning at correct rate, engineers fire four tangential rockets as shown in figure If satellite has a mass of 3600 kg and a radius of 4 m, what is required steady force of each rocket if satellite is to reach 32 rpm in 5 min? End view of cylindrical satellite C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 2 40 The ring of rockets will create a torque with zero net force Overview ⌧net = 4F R ⌧net = I↵ 1 I = M R2 2 ↵= 1 4F R = I↵ = M R2 2 C. B.-Champagne Luis Anchordoqui Monday, October 24, 16 ! t ! 1 ! ) F = MR ⇡ 20 N t 8 t 2 41