Lesson 7 Physics 168 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 2
Continuously varying mass Consider continuous stream of matter moving at velocity which impacts object of mass M that is moving with velocity ~u ~v This impacting particles stick to object increasing its mass by M during time During time velocity ~v changes by t Applying impulse momentum theorem to this system ~F net, ext t = P ~ = P ~ f P i =[(M + M)(~v + ~v )] [M~v + M~u] ~P i t ~v 3
Continuously varying mass (cont d) Rearranging terms ~F net, ext t = M ~v + M(~v ~u )+ M ~v Dividing by t ~F net, ext t = M ~v t + M t (~v ~u )+ M t ~v Taking limit that also means and t! 0 M! 0 ~v! 0 ~F net, ext = M d~v dt + dm (~v ~u ) dt Rearranging terms we obtain Newton s second law for a system that has a continuously changing mass ~F net, ext + dm dt ~v rel = M d~v dt ~v rel = ~u ~v 4
Rocket Propulsion 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 5
Rocket Propulsion 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 R v t uex t and is moving at speed v + v 2 6
Momentum of system at we dropped term Rocket Propulsion t + t is P f =(m R t)(v + v) +R t(v u ex ) = mv + m v vr t R t v + vr t u ex R t mv + m v u ex R t Change in momentum is and R t v which is product of two very small quantities P = P f P i = m v u ex R t P t = m v t u ex R As t approaches zero v/ t approaches derivate dv/dt acceleration 7
Rocket Propulsion For a rocket moving upward near surface of earth F ext = mg Setting dp/dt = F ext = mg gives us rocket equation m dv dt = Ru ex + F ext = Ru ex mg rocket equation or dv dt = Ru ex m g = m 0 Ru ex Rt g ( ) Quantity Ru ex is force exerted on rocket by exhausting fuel This is called thrust F th = Ru ex = dm dt u ex 8
Rocket Propulsion ( ) is solved by integrating both sides with respect to time For a rocket starting at rest at v = u ex ln m0 result is Rt m 0 as can be verified by taking time derivative of v Payload of a rocket is final mass m f after all fuel has been burned t b t =0 m f = m 0 Rt b Burn time is given by or gt t b = m 0 A rocket starting at rest with mass m 0 R m f and payload of m f attains a final speed v f = u ex ln m f m 0 gt b final speed of rocket assuming acceleration of gravity to be constant 9
The Saturn V rocket used in Apollo moon-landing program had: initial mass a burn rate and a thrust Saturn V: America s Moon Rocket m 0 =2.85 10 6 kg 73% of which was fuel = 13.84 10 3 kg/s F th = 34 10 6 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 10
(a) (b) F th = dm dt Saturn V: America s Moon Rocket u ex ) u ex =2.46 km/s m b =0.27m 0 =7.70 10 5 kg m fuel = t b t b = m fuel = m 0 m b = 150 s (c) dv y dt = u ex m 0 dm dt g =2.14 m/s 2 (d) dv y dt = u ex m b dm dt g = 34.3 m/s 2 (e) v y = u ex ln m0 m 0 t gt = 1.75 km/s 11
Rotational Dynamics 12
In purely rotational motion Radius of circle is r Angular Quantities all points on object move in circles around axis of rotation O All points on straight line drawn through axis move through same angle in same time arc length Angle in radians is defined = l r 13
Divergence of laser beam A laser jet is directed at Moon, Beam diverges at an angle What diameter spot will it make on Moon? 380.000 km from Earth =1.4 10 5 14
Divergence of laser beam A laser jet is directed at Moon, Beam diverges at an angle What diameter spot will it make on Moon? 380.000 km from Earth =1.4 10 5 = diameter r EM ) diameter = r EM =5.3 10 3 m 15
Angular Quantities (cont d) Angular displacement = 2 1 Average angular velocity is defined as total angular displacement divided by time! = t Instantaneous angular velocity:! = lim t! 0 t 16
Angular Quantities (cont d) Angular acceleration is rate at which angular velocity changes with time =! 2! 1 t =! t Instantaneous acceleration a = lim t! 0! t 17
Angular Quantities Every point on a rotating body has an angular velocity and a linear velocity They are related v v = r!! 18
Angular Quantities (cont d) Therefore objects farther from axis of rotation will move faster 19
Star tracks in a time exposure of night sky 20
Angular Quantities (cont d) If angular velocity of a rotating object changes, it has a tangential acceleration a tan = r Even if angular velocity is constant each point on object has centripetal acceleration a R = v2 r = (r!)2 r =! 2 r 21
Angular Quantities Here is correspondence between linear and rotational quantities LINEAR TYPE ROTATIONAL RELATION x DISPLACEMENT x = r v VELOCITY! v = r! a tan a tan = r ACCELERATION 22
Angular Quantities (cont d) 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 T = 1 f 23
Rotational Kinetic Energy Kinetic energy of rigid object rotating about fixed axis is sum of kinetic energy of individual particles that collectively make object Kinetic energy of the i-th particle Summing over all particles using K = 1 2 m ivi 2 v i = r i! gives rotational kinetic energy K = X i 1 2 m iv 2 i = 1 2 X i m i r 2 i! 2 = 1 2!2 X i m i r 2 i = 1 2 I!2 I = X i m i r 2 i moment of inertia for axis of rotation Object that has both translational and rotational motion also has both translational and rotational kinetic energy KE = 1 2 Mv2 CM + 1 2 I CM! 2 24
Quantity I = X m i r 2 i Moment of Inertia is called rotational inertia of an object 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 I = Z r 2 dm 25
Estimating moment of inertia 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 26
Estimating moment of inertia I = X m i r 2 i = 2 2 2 1 1 1 3 1 5 3 M 6 L + 3 M 6 L + 3 M 6 L = 35 108 ML2 27
Moment of inertia of a thin uniform road 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 28
Moment of inertia of a thin uniform road Z Z I = r 2 dm = x 2 dm dm = dx = M L dx I = Z x 2 dm = Z L 0 x 2 M L dx = 1 3 ML2 29
Steiner s Theorem 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 I cm be moment of inertia about a parallel-axis through center of mass M Let be total mass of object and distance between two axes Parallel axis theorem states that h I = I cm + Mh 2 30
Steiner s Theorem (cont d) Consider object rotating about fixed axis that does not pass through CM Kinetic energy of such a system is K = 1 2 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 2 I cm! 2 Moment of inertia about axis through cm Total kinetic energy of object is K = 1 2 Mv2 cm + 1 2 I cm! 2 31
Steiner s Theorem (cont d) cm moves along a circular path of radius Substituting 1 2 I!2 = 1 2 Mh2! 2 + 1 2 I cm! 2 Multiplying through this equation by I = Mh 2 + I cm h! v cm = h! 2/! 2 leads to 32
Application of Steiner s theorem M L x A thin uniform rod of mass and length on axis has one end at origin Using parallel-axis theorem, find moment of inertia about parallel to y axis, and through center of rod y 0 axis, which is 33
Application of Steiner s theorem I y = I CM + Mh 2 ) I CM = I y Mh 2 = 1 3 ML2 1 4 ML2 = 1 12 ML2 34
Torque 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 35
Torque (Cont d) Torque is defined as = r? F 36
Lever arm for Lever arm for Lever arm for F A F D F C Torque (Cont d) is distance from knob to hinge is zero is as shown 37
Rotational Dynamics Torque and Rotational Inertia Knowing that F = ma This is for a single point mass What about an extended object? = mr 2 ~F r m As angular acceleration is same for whole object we can write X i, net =( X m i r 2 i ) 38
Various Moment of Inertia 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 - 39
Spinning Cylindrical Satellite 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 40
The ring of rockets will create a torque with zero net force net =4FR net = I I = 1 2 MR2 =! t 4FR = I = 1 2 MR2! t ) F = 1 8 MR! t 20 N 41