16.07 Dynamics Fall 2004 Version 1.2 Lecture D10- Angular Impulse and momentum In addition to the equations of linear impulse and momentum considered in the previous lecture there is a parallel set of equations that relate the angular impulse and momentum. Angular momentum We consider a particle of mass, m, with velocity u, moving under the influence of a force F. The angular momentum about point O is defined as the "moment"of the particle's linear momentum, L, about O n Thus, the particle's angular momentum is given by, HO=r×mu=r×L The units for the angular momentum are kg m/s in the SI system, and slug. ft /s in the English system Using aa right handed cartesian coordinate system, the components of the angular momentum are calculated k Ho=Hi+H,j+H k=m z y x=m(u2y-vya)i+m(uz2-v2r)j+m(vyt-vry)k
J. Peraire 16.07 Dynamics Fall 2004 Version 1.2 Lecture D10 - Angular Impulse and Momentum In addition to the equations of linear impulse and momentum considered in the previous lecture, there is a parallel set of equations that relate the angular impulse and momentum. Angular Momentum We consider a particle of mass, m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. Thus, the particle’s angular momentum is given by, HO = r × mv = r × L . (1) The units for the angular momentum are kg·m2/s in the SI system, and slug·ft2/s in the English system. Using a a right handed cartesian coordinate system, the components of the angular momentum are calculated as HO = Hxi + Hyj + Hzk = m i j k x y z vx vy vz = m(vzy − vyz)i + m(vxz − vzx)j + m(vyx − vxy)k . 1
It is clear from its definition that the angular momentum is a vector which is perpendicular to the plane defined by r and v. Thus, in some occasions it may be more convenient to determine the direction of Ho from the right hand rule, and its modulus directly from the definition of the vector product as H Lo=mur sin a where a is the angle between r and v Rate of Change of Angular Momentum We now want to examine how the angular momentum changes with time. Taking a time derivative of expression(1), we have Ho=ixmv+r×mi Here, we have assumed that m is constant. If O is a fixed point, then r= v and r x mU=0. Thus, we end Applying Newton's second law to the right hand side of the above equation, we have that r x ma=rxF= Mo, where Mo is the moment of the force F about point O. The equation expressing the rate of change f angular momentum is then written as Ho We note that this expression is valid whenever point O is fixed. The above equation is analogous to the equation derived in the previous lecture expressing the rate of change of linear momentum. It states that the rate of change of linear momentum about a fixed point O is equal to the moment about o due to the esultant force acting on the particle. Since this is a vector equation, it must be satisfied for each component independently. Thus, if the force acting on a particle is such that the component of its moment along a given direction is zero, then the component of the angular momentum along this direction will remain constant Example Pendulum Here, we revisit the simple pendulum problem introduced in lecture D6 and re-derive the pendulum equ
It is clear from its definition that the angular momentum is a vector which is perpendicular to the plane defined by r and v. Thus, in some occasions it may be more convenient to determine the direction of HO from the right hand rule, and its modulus directly from the definition of the vector product as HO = mvr sin α , where α is the angle between r and v. Rate of Change of Angular Momentum We now want to examine how the angular momentum changes with time. Taking a time derivative of expression (1), we have H˙ O = r˙ × mv + r × mv˙ . Here, we have assumed that m is constant. If O is a fixed point, then r˙ = v and r˙ × mv = 0. Thus, we end up with, H˙ O = r × mv˙ = r × ma . Applying Newton’s second law to the right hand side of the above equation, we have that r × ma = r ×F = MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is then written as MO = H˙ O . (2) We note that this expression is valid whenever point O is fixed. The above equation is analogous to the equation derived in the previous lecture expressing the rate of change of linear momentum. It states that the rate of change of linear momentum about a fixed point O is equal to the moment about O due to the resultant force acting on the particle. Since this is a vector equation, it must be satisfied for each component independently. Thus, if the force acting on a particle is such that the component of its moment along a given direction is zero, then the component of the angular momentum along this direction will remain constant. Example Pendulum Here, we revisit the simple pendulum problem introduced in lecture D6 and re-derive the pendulum equation using equation (2). 2
There are two forces acting on the suspended mass: the string tension and the weight. The string tension T, is parallel to the position vector r and therefore its moment about O is zero. On the other hand, the weight creates a moment about O which is Mo=-Img sin Ak e angola momentum is given by Ho=T×mU=ler×mlee=ml2ber×ea=m2bk Therefore, the z component of equation(2) gives ml-8=-lmg sin which is precisely the same equation as the one derived in lecture D6 Principle of Angular Impulse and momentum Equation(2) gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time ti and time t2. Equation (2)can then be integrated in time to obtain Modt=/Hodt=(Ho)2-(Ho)I=AH (3) Here,(Ho)1=Ho(t1) and(Ho)2=Ho(t2). The term is called the angular impulse. Thus, the angular impulse on a particle is equal to the angular momentum Equation (3)is particularly useful when we are dealing with impulsive forces. In such cases, it is often possible to calculate the integrated effect of a force on a particle without knowing in detail the actual value of the force as a function of time
There are two forces acting on the suspended mass: the string tension and the weight. The string tension, T, is parallel to the position vector r and therefore its moment about O is zero. On the other hand, the weight creates a moment about O which is MO = −lmg sin θk. The angular momentum is given by HO = r × mv = ler × ml ˙θeθ = ml2 ˙θ er × eθ = ml2 ˙θk. Therefore, the z component of equation (2) gives ml2 ¨θ = −lmg sin θ , or, ¨θ + g l sin θ = 0 , which is precisely the same equation as the one derived in lecture D6. Principle of Angular Impulse and Momentum Equation (2) gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. Equation (2) can then be integrated in time to obtain Z t2 t1 MO dt = Z t2 t1 H˙ O dt = (HO)2 − (HO)1 = ∆HO . (3) Here, (HO)1 = HO(t1) and (HO)2 = HO(t2). The term Z t2 t1 MO dt , is called the angular impulse. Thus, the angular impulse on a particle is equal to the angular momentum change. Equation (3) is particularly useful when we are dealing with impulsive forces. In such cases, it is often possible to calculate the integrated effect of a force on a particle without knowing in detail the actual value of the force as a function of time. 3
Conservation of Angular momentum We see from equation(1)that if the moment of the resultant force on a particle is zero during an interval of time, then its angular momentum Ho must remain constant Consider now two particles ma and mb which interact during an interval of time. Assume that interaction forces between them are the only unbalanced forces on the particles that have a non-zero moment about a fixed point O. Let F be the interaction force that particle mb exerts on particle ma. Then, according to Newtons third law, the interaction force that particle ma exerts on particle mb will be-F. Us (3), we will have that△(Hola=-△(Ho)b,or△Ho=△(Ho)a+△(Ho)b=0. That is, the changes in angular momentum of particles mi and mb are equal in magnitude and of opposite sign, and the total ngular momentum change equals zero. Recall that this is true only if the unbalanced forces, those with non-zero moment about O, are the interaction forces between the particles. The more general situation in which external forces can be present will be considered in future lectures We note that the above argument is also valid in a componentwise sense. That is, when two particles interact and there are no external unbalanced moments along a given direction, then the total angular momentum change along that direction must be zero Ball on a cylinder a particle of mass m is released on the smooth inside wall of an open cylindrical surface with a velocity vo that makes an angle a with the horizontal tangent. The gravity acceleration is pointing downwards. We want to obtain: i)an expression for the largest magnitude of vo that will prevent the particle from leaving the cylinder through the top end, and ii) an expression for the angle B that the velocity vector will form with the horizontal tangent, as a function of b o y The only forces on the particle are gravity and the normal force from the cylinder surface. The moment of
Conservation of Angular Momentum We see from equation (1) that if the moment of the resultant force on a particle is zero during an interval of time, then its angular momentum HO must remain constant. Consider now two particles ma and mb which interact during an interval of time. Assume that interaction forces between them are the only unbalanced forces on the particles that have a non-zero moment about a fixed point O. Let F be the interaction force that particle mb exerts on particle ma. Then, according to Newton’s third law, the interaction force that particle ma exerts on particle mb will be −F. Using expression (3), we will have that ∆(HO)a = −∆(HO)b, or ∆HO = ∆(HO)a + ∆(HO)b = 0. That is, the changes in angular momentum of particles m1 and mb are equal in magnitude and of opposite sign, and the total angular momentum change equals zero. Recall that this is true only if the unbalanced forces, those with non-zero moment about O, are the interaction forces between the particles. The more general situation in which external forces can be present will be considered in future lectures. We note that the above argument is also valid in a componentwise sense. That is, when two particles interact and there are no external unbalanced moments along a given direction, then the total angular momentum change along that direction must be zero. Example Ball on a cylinder A particle of mass m is released on the smooth inside wall of an open cylindrical surface with a velocity v0 that makes an angle α with the horizontal tangent. The gravity acceleration is pointing downwards. We want to obtain : i) an expression for the largest magnitude of v0 that will prevent the particle from leaving the cylinder through the top end, and ii) an expression for the angle β that the velocity vector will form with the horizontal tangent, as a function of b. The only forces on the particle are gravity and the normal force from the cylinder surface. The moment of 4
these forces about O(or, in fact, about any point on the axis of the cylinder) always has a zero component the z direction. That is, (Mo)2=0. To see that, we notice that for any point on the surface of the cylinder r and F are always contained in a vertical plane that contains the z axis. Therefore, the moment must be normal to that plane. Since the moment has zero component in the z direction, (Ho) will be constant Thus, we have that (Ho)2=rmu cos a=constant For part 1), we consider the trajectory for which the velocity is horizontal when z= a and let(wo)limit be he initial velocity that corresponds to this trajectory. It is clear that for any trajectory for which vo has a larger magnitude than(vo)limit, the particle will leave the cylinder through the top end. Thus, for the limit rajectory we have, from conservation of energy and from conservation of angular momentum (Ho)2=rm(uo )limit cosa=rmva Here, va is the magnitude of the velocity for the limit trajectory when 2= a. Eliminating va from these equations we finally arrive at vo limit Therefore, for vo <(vo)limit the mass will not leave the cylinder through the top end For part ii), we also consider conservation of energy m=÷m2-mgb and conservation of angular momentu rmu cos a= rmb cos B Eliminating, Ub from these two expressions we obtain COS o √1+20b/ Example Spinning Mass A small particle of mass m and its restraining cord are spinning with an angular velocity w on the horizontal Irface of a smooth disk, shown in section. As the force Fs is slowly increased, r decreases and w changes the mass is spinning with wo and ro. Determine: i)an expression for w as a function of r, and i) done on the particle by Fs between To and an arbitrary r, and verify the principle of work and
these forces about O (or, in fact, about any point on the axis of the cylinder) always has a zero component in the z direction. That is, (MO)z = 0. To see that, we notice that for any point on the surface of the cylinder, r and F are always contained in a vertical plane that contains the z axis. Therefore, the moment must be normal to that plane. Since the moment has zero component in the z direction, (HO)z will be constant. Thus, we have that (HO)z = rmv0 cos α = constant . For part i), we consider the trajectory for which the velocity is horizontal when z = a and let (v0)limit be the initial velocity that corresponds to this trajectory. It is clear that for any trajectory for which v0 has a larger magnitude than (v0)limit, the particle will leave the cylinder through the top end. Thus, for the limit trajectory we have, from conservation of energy 1 2 m(v0) 2 limit = 1 2 mv2 a + mga , and from conservation of angular momentum (HO)z = rm(v0)limit cos α = rmva . Here, va is the magnitude of the velocity for the limit trajectory when z = a. Eliminating va from these equations we finally arrive at, (v0)limit = r 2ga 1 − cos2 α . Therefore, for v0 ≤ (v0)limit the mass will not leave the cylinder through the top end. For part ii), we also consider conservation of energy 1 2 mv2 0 = 1 2 mv2 b − mgb and conservation of angular momentum, rmv0 cos α = rmvb cos β . Eliminating, vb from these two expressions we obtain, β = cos−1 cos α p 1 + 2gb/v2 0 ! . Example Spinning Mass A small particle of mass m and its restraining cord are spinning with an angular velocity ω on the horizontal surface of a smooth disk, shown in section. As the force Fs is slowly increased, r decreases and ω changes. Initially, the mass is spinning with ω0 and r0. Determine : i) an expression for ω as a function of r, and ii) the work done on the particle by Fs between r0 and an arbitrary r, and verify the principle of work and energy. 5