Chap11. Angular Momentum Level : AP Physics Teacher : Kim 11.1 The Vector Product and Torque (p.335) Properties of the Vector Product For scalar product, we are familiar with +,. Ex) 3+2=5. Numbers are scalar quantities and since all scalars are independent of directions, we can simply use +,. For vector product, there are the dot product(ex. A B=ABcosφ) and cross product(a B=C, where C=ABsinφ) Unlike the dot product, the cross product of two vectors A B will result in another vector C in a new direction perpendicular to the plane of the two vectors A and B. => (right-hand rule) The magnitude of the vector C is determined by the magnitude of vector and the angle between A and B. See figure 11.8 p.333 C τ F B φ r A C=A B where C=ABsinφ F τ =r F where τ=rfsinφ For 2-D, the unit vectors i, j, and k follows the rule i i = j j = k k = 0, i j = j i = k, j k = k j = i, k i = i k = j ex) the magnitude of i i, 1 1 sin0º=0 So if A=A xi + A yj and B=B xi + B yj, then A B = (A xb y A yb x)k Do Example 11.1 p.337. Two vectors lying in the xy plane are given by the equations A=2i +3j and B= i+2j. Find the A B and verify that A B= B A z x i k j y
Torque Revisit. Defining Torque using the Cross Product Torque is the measure of the ability of a force to rotate an object around an axis. Torque is defined as τ = rfsinφ Axis F φ Torque will be maximum when φ=90. That is rotation of the object will occur best(easier) when a force is applied φ=90 r This is the reason why the doorknob is positioned as far away from the axis (hinge of the door) as possible. The torque τ can be expressed using cross product We choose counter-clockwise as positive direction τ = r F => τ = rfsinφ :magnitude The vector r and F are lying on the same plane. The new vector τ is perpendicular to that plane. Do Example 11.2 The Torque Vector A force of F=(2.00i + 3.00j)N is applied to an object that is pivoted about a fixed axis aligned along the z coordinate axis. The force is applied at a point located at r=(4.00i + 5.00j)m. Find the torque τ applied to the object. Q1) A particle located at the position vector r = (i + j)m has a force F = (2i + 3j)N acting on it. The torque about the origin is: a) (1k)N m b) (5k)N m c) ( 1k)N m d) ( 5k)N m e) (2i +3j)N m
11.2 Analysis Model : Nonisolated System(Angular Momentum) see p.338 *~An object in motion has momentum. If the object is rotating about a fixed axis, then the object has angular momentum The instantaneous angular momentum L of a particle relative to the z origin O is defined as the cross product of the particle s instantaneous L = r p position vector r and its instantaneous linear momentum p: L = r p => Since p=mv, the magnitude of the L is L=rmvsinφ, where φ is the angle between r and p. O p r x m φ y We can see that angular momentum is zero when the angle is φ=0 or 180. Note that L is the result between the cross product r and p. L is a vector the has a unique direction which perpendicular to the plane of r and p Linear Momentum p p = mv F = m dv dt = d(mv) = dp dt dt => F = dp dt Angular Momentum L L= Iw = rp τ = dl (see p.338 for derivation) dt Do example 11.3 Angular Momentum of a Particle in Circular Motion p.339 A particle moves in the xy plane in a circular path of radius r. See textbook p.339 for figure. Find the magnitude and direction of its angular momentum relative to an axis through O when its velocity is v. Q2) A car of mass 1000kg moves with a speed of 50m/s on a circular track of radius 100m. What is the magnitude of its angular momentum relative to the center of the race track? a) 3 10 6 kg m 2 /s b) 5 10 6 kg m 2 /s c) 7 10 6 kg m 2 /s d) 9 10 6 kg m 2 /s
**Q3) A particle whose mass is 2kg moves in the xy plane with a constant speed of 3m/s in the x- direction along the line y=5m. What is its angular momentum relative to the origin? Solution for Q3) y v=3m/s 5 r= xi + 5j 0 x The object does not always have to be in a circular motion to have angular momentum. Any object in motion can have angular momentum by defining an origin relative to its motion!!! velocity will be v=3i(m/s) and the displacement vector r= xi + 5j (m) To find angular momentum, L= r p = r mv = (xi + 5j) m(3i) = (xi + 5j) 6i Since i i=0 and j i= k L= 30k(kg m 2 /s) This means that the magnitude of angular momentum is 30(kg m 2 /s)and the direction is in the negative z-axis Do Example 11.4 A System of Objects A sphere of mass m 1 and a block of m 2 are connected by a light cord that passes over a pulley. The radius of the pulley is R, and the mass of the thin rim is M. The spokes of the pulley has negligible mass. The block slides on the frictionless horizontal surface. Find an expression for the linear acceleration of the two objects, using the concepts of angular momentum and torque. R,M m1 m2
11.3 Angular Momentum of a Rotating Rigid Object see p.342 The z-component of angular momentum of a rigid object rotating about a fixed z axis is L z = Iw Do example 11.5 Bowling Ball p.343 Estimate the magnitude of the angular momentum of a bowling ball spinning at 10rev/s. See p.343 for figure. Q4) A thin spherical shell of radius R = 0.50m and mass 15kg rotates about the z-axis through its center and parallel to its axis. When the angular velocity is 5.0rad/s, its angular momentum( in kg m 2 /s) is approximately a) 15 b) 9.0 c) 12.5 d) 19 e) 25 Q5) A solid cylinder of radius R = 1.0m and mass 10kg rotates about its axis. When its angular velocity is 10rad/s, its angular momentum ( in kg m 2 /s) is a) 50 b) 20 c) 40 d) 25 e) 70 Do example 11.6 The Seesaw p.344. Mass of seesaw is M and length is l. i) Find angular momentum and ii) find the magnitude of the angular acceleration when the seesaw makes angle of θ with the horizontal. m d θ m f
11.4 Analysis Model : Isolated System(Angular Momentum) see p.345 Conservation of Angular Momentum If no external Torque is acting on the rotating system, angular momentum of the system does not change If the rotating object is not rigid, like an ice skater spinning on ice, then the angular speed w will change depending the how the skater changes his/her shape. See p.341! L i = I i w i = I f w f =L f Q6) An ice skater spins with arms outstretched at an angular speed of 1.9rev/s. Her rotational inertia(i) at this time is 1.33kg m 2. She pulls in her arms to increase her rate of spin. If her rotational inertia is 0.48kg m 2 after she pulls in her arms, what is her new angular speed? (hint: the skater angular momentum is constant(=conserved) before and after she pulled in her arms) a) 5.26rev/s b) 4.26rev/s c) 3.26rev/s d) 2.26rev/s Do example 11.7 Formation of a Neutron Star p.347 A star rotates with a period of 30 days about an axis through its center. The period is the time interval required for a point on the star s equator to make one complete revolution around the axis of rotation. After the star undergoes a supernova explosion, the stellar core, which had a radius of 1.0 10 4 km, collapses into a neutron star of radius 3.0km. Determine the period of rotation of the neutron star.
Do example 11.8 The merry-go-round p.348 A horizontal platform in the shape of a circular disk rotates freely in a horizontal plane about a frictionless, vertical axle. The platform has a mass M=100kg and a radius R=2.0m. A student whose mass is m=60kg walks slowly from the rim of the disk towards its center. If the angular speed of the system is 2.0rad/s when the student is at the rim, what is the angular speed when she reaches a point r=0.50m from the center? Do example 11.9 Disk and Stick p.349 A 2.0kg disk traveling at 3.0m/s strikes a 1.0kg stick of length 4.0m that is lying flat on nearly frictionless ice. See p.349 for picture! The disk strikes at the endpoint of the stick, at a distance of r=2.0m from the stick s center. Assume the collision is elastic and the disk does not deviate from its original line of motion. Find the translation speed of the disk, the translational speed of the stick, and the angular speed of the stick after the collision. The moment of inertia of the stick about its center of mass is 1.33kg.m 2.