Rotational Motion & Angular Momentum
Rotational Motion Every quantity that we have studied with translational motion has a rotational counterpart TRANSLATIONAL ROTATIONAL Displacement x Angular Displacement Velocity v Angular velocity acceleration a Angular acceleration Mass m Inertia I Momentum p Angular Momentum L Force F Torque
Angular Position s r Arclength Radius (from axis of rotation) Measured in radians all angular quantities will be measured in radians (NOT degrees)
Angular Position In translational motion, position is represented by a point, such as x. x = 3 x 0 5 p/2 linear In rotational motion, position is represented by an angle, such as, and a radius, r. p r 0 3p/2 angular
Displacement Linear displacement is represented by the vector Dx. Dx = 4 0 5 x p/2 linear Angular displacement is represented by D, which is not a vector, but behaves like one for small values. p D 60 3p/2 0 angular
Angular Displacement D 0 Compare to Dx = x x o Which direction is positive (by convention)? Positive Counterclockwise Negative - Clockwise
EXAMPLE: (a) What is the angular position,, if we go around a circle two times? Ans. = s r = 2(2pr) r = 4π (b) Say you go a quarter turn more, what is D? Ans. D = o = 9π 2 4π = π 2 (c) What is the arclength covered in total between Probls. (a) and (b) if the radius of the circle is 3 m? Ans. s = r = (3 m)(4p + π 2 ) =3m(9π 2 ) = 27π 2 m
P.O.D. 1: A Boy on a horse is hunting a goose around a strange world of radius 25 m. The angular separation between the hunter/hunted is a constant 5. What is the angular distance (in m) between the boy/horse and the goose?
Speed and velocity s v T The same particle rotates with an avg. angular velocity given by v T D r = Dθ Dt Linear (Tangential) and angular speeds are related by the equation v = r
Angular Velocity avg D Dt Compare to v avg = x t Units rad/s, rev/s Direction Same as displacement (positive is counterclockwise, negative is clockwise)
EXAMPLE: A space station ring of radius 500 m spins twice in 12 minutes. (a) Find its angular velocity (in rads/s). (b) Find the linear (tangential) velocity (in m/s) of a point on the outer edge of the ring. Ans. (a) Twice means two revolutions = 2(2p) = 4 p radians = θ t = 4π 12 min 60 Τ sec min = 0.017 rad/sec (b) tangential velocity, v T = r = 0.017 (500 m) = 8.73 m/s
P.O.D. 2: A figure skater spins through five revolutions in a time of 2.4 s. (a) Find her angular velocity (in rads/s). (b) Find the linear (tangential) velocity (in m/s) of her foot if it is 0.3 meters away from her body.
Acceleration s v T The same particle rotates with an avg. angular acceleration given by v T D r = D Dt Linear (Tangential) and angular accelerations are related by the equation a = r
Angular Acceleration avg D Dt Compare to a avg = v t Angular Acceleration is how the angular velocity changes with time Units rad/s 2
Centripetal Acceleration F c F g A pendulum is swinging back and forth. At the bottom of the swing the force of gravity is pulling it downwards but it doesn t fall down. This means there must be a force pulling upwards to balance it out. This is the centripetal force. Since F = m a, the center-seeking acceleration is called centripetal acceleration and is given by: v 2 a 2 r c r
Sample Problem A compact disk rotates about an axis according to the formula: (t) = t 2 6 (a) What is the linear speed of a point 20 cm from the center at t = 5 s? (b) What is the linear acceleration at 0.5 m at t = 5 s? Ans. (a) To find the linear speed, use the relationship, where (5) we obtained from the function (t) = t 2 6 (5)= 5 2 6=19 v = r = (19 rad/s)(0.20 m) = 3.8 m/s. (b) To find the linear acceleration, use the relationship a c = 2 r = (19 rad/s) 2 (0.20 m) = 72.2 m/s 2
P.O.D. 3: Diana Prince starts rotating with an angular velocity of 5 rad/s, where the negative sign indicates a clockwise rotation, to transform into Wonder Woman. After 5 seconds she has completed the transformation and her angular velocity is 2 rad/s. (a) Find Wonder Woman s angular acceleration (in rad/s 2 ). (b) Find the centripetal acceleration (in m/s 2 ) of her indestructible bracelets (at = 2 rad/s) if they are 0.8 m away from her body when she is spinning.
Constant Angular Acceleration Our kinematics equations have angular equivalents Just as with their linear counterparts, these only work for constant acceleration
First Kinematic Equation x = v o t + ½ at 2 (linear form) Substitute angle for position. Substitute angular velocity for linear velocity. Substitute angular acceleration for linear acceleration. = o t + ½ t 2 (angular form)
Second Kinematic Equation v = v o + at (linear form) Substitute angular velocity for linear velocity. Substitute angular acceleration for linear acceleration. = o + t (angular form)
Sample Problem A bicycle starts from rest and for 10.0 s has a constant linear acceleration of 0.8 m/s 2 to the right. During this period, the tires do not slip. The radius of the tires is 0.50 m. At the end of the 10.0 s interval what is the angle through which each wheel has rotated? Ans. The angular acceleration can be found from the formula a = r α = a α = 0.8 m s 2 = 1.6 rad/s2 r 0.50 m The angular acceleration should be negative because the tire spins clockwise. To find the angular displacement: (t) = o t + ½ t 2 = 0(10 s) + ½( 1.6 rad/s 2 )(10 s) 2 = 80 rad
P. O. D. 4: an extreme diver rotates at an angular velocity of +3.75 rad/s while doing his first set of flips. When he does his second set of flips, he accelerates and reaches a greater angular velocity of +4.4 rads/s. If his angular acceleration is 1.7 rad/s 2... (a) How long does this flipping part of the dive take (in s)? (b) Find his angular displacement after the time found in (a).
ANGULAR MOMENTUM!
Angular Momentum Angular momentum depends on linear momentum and the distance from a particular point. It is a vector quantity with symbol L. If r and v are then the magnitude of angular momentum w/ resp. to point Q is given by L = rp = mvr. In this case L points out of the page. If the mass were moving in the opposite direction, L would point into the page. Q r v m The SI unit for angular momentum is the kg m 2 / s. (It has no special name.) Angular momentum is a conserved quantity. A torque is needed to change L, just a force is needed to change p. Anything spinning has angular has angular momentum. The more it has, the harder it is to stop it from spinning.
Angular Momentum: General Definition If r and v are not then the angle between these two vectors must be taken into account. The general definition of angular momentum is given by a vector cross product: L = r p This formula works regardless of the angle. From cross products, the magnitude of the angular momentum of m relative to point Q is: L = r p sin = m v r. In this case, by the right-hand rule, L points out of the page. If the mass were moving in the opposite direction, L would point into the page. v r m Q
Moment of Inertia vs. Angular Momentum Any moving body has inertia. (It wants to keep moving at constant v.) The more inertia a body has, the harder it is to change its linear motion. Rotating bodies possess a rotational inertia called the moment of inertia, I. The more rotational inertia a body has, the harder it is change its rotation. For a single point-like mass w/ respect to a given point Q, I = mr 2. For a system, I = the sum of each mass m times its respective distance from the point of interest. r I = mr 2 Q m 1 r 1 Q r 2 m 2 I = m r 2 = m r 2 + m r 2
Moment of Inertia of various shapes
Moment of Inertia Example Two merry-go-rounds have the same mass and are spinning with the same angular velocity. One is solid wood (a disc), and the other is a metal ring. Which has a bigger moment of inertia relative to its center of mass? r r m m Ans. I is independent of the angular speed. Since their masses and radii are the same, the ring has a greater moment of inertia. This is because more of its mass is farther from the axis of rotation. Since I is bigger for the ring, it would more difficult to increase or decrease its angular speed.
Torque & Angular Acceleration Newton s 2 nd Law, as you know, is F net = ma The 2 nd Law has a rotational analog: net = I A force is required for a body to undergo acceleration. A turning force (a torque) is required for a body to undergo angular acceleration. The bigger a body s mass, the more force is required to accelerate it. Similarly, the bigger a body s rotational inertia, the more torque is required to accelerate it angularly. Both m and I are measures of a body s inertia (resistance to change in motion).
Example: The torque of an Electric Saw Motor The motor in an electric saw brings the circular blade up to the rated angular speed of 80.0 rev/s in 240.0 rev. One type of blade has a moment of inertia of 1.41 x 10-3 kg m 2. What net torque (assumed constant) must the motor apply to the blade? Ans. First we need to convert our values into rad/s for calculation purposes. o t 240 rev x 2p = 1508 rads? 80 revs/s x 2p = 503 rad/s 0 rad/s? We can find the angular acceleration from 2 = o2 + 2 Solving for : = 2 2 o = (503 rad/s)2 0 2 = 83.89 rad/s 2 2(1508 rad/s) 2 = I = 1.41 x 10-3 kg m 2 83.89 rad/s 2 = 0.118 N m
PROBLEM 6: A Chinese star of mass 0.025 kg and radius 0.03 m is thrown by Bruce Lee at his adversary. The Chinese star is thrown from rest. If its final angular velocity is 15 revs/s after 3 sec, (a) find the angular acceleration of the Chinese star. (b) Find the torque of the Chinese star (Assume the Chinese star is hoop-shaped).
Linear Momentum vs. Angular Momentum If a net force acts on an object, it must accelerate, which means its momentum must change. Similarly, if a net torque acts on a body, it undergoes angular acceleration, which means its angular momentum changes. Recall, angular momentum s magnitude is given by L = mvr (if v and r are perpendicular) r v m So, if a net torque is applied, angular velocity must change, which changes angular momentum. Proof: net = r F net = r m a = r m Dv = DL t t So net torque is the rate of change of angular momentum, just as net force is the rate of change of linear momentum. continued on next slide
Linear & Angular Momentum (cont.) Here is yet another pair of similar equations, one linear, one rotational. From the formula v = r, we get L = m v r = m ( r) r = m r 2 = I This is very much like p = m v, and this is one reason I is defined the way it is. In terms of magnitudes, linear momentum is inertia times speed, and angular momentum is rotational inertia times angular speed. L = I p = m v
Comparison: Linear & Angular Momentum Linear Momentum, p Tendency for a mass to continue moving in a straight line. Parallel to v. A conserved, vector quantity. Magnitude is inertia (mass) times speed. Net force required to change it. The greater the mass, the greater the force needed to change momentum. Angular Momentum, L Tendency for a mass to continue rotating. Perpendicular to both v and r. A conserved, vector quantity. Magnitude is rotational inertia times angular speed. Net torque required to change it. The greater the moment of inertia, the greater the torque needed to change angular momentum.
Example: Spinning Ice Skater Suppose Mr. Stickman is sitting on a stool that swivels holding a pair of dumbbells. His axis of rotation is vertical. With the weights far from that axis, his moment of inertia is 600 kg m 2 and he is spinning at an angular velocity of 20 rad/s. When he pulls his arms in as he s spinning, the weights are closer to the axis, so his moment of inertia gets to 400 kg m 2. What will be his angular velocity at this moment? Ans. I = L = I 600 kg m 2 (20 rad/s) = 400 kg m 2 12,000 = 400 30 rad/s =
PROBLEM 7: An artificial satellite (m = 1500 kg) is placed into an elliptical orbit about the earth. Telemetry data indicate that its point of closes approach (called the perigee) is r p = 8.37 x 10 6 m from the center of the earth, while its point of greatest distance (called the apogee) is r A = 25.1 x 10 6 m from the center of the earth. The speed of the satellite at the perigee is v p = 8450 m/s. (a) Find its speed v A at the apogee. (b)find its angular momentum at any point in its orbit.
Rotational Kinetic Energy A particle in a rotating object has rotational kinetic energy: K i = ½ m i v i 2, where v i = i r (tangential velocity) The whole rotating object has a rotational kinetic energy given by: 1 K K m r 2 2 R i i i i i 2 1 1 KR m r I 2 i 2 2 2 2 i i
Rotational Kinetic Energy example: A thin walled hollow cylinder (mass = m h, radius = r h ) and a solid cylinder ( mass = m s, radius = r s ) start from rest at the top of an incline. Both cylinders start at the same vertical height h o. All heights are measured relative to an arbitrarily chosen zero level that passes through the center of mass of a cylinder when it is at the bottom of the incline. Ignoring energy losses due to retarding forces, determine which cylinder has the greatest translational speed upon reaching the bottom.
Rotational Kinetic Energy example (cont.): Ans. At the top of the incline the cylinder have only gravitational potential energy. At the bottom of the incline this energy has converted into translational kinetic and rotational kinetic energy. E in = E out GPE in = TKE out + RKE out mgh = ½mv f2 + ½I f 2 The angular velocity can be related to the linear velocity v f by f = v f r Substituting the given values and for the angular velocity: m h gh o = ½m h v f2 + ½I( v f r h ) 2 For the hollow cylinder, the moment of inertia is given by: I = m r 2 Substituting: m h gh o = ½m h v f2 + ½(m h r h2 )( v f r h ) 2 Simplifying: gh o = ½v f2 + ½ r h2 ( v f r h ) 2 gh o = ½v f2 + ½ r h2 v f2 gh o = ½v f2 + ½ v f 2 gh o = v f 2 gh o = v f r h 2
Rotational Kinetic Energy example (cont.): Ans. At the top of the incline the cylinder have only gravitational potential energy. At the bottom of the incline this energy has converted into translational kinetic and rotational kinetic energy. E in = E out GPE in = TKE out + RKE out mgh = ½mv f2 + ½I f 2 The angular velocity can be related to the linear velocity v f by f = v f r Substituting the given values and for the angular velocity: m s gh o = ½m s v f2 + ½I( v f r s ) 2 For the solid cylinder, the moment of inertia is given by: I = ½m r 2 Substituting: m s gh o = ½m s v f2 + ½(½ m s r s2 )( v f r s ) 2 Simplifying: 4 gh o = ½v f2 + ½ ½ r s2 ( v f r s ) 2 gh o = ½v f2 + ¼ r s2 v f2 gh o = ½v f2 + ¼ v f 2 gh o = ¾v f 2 3 gh o= v f r s 2
PROBLEM 8: