advertisement

Physics 207: Lecture 22 Today’s Agenda z Elastic collisions in two dimensions ÍExamples (billiards, nuclear scattering) z Rotational Kinematics ÍAnalogy with one-dimensional kinematics z Kinetic energy of a rotating system ÍMoment of inertia ÍDiscrete particles ÍContinuous solid objects z Parallel axis theorem 1 2-D Elastic Collision: Nuclear Scattering z A particle of unknown mass M is initially at rest. A particle of known mass m is “shot” at it with initial momentum pi . After the particles collide, the new momentum of the shot particle pf is measured. Í Figure out what M is in terms of pi and pf and m. at rest m P M M pi m pf initial final 2 Page 1 2-D Elastic Collision: Nuclear Scattering z z Using momentum conservation: pi = pf + P Í So P2 = (pi -pf )2 pf P pi Using kinetic energy conservation: p i2 p2 P2 = f + 2m 2m 2M and using ⎛ p2 p2 ⎞ P 2 = 2M ⎜ i − f ⎟ ⎝ 2m 2m ⎠ 2 P 2 = (p i − p f ) ⎡ (p − p )2 ⎤ M = m ⎢ i2 f 2 ⎥ ⎢⎣ p i − p f ⎥⎦ 3 2-D Elastic Collision: Nuclear Scattering z ⎡ (pi − pf )2 ⎤ So we find that M = m ⎢ 2 2 ⎥ ⎢⎣ p i − p f ⎥⎦ pf P pi z If we measure pi and pf and we know m we can measure M. Í We can learn about something we can’t see! z This is the basic idea behind a large body of work done in atomic, nuclear and particle physics. 4 Page 2 Rutherford Backscattering z Shoot a beam of α particles (helium nuclei) having known energy Ei into a sample of unknown composition. Measure the energy Ef of the α particles that bounce back out at ~180o with respect to the incoming beam. Ei unknown stuff Ei particle detector (measures energy) 5 Rutherford Backscattering pf ⎡ (p − p )2 ⎤ M = m ⎢ i2 f 2 ⎥ ⎢⎣ p i − p f ⎥⎦ z P pi In the 180o case, this simplifies significantly: ⎡ (p + p )2 ⎤ (v + v f )(v i + v f ) M = m ⎢ i2 f 2 ⎥ = m i − (v i + v f )(v i − v f ) p f ⎥⎦ ⎢⎣ p i M =m (v i (v i + vf ) − vf ) vf = v i 2 2 1 mv f ⎛M − m⎞ Ef ⎟ ⎜ = 2 = 1 mv i 2 ⎝ M + m ⎠ Ei 2 (M − m ) (M + m ) Solving for M: ⎛ ⎞ m⎜1 + Ef ⎟ E i ⎠ M= ⎝ ⎛ ⎞ E f ⎜1 − Ei ⎟⎠ ⎝ 6 Page 3 Rutherford Backscattering z Shoot a beam of α particles (helium nuclei) having known energy Ei into a sample of unknown composition. Measure the energy Ef of the α particles that bounce back out at ~180o with respect to the incoming beam. Ei unknown stuff Ei particle detector (measures energy) ⎛ ⎞ m⎜1 + Ef ⎟ E i ⎠ M= ⎝ ⎛ ⎞ E f ⎜1 − Ei ⎟⎠ ⎝ So we learn about the mass of the nuclei in the unknown stuff. (We learn what the stuff is). 7 Rutherford Backscattering z z For example: Suppose we are shooting α particles that have an initial energy of Ei = 2 MeV at a target made of an unknown material. The α particles return with final energy Ef = 1.1 MeV. What is the weight of the unknown material? Í m(α) = 4 (2 protons, 2 neutrons) So ⎛ ⎞ m⎜1 + Ef ⎟ Ei ⎠ ⎝ M= ⎛ Ef ⎞ ⎜1 − Ei ⎟⎠ ⎝ = ( (1 − ) 1.1 ) 2 4 1 + 1.1 2 M = 27 Aluminum!! (13 protons, 14 neutrons) 8 Page 4 Another example of 2-D elastic collisions: Billiards. z If all we know is the initial velocity of the cue ball, we don’t have enough information to solve for the exact paths after the collision. But we can learn some useful things... 9 Billiards. z Consider the case where one ball is initially at rest. pf pi vcm Pf F initial The final direction of the red ball will depend on where the balls hit. final 10 Page 5 Billiards z We know momentum is conserved: pi = pf + Pf pi2 = (pf + Pf )2 = pf2 + Pf2 + 2 pf • Pf z We also know that kinetic energy is conserved: p 2i p2 P 2 = f + f 2m 2m 2m z p 2i = pf2 + Pf 2 Comparing these two equations tells us that: pf • Pf = 0 Pf pf Therefore, pi and pf must be orthogonal! pi 11 Billiards. z The final directions are separated by 90o. pf pi vcm Pf F initial final 12 Page 6 Billiards. z So, we can sink the red ball without sinking the white ball. 13 Billiards. z z So, we can sink the red ball without sinking the white ball. However, we can also scratch. All we know is that the angle between the balls is 90o. 14 Page 7 Billiards. z Tip: If you shoot a ball spotted on the “dot”, you will sink both balls ! 15 Lecture 22, Act 1 Elastic Collisions in 2-D z A moving ball initially traveling in the direction shown hits an identical but stationary ball. The collision is elastic. ÍDescribe one possible direction of both balls just after the collision. (a) (b) (c) 16 Page 8 Lecture 22, Act 1 Solution z In the first solution, the angle between the balls is not 90o. z In the second solution, there are no downward y components to balance out the upward y components. 17 Lecture 22, Act 1 Solution z The third choice both balances the y components and has 90o between the final direction vectors of the two balls. z As a result, the third choice is the only one of the three that fits all necessary criteria. 18 Page 9 Rotation z Up until now we have gracefully avoided dealing with the rotation of objects. ÍWe have studied objects that slide, not roll. ÍWe have assumed pulleys are without mass. z Rotation is extremely important, however, and we need to understand it! z Most of the equations we will develop are simply rotational analogues of ones we have already learned when studying linear kinematics and dynamics. 19 Lecture 22, Act 2 Rotations z Bonnie sits on the outer rim of a merry-go-round, and Klyde sits midway between the center and the rim. The merry-goround makes one complete revolution every two seconds. ÍKlyde’s angular velocity is: (a) the same as Bonnie’s (b) twice Bonnie’s (c) half Bonnie’s 20 Page 10 Lecture 22, Act 2 Rotations z The angular velocity ω of any point on a solid object rotating about a fixed axis is the same. ÍBoth Bonnie & Klyde go around once (2π radians) every two seconds. (Their “linear” speed v will be different since v = ωr). ω 1 VKlyde = VBonnie 2 21 Rotational Variables. z z Rotation about a fixed axis: ÍConsider a disk rotating about an axis through its center: First, recall what we learned about Uniform Circular Motion: ω= θ ω dθ dt (Analogous to v = dx ) dt 22 Page 11 Rotational Variables... z z z Now suppose ω can change as a function of time: We define the dω d 2θ = 2 angular acceleration: α = dt dt Consider the case when α is constant. Í We can integrate this to find ω and θ as a function of time: α = constant ω = ω 0 + αt θ ω α 1 2 θ = θ 0 + ω 0t + αt 2 23 Rotational Variables... α = constant ω = ω 0 + αt v 1 θ = θ 0 + ω0 t + αt 2 2 Recall also that for a point at a distance R away from the axis of rotation: Íx = θR Ív = ωR And taking the derivative of this we find: Ía = αR θ R x ω z α 24 Page 12 Summary (with comparison to 1-D kinematics) Angular Linear a = constant α = constant ω = ω 0 + αt v = v 0 + at 1 θ = θ0 + ω 0 t + αt 2 2 x = x0 + v 0t + 1 2 at 2 And for a point at a distance R from the rotation axis: x = Rθ v = ωR a = αR 25 Example: Wheel And Rope z A wheel with radius R = 0.4 m rotates freely about a fixed axle. There is a rope wound around the wheel. Starting from rest at t = 0, the rope is pulled such that it has a constant acceleration a = 4 m/s2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2π radians) a R 26 Page 13 Wheel And Rope... Use a = αR to find α: α = a / R = 4 m/s2 / 0.4 m = 10 rad/s2 z Now use the equations we derived above just as you would use the kinematic equations from the beginning of the semester. 1 1 θ = θ 0 + ω 0 t + α t 2 = 0 + 0(10) + (10)(10)2 = 500 rad 2 2 z = 500 rad x 1 rot 2π rad a ≈ 80 rev α R 27 Rotation & Kinetic Energy z z Consider the simple rotating system shown below. (Assume the masses are attached to the rotation axis by massless rigid rods). The kinetic energy of this system will be the sum of the kinetic energy of each piece: m4 m3 r1 m1 ω r4 r3 r2 m2 28 Page 14 Rotation & Kinetic Energy... z 1 K = ∑ mi v i2 i 2 So: K= but vi = ωri 1 1 2 2 2 ∑ m i (ωri ) = ω ∑ m i ri 2 i 2 i v1 which we write as: K= m4 1 I ω2 2 v4 m3 I = ∑ mi ri 2 v2 r2 r3 i Define the moment of inertia about the rotation axis r1 m1 ω r4 m2 v3 I has units of kg m2. 29 Rotation & Kinetic Energy... z The kinetic energy of a rotating system looks similar to that of a point particle: Point Particle K = 1 mv 2 2 v is “linear” velocity m is the mass. Rotating System 1 I ω2 2 ω is angular velocity I is the moment of inertia about the rotation axis. K= I = ∑ mi ri 2 i 30 Page 15 Moment of Inertia z So K = 2 where I = ∑ mi ri 1 I ω2 2 i z Notice that the moment of inertia I depends on the distribution of mass in the system. ÍThe further the mass is from the rotation axis, the bigger the moment of inertia. z For a given object, the moment of inertia will depend on where we choose the rotation axis (unlike the center of mass). z We will see that in rotational dynamics, the moment of inertia I appears in the same way that mass m does when we study linear dynamics! 31 Calculating Moment of Inertia z We have shown that for N discrete point masses distributed about a fixed axis, the moment of inertia is: I= N ∑ mi ri 2 where r is the distance from the mass to the axis of rotation. i =1 Example: Calculate the moment of inertia of four point masses (m) on the corners of a square whose sides have length L, about a perpendicular axis through the center of the square: m m m m L Page 16 32 Calculating Moment of Inertia... z The squared distance from each point mass to the axis is: 2 L2 ⎛L⎞ r 2 = 2⎜ ⎟ = 2 ⎝2⎠ N so I = ∑ mi ri 2 = m i =1 Using the Pythagorean Theorem L2 L2 L2 L2 L2 + m + m + m = 4m 2 2 2 2 2 I = 2mL2 L/2 m m r L m m 33 Calculating Moment of Inertia... z Now calculate I for the same object about an axis through the center, parallel to the plane (as shown): N I = ∑ mi ri = m i =1 2 L2 L2 L2 L2 L2 + m + m + m = 4m 4 4 4 4 4 r I = mL2 m m m m L 34 Page 17 Calculating Moment of Inertia... z Finally, calculate I for the same object about an axis along one side (as shown): N I = ∑ mi ri = mL2 + mL2 + m0 2 + m0 2 2 i =1 r I = 2mL2 m m m m L 35 Calculating Moment of Inertia... z For a single object, I clearly depends on the rotation axis!! I = 2mL2 I = mL2 m m m m I = 2mL2 L 36 Page 18 Lecture 22, Act 3 Moment of Inertia A triangular shape is made from identical balls and identical rigid, massless rods as shown. The moment of inertia about the a, b, and c axes is Ia, Ib, and Ic respectively. ÍWhich of the following is correct: z a (a) Ia > Ib > Ic (b) Ia > Ic > Ib b (c) Ib > Ia > Ic c 37 Lecture 22, Act 3 Moment of Inertia z Label masses and lengths: z Calculate moments of inertia: I a = m (2 L ) + m (2 L ) = 8 mL2 2 2 I b = mL2 + mL2 + mL2 = 3 mL2 I c = m (2 L ) = 4 mL2 2 m a L b So (b) is correct: Ia > Ic > Ib L c m m 38 Page 19 Calculating Moment of Inertia... z For a discrete collection of point masses we found: I= N ∑ mi ri 2 i =1 z For a continuous solid object we have to add up the mr2 contribution for every infinitesimal mass element dm. ÍWe have to do an integral to find I : dm I = ∫ r 2 dm r 39 Moments of Inertia z Some examples of I for solid objects: I = MR 2 R R I= Thin hoop (or cylinder) of mass M and radius R, about an axis through its center, perpendicular to the plane of the hoop. 1 MR 2 2 Thin hoop of mass M and radius R, about an axis through a diameter. 40 Page 20 Moments of Inertia... z Some examples of I for solid objects: 2 I = MR 2 5 Solid sphere of mass M and radius R, R about an axis through its center. I= 1 MR 2 2 Solid disk or cylinder of mass M and radius R, about a perpendicular axis through its center. R 41 Lecture 22, Act 4 Moment of Inertia z Two spheres have the same radius and equal masses. One is made of solid aluminum, and the other is made from a hollow shell of gold. ÍWhich one has the biggest moment of inertia about an axis through its center? (a) solid aluminum (b) hollow gold (c) same hollow solid same mass & radius 42 Page 21 Lecture 22, Act 4 Moment of Inertia z Moment of inertia depends on mass (same for both) and distance from axis squared, which is bigger for the shell since its mass is located farther from the center. ÍThe spherical shell (gold) will have a bigger moment of inertia. ISOLID < ISHELL hollow solid same mass & radius 43 Moments of Inertia... z Some examples of I for solid objects (see textbook, Table 10-1): I= L L 1 ML2 12 Thin rod of mass M and length L, about a perpendicular axis through its center. 1 I = ML2 3 Thin rod of mass M and length L, about a perpendicular axis through its end. 44 Page 22 Parallel Axis Theorem z z Suppose the moment of inertia of a solid object of mass M about an axis through the center of mass, ICM, is known. The moment of inertia about an axis parallel to this axis but a distance D away is given by: IPARALLEL = ICM + MD2 z So if we know ICM , it is easy to calculate the moment of inertia about a parallel axis. 45 Parallel Axis Theorem: Example z Consider a thin uniform rod of mass M and length D. Figure out the moment of inertia about an axis through the end of the rod. D=L/2 M CM IPARALLEL = ICM + MD2 x L We know ICM = So 1 ML2 12 IEND = IEND ICM 1 L 2 1 ML2 + M ⎛⎜ ⎞⎟ = ML2 ⎝2⎠ 12 3 which agrees with the result on a previous slide. 46 Page 23