7A supplemental problems

Transcription

1 7A supplemental problems Claire Zukowski December 11, Kinematics 1.1 Mud on a wheel (F&C 4.7) Problem: A wheel of radius b rolling forward at velocity v 0 throws small clumps of mud from its rim. What is the greatest height above the ground the mud can reach, and at what point on the wheel must the mud be ejected to attain this maximum? Solution: The velocity of mud leaving the rim of the wheel will be the same as the forward velocity v 0. The maximum height at a given angle h(θ) can be found using the kinematics equations (here we define θ as the angle from the vertical) h(θ) = h 0 + (v 0 sin θ). (1.1) g Note that we have set v f to zero since h(θ) is evaluated at the highest point the mud reaches before reversing direction. From geometry we see that h 0 = b + b cos θ, so h(θ) = b + b cos θ + (v 0 sin θ). (1.) g Setting the derivative to zero to maximize, and using θ to denote the maximum angle, we find dh dθ = b sin θ + (v 0 sin θ cos θ ) θ g = 0 cos θ = gb. (1.3) v0 Plugging this into our expression for the height, we find the maximum of all maximum heights, h max = b + v 0 g + gb. (1.4) v 0 1

2 Force and friction.1 Double Atwood (G 4.56) Problem: The double Atwood machine is depicted to the left. Assuming no friction and massless pulleys and cords, find (a) the acceleration of masses m 1, m and m 3, and (b) the tensions T 1 and T in the cords. Solution: Taking the positive direction as up, the force equations for the masses are T 1 m 1 g = m 1 a 1, T 1 m g = m a, T 3 m 3 g = m 3 a 3. (.1) We also have a force equation for the lower pulley, T 1 = T 3. (.) We have four equations and five unknowns, so we need a fifth equation. We can find a constraint equation relating the three accelerations as follows. Let x 1, x and x 3 be the positions of the three masses below the top pulley, and let x P be the position of the bottom pulley below the top pulley. If the pulleys have radius R, the lengths of the top and bottom cords l 1 and l are given in terms of these positions (which in general vary in time) by l 1 = x P + πr + x 3 l = x 1 + πr + x x P. (.3) The lengths are constant, so the constraints can be obtained by taking derivatives, Thus the accelerations are related as ẍ P + ẍ 3 = 0 ẍ P = ẍ 3, ẍ 1 + ẍ = ẍ P = ẍ 3 (.4) a 1 + a = a 3. (.5) Combined with the force equations, we can solve these five equations simultaneously to obtain the tensions and accelerations of all the masses. Note that the trick of differentiating lengths of ropes is very useful in finding constraints relating accelerations in pulley problems.

3 . Accelerating wedge Problem: A block of mass m rests on a wedge of mass M inclined at an angle θ, which sits on a table. A force F is applied horizontally. The coefficient of static friction between the block and wedge is µ s and the coefficient of kinetic friction between the wedge and the table is µ k. (a) Find the maximum force that can be applied such that the mass does not move up the incline (b) Find the minimum force that must be applied such that the mass does not start to move down the incline. Solution: The free body diagrams for the two masses are drawn to the right. Notice that we picked our coordinate system for mass m to be in line with the acceleration rather than tipped. This makes the algebra much simpler since we won t need to break the acceleration into components. Figure 1: The accelerating wedge. Figure : The free body diagrams. For mass the m, the x- and y-components of the force equation are For mass the M, they are N 1 sin θ + f s cos θ = ma N 1 cos θ f s sin θ mg = 0. (.6) F max N 1 sin θ f s cos θ f k = Ma N + f s sin θ N 1 cos θ Mg = 0. (.7) At F = F max the static friction force attains its maximum value, f s = µ s N 1. Also, we know f k = µ k N. Plugging these into the first set of equations we find N 1 = mg cos θ µ s sin θ, a = N 1 m ( sin θ + µ s cos θ). (.8) Plugging these into the last two and rearranging we find [ ] sin θ + µs cos θ F max = (m + M)g cos θ µ s sin θ + µ k. (.9) 3

4 You can check that this answer makes sense in the limit θ 0. Now, for the second case you can save yourself some algebra by noticing that this is the same problem with the replacements f s f s, or equivalently, µ µ. The answer is.3 Massive whirling rope [ ] sin θ µs cos θ F min = (m + M)g cos θ + µ s sin θ + µ k. (.10) Problem: A uniform rope of mass M and length L whirls at a constant angular velocity ω around a pivot point at one end of the rope. Neglecting gravity, find the tension T (r) a distance r from the pivot. Solution: Consider an infinitesimal length of rope r a length r + r/ away from the pivot. The mass of the segment will be m = λ r where λ = M/L is the linear mass density of the rope. The forces on the segment will be an inward pointing force T (r) and an outward pointing force T (r + r). Thus the force equation (replacing r + r/ with r for the radius since they will be the same in the limit r 0) gives T (r + r) T (r) = ( m)rω = λ( r)rω. (.11) Dividing by r and taking the limit as r 0 we obtain dt dr = lim T (r + r) T (r) r 0 r = λrω. (.1) To integrate, move like quantities to the same side and be careful to use dummy indices and appropriate limits of integration, T (r) T 0 dt = λω r 0 r dr (.13) T (r) = T 0 λω r. (.14) To evaluate the arbitrary constant we need impose a boundary condition, which we realize is T = 0 at r = L (since there is no extra rope pulling on it from the other side). Thus we find T 0 = λω L /, so the tension at any point is.4 Capstan (K&K.4, final review) T (r) = λω (L r ). (.15) Problem: A capstan is a device used on ships that controls ropes under great tension. The rope is wrapped around a drum through a certain angle θ (often several times around) and the load pulls with a force T A while the sailor holds the rope with a smaller force T B. Find an expression for T B in terms of T A, the coefficient of static 4

5 friction µ between the rope and the capstan, and θ. Solution: The forces acting on a piece of the rope that covers an angle θ are an outward normal force, tensions T and T + T and a friction force f. Since this is a statics problem the sum of forces is ( ) ( ) θ θ f + (T + T ) cos T cos = 0, N (T + T ) sin ( θ ) T sin ( θ ) = 0. (.16) Using the small angle approximation sin ( θ/) θ/ and cos ( θ/) 1, we find f = T, N = T θ. (.17) Figure 3: The capstan. Figure 4: The capstan force diagram. Setting f = µn at the limit where the system is just barely stationary, we find Taking the limit as θ 0, we find T θ = µt. (.18) dt dθ = µt T = Ae µθ, (.19) where A is a constant. But we know T = T A at θ = 0, so A = T A and 3 Energy conservation T B = T A e µθ. (.0) 3.1 Pendulum hitting pivot point (G 8.86, quiz #) Problem: A small mass m is held horizontally at rest at the end of a massless string of length l. A peg is located a distance h vertically below the other end of 5

6 the string. The mass is released and, when the string is vertical, the string catches on the peg and the mass describes a new circular path with the peg as its center. What is the critical distance h such that the mass just barely completes a half circle? Solution: Let s put the zero of gravitational potential at the lowest point in the pendulum s trajectory. By energy conservation, comparing the initial energy with the energy when the mass completes a half circle, we have mgl = mg(r) + 1 mv (3.1) where v is the velocity at the top and r = l h is the radius of the smaller circle. (Be careful since to not set the velocity to zero at the top point.) Our radial force equation also tells us that at this point, i F i,r = T + mg = mv r. (3.) The critical condition is that T = 0, since otherwise the string would go slack and the mass would not reach the top. In this case we find v = gr. Plugging this into the energy equation, we can solve to find the critical distance, h = 3 l. (3.3) 5 3. Block on sphere Problem: A small mass sits balanced on top of a fixed frictionless sphere of radius R. The system is perturbed and the mass slides down. At what angle from the vertical does the mass lose contact with the sphere? Solution: The radial force equation at an angle θ from the vertical line passing through the center of the sphere is N mg cos θ = mv R, (3.4) where we have taken the outward direction as positive (thus, the centripetal acceleration is negative). The normal forces vanishes (N = 0) at the point the mass loses contact, giving v = g cos θ. (3.5) R We don t know v, but we can solve for it using energy conservation. We have KE = U 1 mv = mgr(1 cos θ). (3.6) 6

7 Combining, we can solve for the angle at which the block loses contact with the sphere, cos θ = 3. (3.7) 4 Gravity 4.1 Gravitation force due to a disk (quiz #) Problem: Find the gravitational field a distance z above a uniform disk of mass M and radius R. Solution: Divide the disk up into infinitesimal rings of mass dm and radius r. We first need to compute the gravitational force due to a single ring on a test mass m a distance z above the xy-plane. By symmetry, all the forces in the x-direction cancel, so we are left with a net y-component of gravity. Since all points on the ring are the same distance r = z + R away from the mass, the y force is just df g = GdMm cos θ = GdMmz (4.1) r (z + r ) 3/ (here cos θ = z/r is the angle between r and the z-axis). Also, the infinitesimal mass element is given by dm = σda = M Mrdr πrdr =. (4.) πr R We can compute the total force by integrating over all rings, F g = df g = R F g = GMm R 0 GdMmz GMmz = (z + r ) 3/ R ( ) z 1 R + z R 0 rdr (z + r ) 3/ ( ŷ direction). (4.3) The field is just F g /m, independent of the mass of our test charge. Let s check the limit z to make sure our answer makes sense. In this limit, z R + z = ( R z ) where we have used a Taylor expansion for small R/z. This gives F g GMm R R z ( ) R (4.4) z = GMm z. (4.5) Thus, in the limit z the gravitational field reduces to that of a point mass, which is what we would expect! 7

8 4. Giant projectile (K&K 9.8, final review) Problem: A projectile of mass m is fired from the surface of the Earth at an angle α from the vertical, at an initial velocity v 0 = GM/R where M and R are the mass and radius of the Earth. Neglecting air resistance as well as the rotation and motion of the Earth, how high does the projectile rise? Solution: Let r denote the maximum distance from the Earth s center. We need two equations since we have two unknowns, r and the final velocity. Since the only force acting on the projectile is gravity which is directed through the Earth s center, there is zero net torque about the center of the Earth. Thus, angular momentum about this point is conserved, mrv 0 sin α = mv f r v f = Rv 0 sin α. (4.6) r Energy is also conserved. This gives, plugging in our values for v 0 and v f, 1 mv 0 GmM }{{} R = 1 R r mv 0 sin α GmM r GmM/R GmM r = 1 GmM (1 + R R r ) sin α r Rr + R sin α = 0 r = R ± 1 4R 4R sin α = R ± R cos α. (4.7) Only the positive root is physical, so we obtain r = R(1 + cos α), in other words the projectile attains a maximum height R cos α. 5 Collisions 5.1 Double ball drop (K&K 4.3) Problem: A ball of a small mass m is placed on top of a ball of mass M m and the two are dropped to the floor from a height H above the ground. Assume that the collision between the two balls (in the instant after the mass M hits the ground) is elastic. How high does mass m rise after collision? Solution: From energy conservation, the two masses reach the ground at velocity v given by (M + m)gh = 1 (M + m)v 0 v 0 = gh. (5.1) 8

9 The problem is easier if we assume the balls are slightly separated when mass M hits the ground. Then, assuming the ground is perfectly reflective, the mass M will rebound at velocity v 0 (positive direction taken downwards) while the mass m continues with velocity v 0. An instant later they collide. Momentum conservation gives mv 0 Mv 0 = mv m + Mv M (5.) and, since the collision is elastic, energy conservation gives 1 mv Mv 0 = 1 mv m + 1 Mv M. (5.3) After some messy algebra, we can solve these two equations for the unknowns v m and v M. In fact, we are only interested in v m, which we find is v m = 3v = 3 gh in the limit m/m 0 (negative means upwards since we are taking down as positive). Using energy conservation one more time, we find 1 mv m = mgh H = 9H. (5.4) Thus the little mass rises much higher than it started out! Note that we could simplify the algebra considerably by using the trick of treating the collision in the comoving, i.e. rest frame of the mass M (you can disregard this trick if you find it confusing). In this frame the mass M is at rest while the mass m moves at velocity v. Thus, given the assumption M m the collision reduces to the problem of a mass bouncing off a wall, so the final velocity of mass m is v. Adding on an additional v to transfer back to the lab frame, we find that v m = 3v, which is what we had before. 5. Pendulum on a cart (quiz #4) Problem: A pendulum consisting of a small mass m attached to a rope of length l hangs from a pole that is fixed to the top of a cart of mass M 1. The cart, which moves with an initial velocity V 0, collides inelastically with a stationary cart of mass M and sticks to it. The small mass is initially also moving at V 0 so that the pendulum is not swinging at first, but it begins to swing up after the collision. Find the smallest initial velocity V 0 such that the ball can go in circles around the top of the pole upon collision. Solution: First consider the case of the mass m on the pendulum swinging at velocity v 1 with the cart at rest. The velocity v at the top of the loop is given by the y- component of the force equation at the top, T + mg = mv. (5.5) l The minimum velocity for the ball to pass the top point without the rope going slack will occur for T = 0 (the same as for the problem with the pendulum hitting the 9

10 pivot point). This gives v = gl. The initial velocity that leads to this velocity at the top can be found using energy conservation, 1 mv 1 = 1 mv + mgl v 1 = 5gl. (5.6) Now consider the colliding cart system, forgetting the pendulum for the moment. Since the collision is completely inelastic we can solve for the final velocity V of the system, M 1 V 0 = (M 1 + M )V V = M 1V 0 M 1 + M. (5.7) Finally, add in the pendulum again. When the carts collide, the mass will continue moving at velocity V 0 while the cart moves at the slower velocity V. The mass will thus have a relative velocity V 0 V compared to the carts. In the frame of the cart the pendulum swings up at velocity V 0 V, which we plug in as our velocity v 1. The minimum initial cart velocity needed can thus be obtained by solving V 0 M 1V 0 M 1 + M = 5gl V 0 = M 1 + M M 5gl. (5.8) 6 Momentum transfer 6.1 Particles bouncing off a wall Problem: A stream of particles of mass m, average separation l and velocity v is incident on a perfectly reflecting wall. The mass per unit length is λ = m/l. What is the force imparted to the surface? Solution: Consider a length L of the stream about to hit the wall (a dummy quantity that will cancel out in our final calculation). The number of particles contained in this length is L/l. Since the wall is perfectly reflecting, the velocity of each particle will change from v to v upon collision, so the net change in momentum of each particle hitting the wall will be mv ( mv) = mv. We also know that the particles will hit the wall during a time t = L v, (6.1) so the average force from each particle will be p/ t = (m/l)v. To obtain the total force imparted to the wall we multiply by the number of particles L/l, F avg = m l v = λv F avg = λv. (6.) 10

11 6. Falling chain Problem: A uniform chain hangs vertically so that the bottom end just barely grazes the top of a horizontal table. If the upper end is released, show that at any instant during the fall the force on the table is three times the weight of the portion on the table. Solution: Let λ be the linear mass density of the chain. Consider an infinitesimal mass segment m. Take the positive direction to be down, and let x be the distance from the top of the chain before it is released to m (the amount the mass has fallen). We can find the velocity v of the segment using energy conservation, KE = U 1 ( m)v = ( m)gx v = gx. (6.3) (Note that U f has a lower potential energy than U i so U is negative.) To find the force let s first calculate the infinitesimal change in momentum, since x = v t. falling chain is p = ( m)v = (λ x)v = λv t (6.4) Thus the force imparted on the table by the p F = lim t 0 t = λv = λxg. (6.5) The total weight will be the sum of the force imparted on the table and the weight of the part of the chain already on the table at that instant, which is just λxg. This gives at any instant, F tot = λxg }{{} force from falling chain 7 Variable mass systems 7.1 Rocket ship + λxg }{{} = 3λxg F tot = 3λxg. (6.6) weight of chain on table Problem: A rocket of initial mass M 0 and starting velocity v 0, subject only to the force of gravity, launches vertically by burning and expelling fuel at a constant relative velocity v rel. In the process, the mass of the rocket decreases. Find the velocity v(t) of the rocket as a function of t, M(t), M 0, v 0 and v rel. Solution: At time t, let the rocket have mass M(t) and velocity v. At time t + t let the rocket have mass M(t) m and velocity v + v, and let the exhaust have mass m and velocity u. The change in momentum between t and t + t is p = (M(t) m)(v + v) + ( m)u M(t)v = M(t) v + (u v) m m v. (7.1) 11

12 To find the force we divide by t and take the limit as t 0. In this limit the last term, which will still be proportional to an infinitesimal quantity even after dividing by t, will vanish. Thus we are left with i F i,ext = d p dt = lim p t 0 t = M(t)d v dm + ( u v) dt }{{} dt v rel (7.) We have to be careful with minus signs. Since the the fuel comes from the rocket, its rate of change of mass is exactly opposite that of the rocket itself, dm dt = dm dt. (7.3) Thus, recalling that F = Mg in a pure gravitational field, and dropping all the vector signs since the motion is purely vertical, we have M dv dt v dm rel dt = Mg dv = v rel dm M gdt. (7.4) We can integrate this equation, if we are careful to use dummy indices and the correct limits. We have v(t) M(t) t dv dm = v rel g dt v 0 M 0 M 0 ( ) M(t) v(t) = v 0 gt + v rel ln M 0 (7.5) where v rel u v is the relative velocity of the fuel with respect to the rocket. Note that for a rocket, v rel will almost always be negative (since in our conventions it is pointing in the same direction as v). The logarithm term will also be negative, since the mass of the rocket is decreasing. Thus, the last term adds a positive contribution to the velocity of the rocket. 7. Fire hydrant Problem: A cart collects water from a fire hydrant. The water flows from the fire hydrant with constant horizontal velocity u and constant mass per unit length λ = dm/dl. Find the equation of motion of the cart by expressing the acceleration, dv/dt, in terms of λ, u, the cart s velocity v(t) and the cart s mass M(t). Solution: At time t, let the cart have mass M(t) and velocity v, and let the water have mass m and velocity u. At time t+ t let the combined cart and water system have mass M + m and velocity v + v. Following the procedure outlined in the rocket problem, we obtain the equation of motion i F i,ext = d p dt = lim p t 0 t = M(t)d v dm ( u v) dt }{{} dt. (7.6) v rel 1

13 Since there are no external forces, this gives (dropping the vector signs since the motion is horizonal) dv dt = v rel dm M dt. (7.7) By the chain rule we also know that dm dt = dm dl dl dt = λv rel. (7.8) Note that dl is the infinitesimal length of water hitting the cart, so that dl/dt is v rel as opposed to a rest frame velocity (intuitively, if I go faster less of the water will hit me). Thus, combining, we obtain the differential equation dv dt 7.3 Cart of sand (K&K 3.11) λ(u v(t)) =. (7.9) M(t) Problem: While moving at constant velocity u, a cart B drops sand vertically at a rate b kilograms per second into a second cart A that starts moving at velocity v 0. Find the velocity v(t) of cart A as a function of time. F i,ext = d p dt = lim p t 0 t = M(t)d v dm ( u v) dt }{{} dt. i v rel (7.10) In this problem there are no external forces, so we find (dropping the vector signs for purely horizontal motion) Solution: The dropped sand will be moving at the same horizontal velocity u as cart B. In the fire hydrant problem, we derived the force equation when mass is added to an object, M(t) dv = (u v)dm = (u v)b. (7.11) dt dt We are told that the rate of change of mass per time is constant, dm dt = dm dt = b M(t) = M 0 + bt. (7.1) We can integrate after moving like terms to both sides of the differential equation, v(t) t dv bdt = v 0 u v 0 M 0 + bt ( ) ( ) u v(t) M0 + bt ln = ln u v 0 M 0 (7.13) 13

14 Flipping the inside of the second logarithm to get rid of the minus sign, we can solve to obtain v(t) = u M 0 M 0 + bt (u v 0). (7.14) This has the expected result as we take t 0, and decreases with time as we expect for a cart with mass being added with time. 8 Rotational motion 8.1 Stick suspended off table (K&K 6.14) Problem: A stick of uniform mass M and length L is suspended horizontally with end B on the edge of a table and the other end A held by hand. At the moment after release, what is the magnitude of the normal force N between the table and the stick? (The moment of inertia of a stick about its center of mass is I CM = ML /1.) Solution: Take the up direction to be positive. The sum of forces in the y-direction is F i,y = N Mg = Ma. (8.1) Let s compute the torque around point B. To do this we must first find the moment of inertia about point B. This is given by the parallel axis theorem, I B = I CM + M(L/) = ML /3. The torque about point B is then i τ i = MgL i = I B α, (8.) giving α = 3g/L, where the negative sign just tells us the rod is rotating clockwise. Now, we know α is related to a by a = αr where r = L/ is the distance from the pivot point to the center of mass (compare: v = ωr and arclength = θr). Thus, a = 3g/4 (again, the negative sign tells us it is accelerating downwards). Plugging this into the force equation we get N = Mg/4. 8. Horizontal yo-yo (K&K 6.7, 6.8) Problem: A yo-yo of mass M and moment of inertia MR / has a radius R and a smaller axle of radius b about which a string is wound. The yo-yo is placed upright on a table and a constant force F is applied to the string from the lower end of the axle. The coefficient of static friction between the yo-yo and the table is µ. (a) If the string is pulled horizontally, what direction will the yo-yo move? What is the maximum value of F for which the yo-yo will roll without slipping? 14

15 (b) If the string is pulled an angle θ above the horizontal, for what value of θ will the yo-yo slide completely? Solution: (a) The yo-yo will move to the right, as we see by noticing that F is the only force that exerts a nonzero (clockwise) torque about the point of contact with the ground. Taking the direction of the force to be positive, the force equations are F i,x = F f = Ma, F i,y = N Mg = 0. (8.3) i Thus N = Mg so the maximum friction force is f max = µn = µmg. The maximum force F max is obtained using this value for the static friction. The torque equation around the center of the yo-yo is τ i = µmgr F max b = Iα. (8.4) i The condition for rolling without slipping is a = Rα. Combining, we find F max µmg = Ma = MRα = R (µmgr F maxb) ( F max 1 + b ) = 3µMg F max = 3µMg R R + b. (8.5) i Our expression verifies that for R > b, the applied force will be exactly larger than the frictional force so the yo-yo will move to the right. (b) The yo-yo will slide completely if α = 0 or Relating to the horizontal force equation, F b = µnr. (8.6) F cos θ µn = Ma (8.7) we find that the critical angle at which the yo-yo slides without rolling is F ( cos θ b ) = Ma cos θ = Ma R F + b R. (8.8) 8.3 Cylinder rolls off cylinder (F&C 8.16) Problem: A circular cylinder of radius a is balanced on top of a fixed cylinder of radius b. If the balance is slightly disturbed, find the angle at which the cylinder of 15

16 radius a leaves the fixed one. Solution: Let s measure angles clockwise from a vertical line passing through the cylinders. Then energy conservation gives KE = U 1 mv + 1 Iω = mgb(1 cos θ). (8.9) For a solid cylinder, the moment of inertia about the center of mass is I = 1 ma. Applying the relation v = ωa for the rotation of the top cylinder, we find v = 4 gb(1 cos θ). (8.10) 3 Now, let s go back to the force equation. Applying Newton s second law in the direction of the outward pointing normal to the bottom cylinder, we have N mgcos θ = mv b. (8.11) Combining these two equations to eliminate the velocity, we find the angle at which the top cylinder falls off, cos θ = 4 7. (8.1) 8.4 Wheel climbs up step (final review) Problem: A solid wheel of mass M and radius R (I = MR /) rolls without slipping at a constant velocity v 0 until it collides inelastically with a step of height h < R. Assume that there is no slipping at the point of impact. What is the minimum velocity v 0 in terms of h and R needed for the wheel to climb the step? Solution: I was rushing on this one so I didn t explain it very well, plus some questions came up that I had to cut off; I ll try to address them here. Let ω 0 and ω be the angular velocity of the wheel before and instantaneously after collision with the step. Before collision the wheel rolls without slipping so v 0 = ω 0 R by the no-slip condition, and after it is just barely starting to roll about the pivot point but has no translational velocity. Note that ω is the velocity right before it starts to rise, and for it to just barely reach the top means that the angular velocity at this final point, as well as the final translational velocity, must be zero. There are no net external torques about the point of contact during the collision, so angular momentum about this point is conserved. The reasoning for this is as follows. Obviously the torque due to f is zero since it is exactly at the point of contact. Also, the force of gravity is exactly cancelled by a normal force, so the net torque due to both forces combined is zero. There was a question about why we can neglect torques after it starts to rise, when the normal force goes away. Thanks for pointing this out, since it is true that the net torque is not zero after it starts to rise! However, we are comparing angular momentum between two points before it starts to rise, so our analysis is kosher. 16

17 Recalling the that angular momentum about a point other than the CM is equal to angular momentum about the CM plus angular momentum of the CM about that point, and using the no-slip condition v 0 = Rω 0, the initial and final angular momenta are L i = Mv 0 (R h) + 1 MR ω }{{} 0 v 0 /R = 3 Mv 0R Mv 0 h, ( ) 1 L f = MR + MR ω = 3 MR ω. (8.13) Equating the two, we find ( L i = L f ω = 1 ) h v0 3 R R. (8.14) Now the turning point such that the ball can just barely climb the step is determined by energy conservation, since this will happen when the rotational kinetic energy instantaneously after collision is exactly equal to the potential energy required to lift a distance h, with no final kinetic energy. The condition is 1 I Cω = Mgh. (8.15) Here we must be careful to use I C = 1 MR + MR, the moment of inertia about the point of contact (calculated using the parallel axis theorem). Plugging in our value for ω, we find the condition becomes Mgh = 3 ( 4 MR 1 ) h v0 3 R R. v 0 = R 3gh 3R h. (8.16) Note that the same procedure carries over for an object with a different I, for instance a sphere. 8.5 Toothed wheel (K&K 6.40, quiz #3) Problem: A wheel with fine teeth is attached to the end of a spring with spring constant k and equilibrium length l. For x > l, the wheel slips freely on the surface, but for x < l the teeth on the wheel mesh with teeth on the ground so that the wheel cannot slip. The wheel is pulled to x = l + b and released. (Assume that the mass of the wheel is concentrated in its rim.) (a) How close will the wheel come to the wall on the first trip? (b) How far out will it go as it leaves the wall? (c) What happens the second time around? 17

18 Solution: Assuming there is no friction beyond the track, we can use energy conservation to find the velocity v 0 right before the wheel hits the track, 1 kb = 1 mv 0 v 0 = b k m. (8.17) There is an impulse force at the track boundary, causing a change in velocity. We can find the velocity directly after collision by noticing that there is zero net torque around the point of contact (both friction and gravity pass through the point of contact, and the spring force is zero because the spring is at its equilibrium length). Notice that some energy is dissipated at the boundary, so we cannot use conservation of energy across this point. Since there is no net torque, angular momentum about an axis through the point of contact is conserved. Denoting by z some axis passing through this point, this is L z = I CM ω + (R MV ) z. (8.18) In words, the angular momentum L about the contact point is L about the center of mass plus L of the center of mass about the contact point. Thus we have L i = L f mrv 0 = Iω + mrv = mr v + mrv = mrv (8.19) R where we have used I = mr since the mass of the wheel is concentrated at its rim, and v = ωr since the wheel rolls without slipping after hitting the teeth. So v = v 0 /. After moving on to the teeth we can again use energy conservation to find the turning point, 1 ka = 1 Iω + 1 mv = mv = 1 4 kb a = b = x 1 = l b. (8.0) After turning around off the track, this time there is no friction at the boundary and hence no torque about the center of rotation, so there is nothing to change the spin motion. Thus the wheel will continue to rotate at the same rate even as it slips. We can use energy conservation to find the next turning point, Thus 1 kd + 1 Iω = 1 mv + 1 Iω kd = mv = 1 4 kb. (8.1) d = b x = l + b. (8.) When the wheel returns to the boundary a third time, it is still spinning in a direction that is now opposite to its direction of motion. By energy conservation its velocity increases up to v = ωr at this point. Again there is no torque about the 18

19 point of contact so angular momentum about an axis through this point is conserved. This time we get canceling contributions from the two parts, L i = L f = Iω mrv = mr ω mr ω = 0. (8.3) There is not enough angular momentum to cross the boundary and the wheel gets stuck the second time around. 8.6 Ladder against wall (K&K 6.41, F&C 8.17, challenge) Problem: A uniform latter of length L leans against a vertical wall supported at an initial angle θ 0. After it is released, find the angle at which the ladder loses contact with the wall. Solution: The forces on the ladder are two normal forces from the ground and wall, and gravity focused at the center of mass. The center of mass is located at a position x CM = L cos θ, y CM = L sin θ (8.4) and its velocity and acceleration can be found by taking time derivatives, ẋ CM = L (sin θ) θ, ẍ CM = L ((cos θ) θ + (sin θ) θ). (8.5) Be careful to remember that we are deriving with respect to time, not θ, so we pick up factors of θ and θ by the chain rule. We can find the horizontal motion of the center of mass using the x-component of Newton s second law, F i,x = N = Mẍ CM (8.6) i where we take M as the mass of the ladder (we expect this will not appear in the final answer). The ladder loses contact with the wall when N = 0, or ẍ CM = 0 (cos θ) θ + (sin θ) θ = 0. (8.7) We don t know θ, θ, but we can determine them from energy conservation, Mgy CM,0 = Mgy CM + 1 M(ẋ CM + ẏ CM) + 1 I θ (8.8) Recalling that I = 1 3 ML for a rod-like object around its center, and plugging in the positions and velocities of the center of mass (also, y CM,0 = (L/) sin θ 0 ) we find θ = (sin θ 0 sin θ). (8.9) 19

20 Taking the derivative with respect to time, we find θ θ = 3(cos θ) θ θ = 3 cos θ. (8.30) Plugging these into 8.7, we find the angle at which the ladder loses contact, ( ) θ = sin 1 3 sin θ 0. (8.31) 9 Oscillations 9.1 Tunnel through the Earth (quiz #4, final review) Problem: A straight hole is drilled obliquely through the Earth (i.e. it does not pass through the Earth s center) and a small rock is dropped down one end. Assume the Earth has constant mass density, with mass M and radius R, and neglect all effects due to its spin and motion. (a) Show that the rock exhibits harmonic motion and find its frequency, ω. (b) How long does does it take until the rock reaches the other side? (c) Show that the orbital period for a satellite in circular orbit just above the Earth s surface is the same as your answer to (a). Solution: (a) Suppose the rock, at some point inside the tunnel, is a distance r away from the center of the Earth. Recall that a test charge inside a spherical shell experiences no net gravitational force, while one outside it experiences the same force as if all the shell s mass were concentrated at its center. Treating the Earth as a superposition of massive shells, the gravitational force experienced by the rock will be due to the mass inside the radius r, concentrated at the center. The total mass inside is M inside = ρv inside = Mr 3 /R 3 since ρ = 3M/4πR 3 and V inside = 4πr 3 /3. The force is directed inwards, so m r = GmM inside r r = GM r. (9.1) R3 In terms of x, the distance from the center of the tunnel to any point inside, and θ the angle between r and x, ẍ = GM r cos θ = GM R3 R r x GM 3 r ẍ = ω x with ω R. (9.) 3 This is just harmonic motion about the center with frequency ω. (b) The total period is T = π/ω. The time to reach the other end will be half of this, T R = π 3 GM. (9.3) 0

21 (c) A satellite of mass m just above the Earth s surface moves according to GmM GM = mω R ω = R R 3 T = π ω = π R 3 GM. (9.4) Thus the satellite returns to its starting point exactly when the rock does. 1