Physics 160

This is a "demo" for the solutions for an older exam over Chapters 5-9 of Resnick, Halliday and Walker.
To make it more useful in that regard, I have added one additional problem, from a different exam, as problem #6.
In the original exam, there were three problems to be done:
One of #1 and #2; one of #3 and #4 (30 points for each), and #5 (40 points) .
Also a (10 point) bonus problem allowed.

1. A ``standard'' pendulum, of length L = 0.6 meters, is hanging from the ceiling of a car, and is free to swing in any way it likes. The car is going around a circle, clockwise, on level ground, of radius 100 meters, at a constant speed of 25 m/s. Assuming that the situation is in equilibrium, so that the pendulum hangs at a constant angle relative to the vertical, what is that angle? Is this angle such that the pendulum leans toward or away from the center of the circle?
   

   Since the car, and therefore also the bob (mass) on the end of the pendulum is going around in a circle, there must be some combination of forces that creates the necessary centripetal acceleration. For the car, presumably, that force is provided by the deformation of the tires. However, for the pendulum bob, we have a more direct way of determining that force, since there are only two forces acting on it:

   1. the force due to gravity, acting straight downward, and
   2. the tension in the rod holding the mass, acting upward at some angle, to be determined.
   We agree to take as coordinate axes the horizontal and vertical directions, with the usual names, i.e.,
   1. x-direction is horizontal, inward toward the center of the circle, and
   2. y-direction is vertical, pointing upward.
   As the rod is at some angle, then it must have both horizontal and vertical components, T_x and T_y.
   Applying Newton's 2nd Law to the pendulum bob, we have
   1. Vertical forces: T_y - mg = m a_y = 0,
      where I have inserted a zero value for the vertical acceleration, as one expects.
   2. Horizontal forces: T_x = m a_x = m v²/R,
      where I have set the horizontal force equal to the centripetal force necessary to cause the pendulum to go around in the same circle as the car.
   We may now solve these equations for the two components of the tension, namely T_y = mg , and T_x = mv²/R. The tangent of the angle with the vertical will, by ordinary trigonometry, be given by the ratio of these two components, i.e., tan() = T_x/T_y = v²/(Rg) = 25²/[100(9.8)] = 625/980 ,
   so that we find that the angle is = arcTan(625/980) = 32.5° . It is relevant that both the components found above are positive, so that the vector that points, along the rod, from the mass to the ceiling of the car, points up and toward the center of the circle.
   Therefore the mass will have to be leaning away from the center, so that the rod supporting it can point toward the center.
   

2. A pendulum of length L and mass m is released to swing from an initial angle of ₀. It swings down to zero angle and then back up to an angle of -₀, since we assume that all the bearings involved are frictionless.
   1. How much work is done by gravity on the downswing?
   2. How much work is done, on the downswing, by the tension in the rod holding the mass?
   3. What is the speed of the mass at the bottom of its swing?
   4. What is the net effect of the work done by gravity on the downswing? In light of your answer to this question, how much work is done by gravity on the upswing, and what is its net effect?
   
   1. At the bottom of the swing, the pendulum bob is a distance L below its support (on the ceiling). At the top of its swing, the bob sticks out at an angle ₀, so that the (vertical) distance from the support is only L cos(₀).
      It is true that the motion of the bob during the downswing was along the arc of a circle; however, as gravity always acts downward, it is only doing effective work for that component of the motion which is vertical, i.e., either downward or upward.
      Therefore, during the downswing, gravity has lowered the bob by the difference of these two, i.e., a displacement (downward) of L{1-cos(₀)}. During that lowering, a constant gravitational force acted, of amount, mg, also downward. Therefore, gravity will have done a positive amount of work on the mass: W_gravity = mgL{1-cos(₀)} .
   2. During the swinging motion of the pendulum bob, the direction of the tension is, however, NOT constant. Instead it always points inward toward the center of the circle. As it points radially inward, and the motion is always tangential along the circle, the two are always perpendicular, so that the tension force does NO work during the swing.
      This is an example of the general principle which says that work is not done by any force causing motion to occur in a circle of constant radius. In the general case, as well, the reason is the same, namely that the force and the displacement are always perpendicular. However, one may also understand this fact by noting that the (squared) speed, v², of a particle moving in a circle of constant radius is always the same, so that the kinetic energy must remain the same.
   3. As work is done by gravity, on the downswing, it must therefore be translated into an energy of the pendulum bob, in particular into kinetic energy, K. Knowing that the speed at the top of the swing was zero, we have ½ m v² - ½ m (v₀)² = K_f - K_i = W = mgL{1-cos(₀)}
      from which we find v² = 2gL{1-cos(₀)}, or
      v = {2gL[1-cos(₀)]}^½ .
   4. As already noted, the positive work done by gravity, on the downswing, caused a positive increase in the kinetic energy. On the upswing, of course, the vertical displacement will be upward, in the opposite direction to the force of gravity, implying that the work done by gravity will be negative. It will therefore cause a negative change in the kinetic energy, returning it back to a value of zero.W_gravity on the upswing = -2gL{1-cos(₀)}.
   

3. Objects moving under the gravitational influence of our sun (in any other way than simply falling straight in) have an effective gravitational potential energy of the following form U = U(r) = -a/r + L/(2 r²) , where r is the distance between the object and the (center of the) sun, and a and L are (positive) constants for any particular object.
   Since this object is not falling straight in, the angle it makes with respect to some initial line between it and the sun changes with time. We are ignoring this motion at the moment, concentrating on the change of distance between the object and the sun, given by r. Therefore the total energy being discussed is just the kinetic energy due to radial motion, i.e., ½ m (dr/dt)², and the potential energy given by U(r).
   1. Please make a sketch of the potential energy function in question.
   2. For fixed values of the constants a and L, at what distance from the sun will there be a purely circular motion, i.e., one for which the distance r does not change?
   3. Describe the behavior of such an object which had exactly total energy, E = 0, and was initially headed inward toward the sun?
   4. Describe other motions the object is allowed to have, and the associated values of its total energy.
   5. What are the minimum and maximum allowed values for the total energy of this object?
   
   1. As a and L are both positive (non-zero), then the function in question has
      1. a single zero, where r = r₀ = L/(2a), and
      2. a single extremum (a minimum) where its derivative is zero, which is at r = r₁ = L/a, and
      3. is very, very small and negative, when r becomes very, very large (positively),
         [This is because for very, very large values of r, the +L/(2r²) term will be very much smaller than the -a/r term, so that the L-term may be ignored. The remaining one, the -a/r term is negative, giving the result.]
      4. the function is very, very large and positive, when r becomes very, very small (and positive).
         [Again, this is because for very, very small values of r, the -a/r term may be ignored relative to the much larger +L/(2r²) term.]
      5. This data then results in the attached curve.
      
   2. Knowing that the plot in question has only one minimum, we know that if the object in question had that minimum value of the energy, at that location, then it would have to simply stay at that single value of r, which would then constitute circular motion.
      Therefore we can assert that for fixed values of a and L, there is circular motion at r = L/a, and E = E₀ = U(L/a) = -a²/(2L) .
   3. Looking at the plot of U(r), we see that it asymptotically approaches E=0, from below, as r approaches infinity. Therefore, an object headed inward, with E=0, would come inward toward the sun, speeding up as it did so, until it came to the minimum point, r=L/a. At that point it continues inward, but now slowing down, until it arrives at the turning point, i.e., the zero of U(r), which we already know is r=L/(2a), where it stops and turns around, now heading outward, going on outward, to infinity. That is to say, it never returns again.
      A simple English phrase for such an orbit is a comet that never returns.
   4. As to other possible motions, there appear two types:
      1. If the energy E is less than 0, but greater than E₀ = -a²/(2L), then the object will have two turning points, between which there will be oscillatory motion forever---all the while as it circles around the sun as well. These two turning points may be thought of as the distances of closest and furthest approach from the sun. This is an elliptical orbit.
      2. If the energy E is greater than 0, then the orbit is much like the one for E=0, in the sense that it comes inward, finds a turning point [nearer the sun than L/(2a)], and then exits the solar system forever. The other difference is simply that it will always have a greater speed than the corresponding E=0 trajectory.
   5. The minimum allowed energy is surely E₀.
      There is no maximum to the allowed energy values.
   

4. A mass of 2 kilograms moves across a level, frictionless floor until it encounters a spring situated at rest upon the floor. Having struck the spring, the mass begins to compress the spring until it reaches some maximum compression and the mass comes to rest. Given that the floor under the spring is rough, rather than frictionless, with a coefficient of friction of µ = .05, and that the spring has a spring constant of k=300 N/m, and that the spring is compressed a distance of 10 centimeters, what was the initial speed of the mass?
   

   As the spring had a non-zero initial speed, it had a positive initial kinetic energy,

   K₀ = ½ m v² .

   When it encounters the spring, it exerts a force on the spring, thereby doing work on the spring, which depletes its original kinetic energy. This change amounts to a storing of that kinetic energy into the potential energy of the spring.
   As well, the frictional resistance of the floor under the spring decreases its kinetic energy, until the two processes reduce its kinetic energy to zero.
   We can summarize the situation as follows:

   K_f + U_f - K_i - U_i = W_f ,
   0 + ½k x² - ½m v² - 0 = -f x = - µ N x = -µ mg x,
   so that v² = (k/m) x² + 2µ g x = (300/2)(0.1)² + 2(.05)(9.8)(0.1) = 1.5+.098 = 1.598,
   or v = 1.264 m/s .
   

5. A large block, of mass M, sits on a frictionless floor. On its left side, another, smaller block, of mass m, has been placed against it, part-way up the side of the larger block, so that the smaller one does not touch the floor, as indicated in the drawing below. An external force, F, is being exerted on the left side of the smaller block, to hold it in that position. (The coefficients of friction between the two blocks are µ_s= 0.20, and µ_k = 0.15. Also take M=80 kg, m=20 kg.)
   1. Please make your own drawing, on your paper, and indicate all forces acting on each of the two blocks.
   2. The external force F causes both blocks to accelerate toward the right. What is the minimum value of F that will allow the two blocks to accelerate (rightward) together without the smaller one slipping?
   
   1. On the small block, there are 4 different forces acting:
      1. The external force, F, pushing horizontally towards the right, which we take as the positive direction, so that we write it as Fx,
      2. the normal force, N, exerted by the larger block on the small one, also horizontal, but of course in the negative direction, -Nx [This is of course in response to the small block pushing on the larger one, following Newton's Third Law.],
      3. its weight, -mgy, due to gravity, which is vertical and downward, and
      4. the frictional force between the blocks, which must surely act upwards, in order to prevent sliding between the two, f = f_sy .
   2. On the large block, there are also 4 forces:
      1. The (horizontal) ``pushing'' force exerted by the small block on the larger one, which we can write as N = Nx,
      2. its weight, -Mgy, acting downward,
      3. the normal force of the floor pushing upward on it, which we can denote by +ny, and
      4. the Newton's-Third-Law response force of the friction between the two blocks, which is of course holding the small block up. Therefore, this response force is downward, and of equal magnitude, for a force -fy.
   3. We now write down the results of Newton's Second Law, for each block, separately for horizontal and for vertical forces. We assume that they move together, with acceleration a in the horizontal direction only.
      1. small Block, vertical: f - mg = m a_y = 0,
         so that mg = f , which may not be larger than µ_s N .
      2. small Block, horizontal: F - N = m a, or F = ma + N .
      3. large Block, vertical: n - f - Mg = m a_y = 0, so that n = f+Mg,
      4. large Block, horizontal: N = M a .
   4. As we want the minimum F, we first use the two horizontal equations to write that F = ma + N = ma + Ma = (m+M)a . We see then that this will imply the minimum value for a, which will then imply a minimum value for N.
      However, we must have that N is sufficiently large to allow the friction to cancel out the weight of the small block. This gives us µ_s Ma _min = µ_s N_min = mg ,
      or the value a_min = (mg)/(µ M) . We may now insert this value if the equation for F above, to give F_min = (m+M)a_min = (m+M)(mg)/(µM) = (20+80)(20)(9.8)/[(0.20)(80)] = 1225 Newtons.
   

   Bonus Problem, Points = +10

   In problem 5, just above, describe the behavior of the system if the external force F has only half the value of the minimum calculated there. In particular, ``describe'' means that you should calculate the acceleration vectors for both blocks, under the behavior of this force.

   

   Given that F = ½(1225 N) = 612.5 N, we immediately can determine the horizontal acceleration:

   a = F/(M+m) = 612.5/(20+80) = 6.125 m/s² . However, since the force F is less than the minimum value required to prevent sliding of the small block, then it must indeed be sliding downward. Therefore, we will have to change the formulae above, to allow a non-zero value for its downward acceleration. The new formula is f_k - mg = m a_y .
   However, under such circumstances, we no longer have an inequality relation for the frictional force between the blocks. Instead, it is operating at its maximum allowed value, which is then f_k = µ_k N = µ_k M a .

   Therefore, we determine that

   f_k = (0.15)(80)(6.125) = 73.5 N. which allows us to determine the sliding acceleration a_y = f_k/m - g = µ_k(M/m) a - g = [µ_k/(2µ_s) - 1] g
   = (0.15)4(6.125) - 9.8 = 3.68- 9.80 = - 6.12 m/s² .
   

6. A large fish eagle is flying low over the surface of a lake, at a speed of 5 m/sec. It sees a fish in the water, swimming in the opposite direction at only 2 m/sec. The bird immediately grabs the fish in its claws and procees along in the same direction---before eventually turning and flying back to its nest for supper. The bird has a mass of 10 kg and the fish 4 kg.
   1. What is the speed of the bird after it has grabbed the fish?
   2. What is the change in kinetic energy of the system, i.e., the bird and the fish?
   

   Initially there are two components of the system, the bird and the fish. Therefore, the initial momentum is

   P = p_bird + p_fish = 10(5)-4(2) = 42 kg-m/s. The initial kinetic energy is K_i = ½m_bird(v_bird)² + ½m_fish(v_fish)² = ½{10(5)² + 4(2)²} = 133 J.

   Afterward the two fly off together, as a single component, with mass of 14 kg. Since there are no (relevant) external forces, momentum is conserved. Since the final momentum is 14v, this conservation law tells that

   v = 42/14 = 3 m/s . This is then sufficient information to calculate the final kinetic energy, as K_final = ½(m_fish+m_bird) v² = ½14(3)² = 63 J.
   Therefore the change is K = 63 - 133 = -70 J. This loss of energy is of course due to the work required to grab and hold onto the fish!