Physics H7A, Fall 2011 Homework 6 Solutions 1. (Hooke s law) (a) What are the dimensions (in terms of M, L, and T ) of the spring constant k in Hooke s law? (b) If it takes 4.00 J of work to stretch a Hooke s-law spring 10.0 cm from its unstressed length, determine the extra work required to stretch it an additional 10.0 cm. (a) [k] = [F/x] = M/T 2 (b) The work done by your hand in stretching a spring from x 1 to x 2 is W x1 x 2 = U(x 2 ) U(x 1 ) = 1 2 k(x2 2 x 2 1). So if W 0 10 = 1 2 k(102 ) = 4 J, then W 10 20 = 1 2 k(202 10 2 ) = 12 J. 2. (K&K Problem 2.31) Find the frequency of oscillation of mass m suspendended by two springs having constants k 1 and k 2, in each of the configurations shown. Ans. clue. If k 1 = k 2 = k, ω a = k/2m, ω b = 2k/m 1
Two springs in series behave as a single spring with spring constant k a = (1/k 1 + 1/k 2 ) 1, while two springs in parallel behave as a single spring with k b = k 1 + k 2. Thus ka ω a = m = 1 (1/k1 + 1/k 2 )m, ω kb b = m = k1 + k 2 m. 3. (Pencil dive) A 70-kg diver steps off a 10-m tower and drops straight down into the water. If he comes to rest 5 m beneath the surface of the water, determine the average resistance force exerted by the water on the diver. The diver hits the water with speed v = 2gh = 14 m/s. To come to rest after x = 5 m, he feels an average upward acceleration a avg = v 2 /2 x = 19.6 m/s 2 = 2g. Considering that gravity still pulls him down with force mg while in the water, the average resistance force from the water must be F w = 3mg = 2058 N. 4. (Pendulum in the wind) A ball having mass m is connected by a strong string of length L to a pivot point and held in place in a vertical position. A wind exerting a constant force of magnitude F is blowing from left to right as in the figure below. (a) If the ball is released from rest, show that the maximum height H reached by the ball, as measured from its initial height, is H = 2L 1 + (mg/f ) 2. Check that this result is valid both for cases when 0 H L and for L H 2L. (b) Compute the equilibrium height H eq of the ball (the height at which the net force on the ball is zero). (c) Could the equilibrium height ever be larger than L? Explain. 2
(a) The wind force F here is constant, so by analogy with the gravitational force mg we can assign a potential energy U w = F x to the ball, where x is the horizontal displacement of the ball relative to some arbitrary origin (which we will take to be starting position of the ball). Then the total initial energy is E 0 = U w0 + U g0 = F x 0 + mgy 0 = 0, while the final energy at its maximum height is E = F x + mgh, and energy conservation implies F x = mgh. From the geometry of the figure we have ( ) mgh 2 L 2 = x 2 + (L H) 2 = + L 2 2LH + H 2 = F hence H = 2L/[1 + (mg/f ) 2 ]. (b) At equilibrium the net force on the ball is zero, so [ 1 + F net = F ˆx mgŷ + T ( sin θˆx + cos θŷ) = 0, ( mg ) ] 2 H 2 2LH + L 2, F where T is the tension in the string and θ is the angle the string makes with respect to the vertical. Setting the x and y components separately to zero, we get F = T sin θ and mg = T cos θ, hence tan θ = F/mg. The equilibrium height is therefore ( H eq = L(1 cos θ) = L 1 1 ) ( ) ( ) 1 1 = L 1 = L 1. sec θ 1 + tan 2 θ 1 + (F/mg) 2 (c) The equilibrium height can never be greater than L, because if the ball is higher than L then there is no upward force to balance the downward force of gravity. 5. (K&K Problem 3.8) A 50-kg woman jumps straight into the air, rising 0.8 m from the ground. What impulse does she receive from the ground to attain this height? v 0 = mgh, p = mv 0 = m mgh = 990 kg m/s 6. (Tennis volley) A tennis player receives a shot with the ball (0.060 kg) traveling horizontally at 50.0 m/s and returns the shot with the ball traveling horizontally at 40.0 m/s in the opposite direction. (a) What is the impulse delivered to the ball by the racquet? (b) What work does the racquet do on the ball? (a) p = mv f mv i = (0.060 kg)(40.0 m/s) (0.060 kg)( 50.0 m/s) = 5.4 kg m/s (b) W = K = 1 2 mv2 f 1 2 mv2 i = 27.0 J 7. *(Leaky bucket) At t = 0, a massless bucket contains a mass M of sand. It is connected to a wall by a massless spring with constant tension T (that is, independent of length). The ground is frictionless, and the initial distance to the wall is L. At later times, let x be the distance from the wall, and let m be the mass of sand in the bucket. The bucket is released, and on its way to the wall, it leaks sand at a rate 3
dm/dx = M/L. In other words, the rate is constant with respect to distance, not time; and it ends up empty right when it reaches the wall. Note that dx is negative, so dm is also. (a) What is the kinetic energy of the (sand in the) bucket, as a function of x? What is its maximum value? (b) What is the magnitude of the momentum of the bucket, as a function of x? What is its maximum value? (a) The leaky bucket can be considered as a simple example of a rocket, in which the exhaust is expelled with relative velocity u = 0 (because the sand leaks out, instead of being propelled backward). Using Equation 3.19 from K&K, we have T = m(t)(dv/dt), where m(t) is the mass of the sand left in the bucket at time t. We can now change independent variables from t to x, using the given relation m(x) = Mx/L along with the identity dv dt = dv dx dx dt = v dv dx, to obtain the differential equation T = Mx L v dv dx = dx x = M LT vdv. Integrating from the initial point (x 0 = L, v 0 = 0) to a final point (x, v), we have ln x L = M 2LT v2. Therefore the kinetic energy as a function of position is with maximum K max = T L/e. K(x) = 1 2 m(x)v(x)2 = T x ln x L, (b) The momentum of the bucket as a function of position is p(x) = m(x)v(x) = x 2T M L ln x L (notice the minus sign since ln x L < 0 for x < L), with maximum p max = MT L/e. 4
8. *(Speedy travel) A straight tube is drilled between two points on the earth, as shown in the figure below. An object is dropped into the tube. What is the resulting motion? How long does it take to reach the other end? Ignore friction, and assume (incorrectly) that the density of the earth is constant (ρ = 5.5 g/cm 3 ). Let a be the distance from the center of the earth to the tube, and let x be the horizontal displacement of the object along the tube (measured from the midpoint) at any given time. At the position x, the object lies a distance r = a 2 + x 2 from the center of the earth. The mass that lies within this radius is m(r) = M E (r/r E ) 3, so the strength of the gravitational force is F = F = Gm(r)m obj r 2 = GM Em obj RE 3 r. But this gravitational force on the object is directed radially inward, and since the object is constrained to travel within the tube, only the horizontal component of the gravitational force contributes to the motion. This component is F x = F sin θ = F x r = GM Em obj RE 3 x. (The minus sign tells us the force acts to the left when x is to the right, and vice versa.) Thus, the x component of Newton s second law (F = ma) gives GM Em obj R 3 E x = m obj a x = ẍ + GM E RE 3 x = 0. This is the generic equation of motion for simple harmonic oscillation (ẍ + ω 2 x = 0), so the object will undergo SHO with period 9. *(Mine shaft) T = 2π ( R 3 ) 1/2 ω = 2π E. GM E (a) If the earth had constant density, the gravitational force would decrease linearly with radius as you descend in a mine shaft. However, the density of the earth is not constant, and in fact the gravitational force increases as you descend. Show 5
that the general condition under which this is true is ρ c < (2/3)ρ avg, where ρ avg is the average density of the earth, and ρ c is the density of the crust at the surface. (The values for the earth are ρ c 3 g/cm 3 and ρ avg 5.5 g/cm 3.) (b) Assuming that the density of a planet is a function of radius only, what should ρ(r) look like if you want the gravitational force to be independent of the depth in a mine shaft, all the way down to the center of the planet? 6
(a) At the surface of the earth (r = R), the strength of the gravitational force on an object of mass m obj is F (R) = GMm obj R 2, where M = 4π 3 R3 ρ avg is the total mass of the earth. If you descend a small vertical distance R down a mine shaft, then the strength of the gravitational force will be F (R R) = G(M M)m obj (R R) 2, where M = 4π 3 [R3 (R R) 3 ]ρ c is the mass of spherical shell between radii R R and R (which lies within the crust). Using the general result that (1 + x) n 1 + nx for x 1, we can expand the above expression for F (R R) to first order in the small quantity R/R. In the end we find F (R) F (R R) = GMm ( obj R 2 3 ρ ) c R 2 ρ avg R. Thus, for ρ c < (2/3)ρ avg, we have F (R) < F (R R), i.e. the force increases as you go down the mine shaft. (b) The gravitational force at a radius r from the center of the planet is proportional to m(r)/r 2, where r m(r) = 4π ρ(r )r 2 dr 0 is the mass within radius r. To make this force independent of r, we must have m(r) = Ar 2 for some constant A. If the radius of the planet is R and its total mass M, then m(r) = M implies A = M/R 2. To find the density, we differentiate the above expression for m(r), using the fundamental theorem of calculus: On the other hand dm/dr = (M/R 2 )2r, so dm dr = 4πρ(r)r2. ρ(r) = M 1 2πR 2 r. (Note that even though the density goes to infinity at the center, the total mass is still finite.) 7