PHYSICS 126 Fall 2010 Midterm 1

Transcription

1 PHYSICS 16 Fall 010 Midterm 1 Name: SOLUTIONS Student ID: Section Number: Closed book with one sheet of notes and a calculator. Answer the questions in spaces provided on each sheet. If you run out of room for an answer, continue on the back. Question Points Score Total 110 Midterm I a Page 1 of 6

2 5 1. An L = m long boom with a mass of m = 10 kg is fixed to a wall by a pivot which allows free rotation. It is held at an angle of θ = 45 with respect to the wall by a cable which is also at an angle of θ = 45 with respect to the wall. The end of the boom supports a mass with M = 100 kg. Use the figure below to label the forces applied to the boom, then find the vertical force applied by the wall to the boom. (Hint: All the angles labeled θ are equal to 45 degree.) F_3 θ F_1 F_4 θ M θ F_5 L/ F_ The boom is in equilibrium so it will obey the equilibrium conditions Fx = 0 = F 5 F 3 cosθ (1) Fy = 0 = F 4 +F 3 sinθ F F 1 () L τz = 0 = F 3 F L sinθ F 1Lsinθ (3) where the torque has been calculated about an axis passing through the pivot. We are asked to find F 4, and we know from the givens that F 1 = Mg and F = mg. Solving equations 1 and for F 3 and F 4 gives F 3 = Mgsinθ +mgsinθ (4) F 4 = Mg +mg Mgsin θ mgsin θ = Mg(1 sin θ)+mg(1 sin θ) (5) so we find, ( ) ( ) F 4 = (100 kg)(9.8 m/s ) 1 +(10 kg)(9.8 m/s ) 1 (6) F 4 = 1 (10 kg)(9.8 m/s ) (7) Midterm I a Page of 6

3 . A cylindrical tank that is 55 m in diameter and 60 m deep is filled with water (density 1000 kg/m 3 ). 5 (a) What is the gauge pressure on the bottom of the tank? The gauge pressure is given P = ρgh, where ρ is the density (1000 kg/m 3 ), g is the gravitational acceleration (9.8 m/s ) and h is the depth (60 m). Therefore P = ρgh = (1000 kg/m 3 )(9.8 m/s )(60 m) (8) 5 (b) What is the mass of the water in the tank? The mass of the water in the tank is given by m = ρv where ρ is the density (1000 kg/m 3 ), and V is the volume ( 1 4 πd h). The mass is m = 1 4 d hρ = 1 4 (55 m) (60 m)(1000 kg/m 3 ) (9) 10 (c) What is the net force on the floor of the tank from the water? The net force on the bottom of the tank due to the water can be calculated in two ways: 1. The net force is F = PA where P is the pressure due to the water, and A is the area of the tank bottom ( 1 4 d ). Therefore F = PA = 1 4 d ρgh = 1 4 (55 m) (1000 kg/m 3 )(9.8 m/s )(60 m) (10). The net force is F = mg where m is the mass and m = ρv. Therefore F = mg = 1 4 d hρg = 1 4 (55 m) (1000 kg/m 3 )(9.8 m/s )(60 m) (11) Midterm I a Page 3 of 6

4 0 3. The Challenger Deep located near the Mariana Islands in the deepest known point in theocean. Ithasadepthof11,000 m(almost7miles)andamaximummeasuredpressure of 110 MPa. If the density of sea water at the surface of the ocean is kg/m 3, what is the density of water at the bottom of the deep assuming that it has a constant bulk modulus of 10 9 N/m? The density of a fluid is given by ρ = m/v. If we think about a certain mass of water, m, on the surface with initial volume V o so that ρ 0 = m/v 0, and then move the water to the bottom of the Challenger Deep, it s volume will change according to Hooke s law, V/V o = P/B where P is the change in pressure ( P = 110 MPa), and B is the bulk modulus. The final volume, V b will be given by V b = V 0 + V = V 0 (1+ V V 0 ) = V 0 (1 P B ) (1) This means that the density at the bottom will be given by ρ b = m V b = m V 0 ( 1 P B ) = ρ o ( 1 P B ) (13) Therefore ρ b = (105 kg/m 3 ) ( ) (14) 1 ( Pa) ( 10 9 N/m ) Midterm I a Page 4 of 6

5 4. A large horizontal pipe with diameter 10 cm (the cross sectional area is m ) contains water flowing at a speed of 0.3 m/s. The pipe branches into four pipelets, each of diameter cm (cross sectional area is m each, or a total of m ). The four pipelets rise 3.5 m and then run horizontally. 10 (a) What is the speed of the fluid in the horizontal portion of each pipelet? The mass continuity equation says that dm/dt in a closed system is constant throughout the system. This implies that ρva is constant where ρ is the density, v is the fluid velocity, and A is the cross-sectional area. For an incompressible fluid va is constant, therefore v 1 A 1 = v A (15) where v 1 and A 1 are the velocity of the fluid and area of the large horizontal pipe, and v and A are the velocity of the fluid and area total area of the four pipelets. This means that v = v 1A 1 = (0.3 m/s)( m ) A m ) = 1.87 m/s (16) 15 (b) If the pressure in the large pipe is P = 50 kpa, what is the pressure in the pipelets? The pressure can be found using Bernoulli s equation P ρv 1+ρgy 1 = P + 1 ρv + ρgy where P 1 = 50 kpa, v 1 = 0.3 m/s, and y 1 = 0 are the pressure, velocity and height of the fluid in the large pipe and P, v = 1.87 m/s and y = 3.5 m are the pressure, velocity and height of the fluid in the pipelets. Solving Bernoulli s equation for P gives P = P 1 1 ρ(v v 1) ρg(y y 1 ) (17) ( 1 [ P = (50 kpa) (1000 kg/m 3 ) (1.87 m/s) (0.3 m/s) ] (9.8 m/s )(3.5 m)) (18) Midterm I a Page 5 of 6

6 5. A kg mass on a string (i.e. a pendulum) has a natural angular frequency of ω 0 =.6 rad/s and is placed in an environment with a damping force proportional to the velocity. 5 (a) What is the period of oscillation? 1 1. The undamped period is equal to T = π ω 0 so the period is T =. The damped period is equal to T = π ω 5 (b) What is the length of the string? where ω is found in part c. π (.6 rad/s) The natural angular frequency, ω 0 of a pendulum is given by ω 0 = g where l is l the length of the string, and g is the acceleration due to gravity. Therefore l = g ω 0 = (9.8 m/s ) (.6 rad/s) (19) 10 (c) If the amplitude is reduced to one tenth of the original in 0 s, what is the angular frequency of the damped motion? The angular frequency of a damped harmonic oscillator (of which a damped pendulum is an example) is given by ω = ω0 γ where gamma can be defined by the behavior of the amplitude A(t) as a function of time, A(t) = A 0 e γt. We know that A(0 s) = 1 A 10 0, so we can calculate γ = ln(a(t)/a 0) t = ln(0.1) (0 second) (0) and ( ) ω = ω0 γ ln(0.1) = (.6 rad/s) (1) (0 s) 1 Due to a mistake in how part a was posed so it can be answered in one of two ways. The question was meant to elicit the undamped period, but the damped period is a valid answer. Midterm I a Page 6 of 6