One Version Homework 8 Heinz 81204 Oct 14, 2004 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. The due time is Central time. Billiard Collision 02 09:04, trigonometry, multiple choice, > 1 min, normal. 001 (part 1 of 1) 10 points Assume an elastic collision (ignoring friction and rotational motion). A queue ball initially moving at 8 m/s strikes a stationary eight ball of the same size and mass. After the collision, the queue balls nal speed is 4.6 m/s . Before 8 m /s
1
Subtracting the equation (1) for the conservation of energy we have 2 p qf p ef = 0 . Dividing equation (3) by 2 m v qf v ef = 0 , (4) (3)
4.6 m/ s
yields three possibilities 1) vef = 0, where mq misses me . 2) vqf = 0, where a head-on collision results. 3) vef and vef are + = 90 . Most pool players know that the queue ball and the target ball scatter at 90 to one-another after a two-body collision (to a close approximation). Use this third possibility.
4.6 m
Before 8 m /s After Find the queue balls angle with respect to its original line of motion. Correct answer: 54.9004 . Explanation: Let : vqi = 8 m/s and v q f = 4. 6 m /s . Given mq = me = m, pei = 0, p m v , and p p p2 . Conservation of energy (and multiplying by 2 m) gives p2f + p2f = p2i . q e q (1) y
/s
54.9
90 rewriting then
2 2 v qi v qf
6 After .55 m/ The impact parameter is y . s Dividing by the mass, Eq. 1 gives us
2 2 2 v qi = v qf + v ef , 2 2 2 v ef = v qi v qf ,
v ef = =
(8 m/s)2 (4.6 m/s)2
Conservation of momentum (specically, pqf + pef = pqi , and squaring) gives (pqf + pef ) (pqf + pef ) = pqi pqi . Carring out the scalar multiplication term by term gives p qf p qf + p ef p ef + 2 p qf p ef = p qi p qi . Rewriting in a simplied form p2f + p2f + 2 pqf pef = p2i . q e q (2)
= 6.54523 m/s , and v ef = arctan v qf 6.54523 m/s = arctan 4.6 m/s = 54.9004 , also vef = vqi sin = (8 m/s) sin(54.9004 ) = 6.54523 m/s , and vqf = vqi cos
Version One Homework 8 Heinz 81204 Oct 14, 2004 = (8 m/s) cos(54.9004 ) = 4.6 m/s , nally = 90 = 35.0996 y 2 R cos = R R = 2 cos(54.9004 ) = 1.15 . Car Collision 09:05, trigonometry, numeric, > 1 min, normal. 002 (part 1 of 1) 10 points Two cars, one of mass 1000 kg, and the second of mass 2000 kg, are moving at right angles to each other when they collide and stick together. The initial velocity of the rst car is 10 m/s in the positive x direction and that of the second car is 16 m/s in the positive y direction. What is the magnitude of the velocity of the wreckage of the two cars immediately after the collision? Correct answer: 11.1754 m/s. Explanation: The collision in this case is a completely inelastic collision. The cars have the same speed v after the collision. To nd v , we break the equation of conservation of linear momentum into x and y components: m 1 v 1 = (m 1 + m 2 ) v x m 2 v 2 = (m 1 + m 2 ) v y so that m1 v1 vx = m1 + m 2 (1000 kg) (10 m/s) = 3.33333 m/s = 1000 kg + 2000 kg m2 v2 vy = m1 + m 2 (2000 kg) (16 m/s) = = 10.6667 m/s 1000 kg + 2000 kg Therefore, v= = vx2 + vy2 (3.33333 m/s)2 + (10.6667 m/s)2 004 (part 2 of 2) 5 points
2
Rigid System Rotating 03 10:05, trigonometry, numeric, > 1 min, normal. 003 (part 1 of 2) 5 points The gure below shows a rigid 3-mass system which can rotate about an axis perpendicular to the system. The mass of each connecting rod is negligible. Treat the masses as particles. The x-axis is along the horizontal direction with the origin at the left-most mass 2 kg. 2 kg 4 kg 5 kg
3m
x
3m
The masses are separated by rods of length 3 m, so that the entire length is 2 (3 m). Determine the x-coordinate of the center of mass for the three-mass system with respect to the origin. Correct answer: 3.81818 m. Explanation: First nd the center of mass. Dene the origin to coincide with the far left mass M .
XCM =
m i xi mi (2 kg)(0 ) = (2 kg) + (4 kg) + (5 kg) (4 kg)(1 (3 m) + (2 kg) + (4 kg) + (5 kg) (5 kg)(2 (3 m) + (2 kg) + (4 kg) + (5 kg) 14 kg = (3 m) 11 kg = 3.81818 m .
= 11.1754 m/s
Version One Homework 8 Heinz 81204 Oct 14, 2004 Note: The position x0 about which the rigid body rotates is not necessarily to scale. x0 2 kg 4 kg 5 kg
3
L
x
L
005 (part 1 of 1) 10 points Assume: The masses are point particles; e.g., neglect the contribution due to moments of inertia about their center of mass. Three spherical masses are located in a plane at the positions shown in the gure below. A has mass 9.26 kg, B has mass 59.3 kg, and C has mass 27 kg. Three Masses in a Plane C
Find the moment of inertia of the 3-mass system about a rotation axis perpendicular to the x-axis and passing through the point (3 m) x0 = , with respect to the origin at the 5 left-most mass 2 kg. Correct answer: 169.56 kg m2 . Explanation: The moment of inertia of a system of point 2 particles is given by I = mi ri . Label the three moments of inertia as Ileft , Imiddle , and Iright . Remembering that the distances ri are with respect to the axis of rotation, we have Ileft = (2 kg) Imiddle = (4 kg) Iright = (5 kg) Hence I = Ileft + Imiddle + Iright 1 (3 m)2 = (2 kg) 25 16 + (4 kg) (3 m)2 25 81 (3 m)2 + (5 kg) 25 471 kg (3 m)2 = 25 = 169.56 kg m2 . Three Masses 02 10:05, trigonometry, numeric, > 1 min, wording-variable. 1 (3 m) 5 4 (3 m) 5 9 (3 m) 5
2
10 9 8 y Distance (m) 7 6 5 4 3 2 1 0
A
B
2
2
.
3 4 5 6 7 8 9 10 x Distance (m) Figure: Drawn to scale. Calculate the moment of inertia (of the three masses) with respect to an axis perpendicular to the xy plane and passing through x = 7 m and y = 7 m . Your answer must be within 3%. Correct answer: 1936.13 kg m2 . Explanation: Basic Concepts: The moment of inertia is I=
i 2 mi ri , 2 m i (x 2 + y i ) . i i
0
1
2
(1) (2)
=
Solution: Using the knot (7 m, 7 m) as the coordinate origin (see gure below), we have xa = (3 m) (7 m) = 4 m , ya = (5 m) (7 m) = 2 m , ra =
2 x2 + ya = 4.47214 m , a
Version One Homework 8 Heinz 81204 Oct 14, 2004 xb = (8.5 m) (7 m) = 1.5 m , yb = (2 m) (7 m) = 5 m , rb =
2 x2 + yb = 5.22015 m , b
4
xc = (8 m) (7 m) = 1 m , yc = (9 m) (7 m) = 2 m , rc =
2 x2 + yc = 2.23607 m , c
3 2 1 y Distance (m) 0 -1 -2 -3 -4 -5 -6
Three Masses in a Plane C
Decelerated Grinding Wheel 10:06, trigonometry, numeric, > 1 min, normal. 006 (part 1 of 2) 5 points The motor driving a grinding wheel with a rotational inertia of 0.1 kg m2 is switched o when the wheel has a rotational speed of 20 rad/s. After 6 s, the wheel has slowed down to 16 rad/s. What is the absolute value of the constant torque exerted by friction to slow the wheel down? Correct answer: 0.0666667 N m. Explanation: We have t = L = (I ) , so that B | | = 3 I | 1 0 | t 1 (0.1 kg m2 ) (20 rad/s 16 rad/s) = 6s = 0.0666667 N m .
A
-7 -7 -6 -5 -4 -3 -2 -1 0 1 x Distance (m) Figure: Drawn to scale. Using Eqs. 1, we have I=
i 2 mi ri ,
2
(1)
2 2 2 = m a ra + m b rb + m c rc = (9.26 kg) (4.47214 m)2 + (59.3 kg) (5.22015 m)2 + (27 kg) (2.23607 m)2 = 1936.13 kg m2 . Alternate Solution: Eqs. Using 2, we have 2 (2) I= m i [x 2 + y i ] , i i 2 2 = m a [x 2 + y a ] + m b [x 2 + y a ] a b 2 + m c [x 2 + y a ] c = (9.26 kg) [(4 m)2 + (2 m)2 ] + (59.3 kg) [(1.5 m)2 + (2 m)2 ] + (27 kg) [(1 m)2 + (2 m)2 ] = 1936.12 kg m2 .
007 (part 2 of 2) 5 points If this torque remains constant, how long after the motor is switched o will the wheel come to rest? Correct answer: 30 s. Explanation: When the wheel comes to rest, its angular speed is 2 = 0; hence t 2 = I (0 2 ) I 0 = | | (0.1 kg m2 ) (20 rad/s) = (0.0666667 N m) = 30 s .
Serway CP 08 31
Version One Homework 8 Heinz 81204 Oct 14, 2004 10:07, trigonometry, numeric, > 1 min, normal. 008 (part 1 of 3) 4 points A model airplane whose mass is 0.75 kg is tethered by a wire so that it ies in a circle 30 m in radius. The airplane engine provides a net thrust of 0.8 N perpendicular to the tethering wire. Find the torque the net thrust produces about the center of the circle. Correct answer: 24 N m. Explanation: Given : m = 0.75 kg , T = 0.8 N , and r = 30 m . The torque is = rT = (30 m) (0.8 N) = 24 N m . 009 (part 2 of 3) 3 points Find the angular acceleration of the airplane when it is in level ight. Correct answer: 0.0355556 rad/s2 . Explanation: The angular acceleration is = I = m r2 24 N m = (0.75 kg) (30 m)2 = 0.0355556 rad/s2 . 010 (part 3 of 3) 3 points Find the linear acceleration of the airplane tangent to its ight path. Correct answer: 1.06667 m/s2 . Explanation: Finally, the linear acceleration is: a = r = (30 m) 0.0355556 rad/s2 = 1.06667 m/s2 . Let : Mb = 50 kg , Vertical Height (m)
5
Boom and Weight 04 12:03, trigonometry, numeric, > 1 min, normal. 011 (part 1 of 1) 10 points A weight (with a mass of 50 kg) is suspended from a point near the right-hand end of a uniform boom with a mass of 50 kg . To support the uniform boom a cable runs from this same point to a wall (the left-hand vertical coordinate in the gure) and by a pivot on the same wall at an elevation of 3.5 m . The acceleration of gravity is 9.8 m/s2 . Boom and Weight 10 9 8 7 6 5 4 3 2 1 0 0 1 50 kg 50 kg
T
2 3 4 5 6 7 8 9 10 Horizontal Distance (m) Figure: Drawn to scale. Calculate the tension T in the cable. Your answer must be within 3%. Correct answer: 1384.97 N. Explanation: Basic Concepts: The Static Equilibrium Conditions are Tx = 0 , Ty = 0 , = 0, and or (1) (2) (3) (4)
W b xb + W w x = T y x .
Version One Homework 8 Heinz 81204 Oct 14, 2004 Mw = 50 kg , = 9 m boom length , xb = = 4. 5 m , 2 x = 7 m, y = 3. 5 m , h = 8.5 m , from the gure hy = arctan x 8. 5 m 3. 5 m = arctan 7m = 35.5377 . Force Scale is 161 N/m 10 9 Vertical Height (m) 8 7 6 5 4 3 (0, y ) 2 1 0 -1 2 3 4 5 6 7 8 9 10 Horizontal Distance (m) Solution: Let the cable make an angle with the horizontal boom, then using Eq. 5 and the gure, we have Ty hy = Since sin = , T (h y ) 2 + x 2 Ty = T (h y ) (h y ) 2 + x 2 , (6) -2 0 1 11.2 kg
55
6
Solving for T , we have [h y ] 2 + x 2 x [h y ] = [(50 kg) (4.5 m) + (50 kg) (7 m)] (9.8 m/s2 )
T = [W b x b + W w x]
(5)
[(8.5 m) (3.5 m)]2 + (7 m)2 (7 m) [(8.5 m) (3.5 m)] = 1384.97 N .
(0, h)
T
Tx Wb
Ty
(x, y ) Ww
Knot in Equilibrium 12:03, trigonometry, numeric, > 1 min, normal. 012 (part 1 of 1) 10 points The system shown in the gure is in equilibrium. A 11.2 kg mass is on the table. A string attached to the knot and the ceiling makes an angle of 55 with the horizontal. The coecient of the static friction between the 11.2 kg mass and the surface on which it rests is 0.38. The acceleration of gravity is 9.8 m/s2 .
m
Solving Eq. 4 for Ty , we have [W b x b + W w x] , x and substituting Ty from Eq. 6, we have Ty = T (h y ) (h y ) 2 + x 2 = [W b x b + W w x] . x
What is the largest mass m can have and still preserve the equilibrium? Correct answer: 6.0782 kg. Explanation:
Let :
M = 11.2 kg , m = 6.0782 kg , = 55 .
and
Version One Homework 8 Heinz 81204 Oct 14, 2004 Basic Concepts: The net force exerted on an object in equilibrium equals zero. Solution: For the system to remain in equilibrium, the net forces on both M and m should be zero; and thus the tension in the rope has an upper bound value Tmax Tmax cos = M g , so (1) M g Tmax = cos (0.38) (11.2 kg) (9.8 m/s2 ) = cos(55 ) = 72.7171 N . For m to remain in equilibrium Tmax sin = mmax g , so Tmax sin mmax = g (72.7171 N) sin(55 ) = (9.8 m/s2 ) = 6.0782 kg . (2)
7
Explanation: Because of translational and rotational equilibrium, m2 = T m1 m g 20 N 0.75 kg 0.1 kg = 9.8 m/s2 = 1.19082 kg Tl g m1 l1 50m l2 = m2 1 20 N 40 cm = [ 1.19082 kg 9.8 m/s2 0.75 kg 6 cm 50 0.1 kg ] = 60.5741 cm
Alternative Solution: Equations 1 and 2 come directly from the free-body diagram for the knot. Dividing Eq. 2 by Eq. 1, we have tan = mmax mmax , so M = M tan = (0.38) (11.2 kg) tan(55 ) = 6.0782 kg .
Serway CP 08 24 12:03, trigonometry, numeric, > 1 min, normal. 014 (part 1 of 3) 4 points A 15 m, 500 N uniform ladder rests against a frictionless wall, making an angle of 60 with the horizontal.
15 m (3)
4m Meter Stick in Equilibrium 12:03, trigonometry, numeric, < 1 min, normal. 013 (part 1 of 1) 10 points A(n) 0.1 kg meter stick is supported at its 40 cm mark by a string attached to the ceiling. A(n) 0.75 kg mass hangs vertically at the 6 cm mark. A second mass is attached at another mark to keep it horizontal and in rotational and translational equilibrium. The acceleration of gravity is 9.8 m/s2 . If the tension in the string attached to the ceiling is 20 N, determine the mark at which the second mass is attached. Correct answer: 60.5741 cm.
500 N 800 N
60
Note: Figure is not drawn to scale. Find the horizontal force exerted on the base of the ladder by Earth when a(n) 800 N re ghter is 4 m from the bottom. Correct answer: 267.506 N. Explanation: Given : L = 15 m , = 4 m, = 60 , W = 500 N , Wp = 800 N .
and
Version One Homework 8 Heinz 81204 Oct 14, 2004 Nw
8
L sin Ng
f cos L cos 2
P ivot
Wp W
016 (part 3 of 3) 3 points If the ladder is just on the verge of slipping when the reghter is 9 m up from the bottom, what is the coecient of static friction between ladder and ground? Correct answer: 0.324204 . Explanation: Given : max = 9 m
L cos
Using the rotational equilibrium equation already derived, Wp W + L tan 2 tan 500 N (800 N) (9 m) + = (15 m) tan 60 2 tan 60 = 421.466 N .
Applying rotational equilibrium with the pivot at the point of contact with the ground, L cos = Wp cos + W 2 Nw L sin = 0 2 Wp cos + W L cos = 2 Nw L sin Nw = Wp cos W cos + L sin 2 sin Wp W = + L tan 2 tan (800 N) (4 m) 500 N = + (15 m) tan 60 2 tan 60 = 267.506 N .
Nw =
Then the maximum friction is fmax = Ng = Nw Nw = Ng 421.466 N = 1300 N = 0.324204 .
Applying translational equilibrium horizontally, f = Nw = 267.506 N , in the positive x direction. 015 (part 2 of 3) 3 points What is the vertical force? Correct answer: 1.3 kN. Explanation: Applying translational equilibrium vertically, Ng = W p + W = (800 N + 500 N) = 1.3 kN , in the positive y direction. 1 kN 1000 N