06 homework RAMSEY, TAYLOR Due: Oct 9 2006, 4:00 am
1
Mechanics - Basic Physical Concepts
Math: Circle: 2 r, r 2 ; Sphere: 4 r2 , (4/3) r3 b2 Quadratic Eq.: a x2 + b x + c = 0, x = -b 2 a-4 a c
Cartesian and polar coordinates: y x = r cos , y = r sin , r 2 = x2 + y 2 , tan = x Trigonometry: cos cos + sin sin = cos( - ) sin + sin = 2 sin + cos - 2 2 cos + cos = 2 cos + cos - 2 2 sin 2 = 2 sin cos , cos 2 = cos2 - sin2 1 - cos = 2 sin2 2 , 1 + cos = 2 cos2 2 Vector algebra: A = (Ax , Ay ) = Ax^ + Ay i ^ Resultant: R = A + B = (Ax + Bx , Ay + By ) Dot: A B = A B cos = Ax Bx + Ay By + Az Bz Cross product: ^ = k, k = ^, k ^ = i ^ ^ ^ ^ i ^ i ^ ^ i Ax Bx ^ Ay By ^ k Az Bz
B F s = F s cos = F s A F ds
(in Joules)
Effects due to work done: Fext = m a - Fc - fnc Wext |AB = KB - KA + UB - UA + Wdiss |AB B Kinetic energy: KB -KA = A m ads, K = 1 m v 2 2
B K (conservative F ): UB - UA = - A F ds 1 k x2 Uspring = 2 Ugravity = m g y, From U to F : Fx = - U , Fy = - U , Fz = - U x y z
Fspring = - U = -k x x U = 0, 2 U > 0 stable, < 0 unstable Equilibrium: x x2 W Power: P = ddt = F v = F v cos = F v (Watts)
Fgravity = - U = -m g, y
Collision
C =AB =
d d 1 Calculus: dx xn = n xn-1 , dx ln x = x , d sin = cos , d cos = - sin , d d d dx const = 0
C = A B sin = A B = A B , use right hand rule
Measurements
Impulse: I = p = pf - pi tif F dt Momentum: p = m v x Two-body: xcm = m1m1 +m2 x2 1 +m2 pcm M vcm = p1 + p2 = m1 v1 + m2 v2 Fcm F1 + F2 = m1 a1 + m2 a2 = M acm K1 + K2 = K1 + K2 + Kcm Two-body collision: pi = pf = (m1 + m2 ) vcm vi = vi - vcm , vi = vi + vcm Elastic: v1 - v2 = -(v1 - v2 ), vi = -vi , vi = 2 vcm - vi Many body center of mass: rcm = Force on cm: Fext = dp = M acm , dt
mi r i = mi r dm mi
t
Dimensional analysis: e.g., 2 F = m a [M ][L][T ]-2 , or F = m v [M ][L][T ]-2 r N (a x + b) = a N x + b N Summation: i i=1 i=1 i
p=
pi
Rotation of Rigid-Body
Motion
One dimensional motion: v = d s , a = d v dt dt s -s v -v Average values: v = tf -tii , a = tf -tii f f One dimensional motion (constant acceleration): v(t) : v = v0 + a t +v s(t) : s = v t = v0 t + 1 a t2 , v = v02 2 2 = v2 + 2 a s v(s) : v 0 Nonuniform acceleration: x = x0 + v0 t + 1 a t2 + 2 1 j t3 + 1 s t4 + 1 k t5 + 1 p t6 + . . ., (jerk, snap,. . .) 6 24 120 720 Projectile motion: trise = tf all =
ttrip v0y 2 = g 1 g t2 , R = v t h = 2 f all ox trip 2 1 Circular: ac = v , v = 2 T r , f = T (Hertz=s-1 ) r 2 + a2 Curvilinear motion: a = at r
Isphere = 2 M R2 , Ispherical shell = 2 M R2 5 3 I = M (Radius of gyration)2 , I = Icm + M D2 Kinetic energies: Krot = 1 I 2 , K = Krot + Kcm 2 Angular momentum: L = r m v = r m r = I Torque: = d L = m d v r = F r = I d = I dt dt dt Wext = K +U +Wf , K = Krot + 1 m v 2 , 2 P =
s Kinematics: = r , = v , = at r r 2 Moment of inertia: I = mi ri = r2 dm 2 2 Idisk = 1 M R2 , Iring = 1 M (R1 + R2 ) 2 2 1 M 2, I 1 M (a2 + b2 ) Irod = 12 rectangle = 12
Rolling, angular momentum and torque
I Rolling: K = 1 Ic + M R2 2 = 1 Rc + M v 2 2 2 2 Angular momentum: L = r p, L = r p = I
Relative velocity: v = v + u
Law of Motion and applications
Force: F = m a, Fg = m g, F12 = -F21
2 Circular motion: ac = v , v = 2 T r = 2 r f r Friction: Fstatic s N Fkinetic = k N
Torque: = d L = r d p = r F , = r F = I dt dt 1 Gyroscope: p = d = L d L = L = m g h I dt dt
Static equilibrium
Equilibrium (concurrent forces):
i Fi = 0
Energy
Work (for all F): W = WAB = WB - WA
i = 0 Fi = 0, about any point Subdivisions: rcm = mA rAcm +mB rBcm mA +mB Elastic modulus = stress/strain stress: F/A strain: L/L, x/h, -V /V
homework 06 RAMSEY, TAYLOR Due: Oct 9 2006, 4:00 am Version number encoded for clicker entry: V1:1, V2:4, V3:3, V4:5, V5:2. Question 1 part 1 of 3 10 points A block starts at rest and slides down a frictionless track (as shown in the figure below). It leaves the track horizontally, flies through the air, and subsequently strikes the ground. The acceleration of gravity is 9.81 m/s2 . 422 g 3.6 m
2 (3) (4)
Ug = m g h W = mg .
Choosing the point where the block leaves the track as the origin of the coordinate system, x = vx t 1 h2 = - g t2 2 since axi = 0 m/s2 and vyi = 0 m/s. Solution: 1 2 m vx = -m g h1 2
2 vx = -2 g (h - h2 ) = -2 g h1
(5) (6)
v
2.1 m
x What is the speed v of the ball when it leaves the track? Correct answer: 5.42494 m/s (tolerance 1 %). Explanation: Let : x = 3.54965 m , g = 9.81 m/s2 , m = 422 g , h1 = -1.5 m , h2 = -2.1 m , h = h1 + h2 , and vx = v . m h1 h
9.81 m/s2
vx = =
-2 g h1 -2 (9.81 m/s2 ) (-1.5 m)
(4)
= 5.42494 m/s . Question 2 part 2 of 3 10 points What is the horizontal distance x the block travels in the air? Correct answer: 3.54965 m (tolerance 1 %). Explanation: At y = -h2 = -2.1 m (below the jump off height), 1 h2 = - g t 2 2 t= -2 h2 . g (5) (6)
h2
g
v
Basic Concepts: chanical Energy
3.5 m Conservation of Me(1)
Since x = vx t , and using Eq. 4 and 6, we have x = vx t = =2 -2 g h1 h 1 h2 -2 h2 g
Ui = U f + K f + W . since vi = 0 m/s. 1 K = m v2 2
(2)
= 2 (-1.5 m) (-2.1 m) = 3.54965 m .
homework 06 RAMSEY, TAYLOR Due: Oct 9 2006, 4:00 am The acceleration of gravity is 9.81 m/s2 . 5.09 kN/m 574 g 2.3 m 4.7 cm
3
v
Explanation: Basic Concept: Kf = U i , since vi = 0 m/s and hf = 0 m. Solution: 1 2 m vf = m g h 2 vf = -2 g h = -2 (9.81 m/s2 ) (-3.6 m) = 8.40428 m/s , where h = h1 + h2 . Alternate Solution: (7) (8)
3.03 m What is the spring constant k ? Correct answer: 5087.6 N/m (tolerance 1 %). Explanation: Let : g = 9.81 m/s2 , m = 0.574 kg , x = 3.03 m , vx = v , and d = 0.047 m . 4.42 m/s
vy = = vf = =
-2 g h2 -2 (9.81 m/s2 ) (-2.1 m) so
2 2 vx + vy
(9)
h
(10) Basic Concepts: chanical Energy
x Conservation of Me(1)
(5.42494 m/s)2 + (6.41888 m/s)2
= 8.40428 m/s . Question 4 part 1 of 3 10 points A block is pushed against the spring with spring constant 5.09 kN/m (located on the left-hand side of the track) and compresses the spring a distance 4.7 cm from its equilibrium position (as shown in the figure below). The block starts at rest, is accelerated by the compressed spring, and slides across a frictionless horizontal track (as shown in the figure below). It leaves the track horizontally, flies through the air, and subsequently strikes the ground.
Ui = U f + K f since vi = 0 m/s. K= Us = 1 m v2 2
1 k d2 (3) 2 Choosing the point where the block leaves the track as the origin of the coordinate system, x = vx t 1 y = - g t2 2 since axi = 0 m/s2 and vyi = 0 m/s.
= 6.41888 m/s ,
d
(2)
g
k
m
What is the speed of the block when it hits the ground? Correct answer: 8.40428 m/s (tolerance 1 %).
9.81 m/s2
Question 3 part 3 of 3 10 points
homework 06 RAMSEY, TAYLOR Due: Oct 9 2006, 4:00 am Solution: Using Eqs. 1, 2, and 3, we have 1 1 2 m vx = k d 2 (4) 2 2 m 2 v d2 x 1 h = - g t2 2 x = vx t . k= Using Eq. 5 and substituting t = 7, we have
2
4
Alternative Solution: At y = h = -2.3 m (below the jump off height), 1 h = - g t2 2 t= -2 h . g (10)
(5) (6) (7) x from Eq. vx
Since x = vx t , therefore using Eq. 5 and 7, we have vx = x t -g 2h -(9.81 m/s2 ) 2 (-2.3 m)
x 1 , so h=- g 2 vx g x2 2 vx = - . (8) 2h 2 Using Eq. 5 and substituting vx from Eq. 8, we have m 2 k = 2 vx d m g x2 = 2 - (9) d 2h 0.574 kg = (0.047 m)2 (9.81 m/s2 ) (3.03 m)2 - 2 (-2.3 m) = 5087.6 N/m . Question 5 part 2 of 3 10 points What is the speed v of the block when it leaves the track? Correct answer: 4.42485 m/s (tolerance 1 %). Explanation: Using Eq. 4, we have k d2 = m k d2 vx = m
2 vx
=x
= (3.03 m)
= 4.42485 m/s .
Question 6 part 3 of 3 10 points What is the total speed of the block when it hits the ground? Correct answer: 8.04396 m/s (tolerance 1 %). Explanation: Basic Concept: Kf = U i , since vi = 0 m/s and hf = 0 m. 1 2 Solution: Using m vy = m g h , we have 2 vy = = vf = = -2 g h -2 (9.81 m/s2 ) (-2.3 m) so
2 2 vx + v y
(11)
= 6.71759 m/s , m)2
=
(5087.6 N/m) (0.047 0.574 kg
(4.42485 m/s)2 + (6.71759 m/s)2
= 4.42485 m/s .
= 8.04396 m/s .
homework 06 RAMSEY, TAYLOR Due: Oct 9 2006, 4:00 am Question 7 part 1 of 3 10 points A block of mass m is pushed a distance D up an inclined plane by a horizontal force F . The plane is inclined at an angle with respect to the horizontal. The block starts from rest and the coefficient of kinetic friction is k .
D
5
3. W = N D cos 4. W = (N - m g cos - F sin ) D 5. W = N D sin 6. W = (m g cos + F sin - N ) D 7. W = 0 correct 8. W = -N D
F
m
k
Explanation: The normal force makes an angle of 90 with the displacement, so the work done by it is zero. Question 9 part 3 of 3 10 points What is the final speed of the block? 1. v = 2. v = 3. v = 4. v = 5. v = 6. v = correct 7. v = 8. v = 2 (F cos - k N ) D m 2 (F cos + m g sin - k N ) D m 2 (F cos - m g sin + k N ) D m 2 (F cos + m g sin ) D m 2 (F sin + k N ) D m 2 (F cos - m g sin - k N ) D m 2 (F sin - k N ) D m 2 (F cos - m g sin ) D m
If N is the normal force, what is the work done by friction? 1. W = +k (N + m g cos ) D 2. W = 0 3. W = +k (N - m g cos ) D 4. W = -k (N - m g cos ) D 5. W = -k (N + m g cos ) D 6. W = -k N D correct 7. W = +k N D Explanation: The force of friction a has magnitude Ff riction = k N . Since it is in the direction opposite to the motion, we get Wf riction = -Ff riction D = -k N D. Question 8 part 2 of 3 10 points What is the work done by the normal force N? 1. W = (N + m g cos + F sin ) D 2. W = N D
Explanation: The work done by gravity is Wgrav = m g D cos(90 + ) = -m g D sin .
homework 06 RAMSEY, TAYLOR Due: Oct 9 2006, 4:00 am The work done by the force F is WF = F D cos . From the work-energy theorem we know that Wnet = K , WF + Wgrav + Wf riction Thus 4. None of these. vf = 2 (F cos - m g sin - k N ) D . m Question 10 part 1 of 1 10 points A student weighing 702 N climbs at constant speed to the top of an 6 m vertical rope in 11 s. What is the average power expended by the student to overcome gravity? Correct answer: 382.909 W (tolerance 1 %). Explanation: Let : F = 702 N , d = 6 m , and t = 11 s . The power expended is W Fd P = = t t (702 N) (6 m) = 11 s = 382.909 W . Question 11 part 1 of 1 10 points An elevator of mass m is initially at rest on the first floor of a building. It moves upward, and passes the second and third floors with a constant velocity, and finally stops at the fourth floor. The distance between adjacent floors is h. 5. W = 4 m g h 6. W = -4 m g h 1 2 = m vf . 2
6
What is the net work done on the elevator during the entire trip, from the first floor to the fourth floor? 1. W = 3 m g h 2. W = 0 correct 3. W = -3 m g h
Explanation: The key to answering this question is finding the net work done on the elevator. The work-energy theorem states that the net work, W , done on an object is related to the change in kinetic energy of the object, K, by W = K. The elevator starts at rest at the first floor, thus Ki = 0. The elevator ends at rest at the fourth floor and so Kf = 0, and so K is zero! Another way to think about this is that the work done on the elevator by the elevator motor is the exact negative of the work done on the elevator by gravity. Adding these two together to find the net work done on the elevator gives zero! Question 12 part 1 of 1 10 points A bullet with a mass of 6.86 g and a speed of 240 m/s penetrates a tree horizontally to a depth of 5.4 cm. Assume that a constant frictional force stops the bullet. Hint: Try energy considerations. Calculate the magnitude of this frictional force. Correct answer: 3658.67 N (tolerance 1 %). Explanation: Let : m = 6.86 g , v = 240 m/s ,
and
homework 06 RAMSEY, TAYLOR Due: Oct 9 2006, 4:00 am d = 5.4 cm . We can use conservation of energy to relate the initial kinetic energy of the bullet to the work done by the frictional force. 1 m v2 = f s 2 Solving for the frictional force, f , m v2 f= 2d (0.00686 kg) (240 m/s)2 = 2 (0.054 m) = 3658.67 N . Question 13 part 1 of 2 10 points Consider the loop-the-loop setup, where the mass is released from rest at C with height h. The mass is sliding along a frictionless track, and the radius of the loop is R. C m A R h D
7
Explanation: The centripetal force at point A is given by
2 vA Fc = m ac = m (2 g) = m R
1 2 m vA = m g R . 2 From the conservation of energy, = KA = UC + K C = U A + K A , or m g h + 0 = m g (2 R) + 1 2 m vA = 3 m g R 2
= h = 3 R . Question 14 part 2 of 2 10 points Consider a new situation where h = 4 R. The speed at the bottom of the track is 1. v = 2. v = 3. v = 7gR. 5gR. 4gR. 3gR. 8 g R . correct 10 g R . 9gR. 2gR. gR. 6gR.
B Determine the height h such that the centripetal acceleration at A is ac = 2 g . 1. h = 4 R 1 R 2 1 3. h = 4 R 2 2. h = 2 4. h = 3 R correct
4. v = 5. v = 6. v = 7. v = 8. v = 9. v = 10. v =
5. h = 5 R 6. h = 2 R 7. h = 3 1 R 2
Explanation: The total energy at C and at B are given respectively by EC = U C + K C = m g h + 0 = m g h
homework 06 RAMSEY, TAYLOR Due: Oct 9 2006, 4:00 am 1 1 2 2 m vB = m vB . 2 2 Conservation of energy gives EB = UB +KB = m g(0)+ EC = m g h = m g (4 R) = EB = = vB = 8gR. 1 2 m vB 2
8
smallest. This means that she will be moving fastest at the bottom of the swing. By conservation of energy, Etop = Ebot Ktop + Utop = Kbot + Ubot 1 1 2 2 m vtop + m g htop = m vbot + m g hbot . 2 2 Since vtop = 0, 1 2 m vbot = m g (htop - hbot ) so 2 vmax = vbot = = 2 g [htop - hbot ] 2 (9.8 m/s2 )[(1.9 m) - (0.6 m)]
Question 15 part 1 of 2 10 points A girl swings on a playground swing in such a way that at her highest point she is 1.9 m from the ground, while at her lowest point she is 0.6 m from the ground. The acceleration of gravity is 9.8 m/s2 .
= 5.04777 m/s . Question 16 part 2 of 2 10 points Given: In the following choices, htop is the height of the highest point and hbot is the height of the lowest point. At what height above the ground will the girl be moving at a speed half of her maximum speed? Correct answer: 1.575 m (tolerance 1 %). Explanation: Let h1/2 = the height, where v = v1/2 = 1 vmax . 2
4.5 m
1.9 m 0.6 m What is her maximum speed? Correct answer: 5.04777 m/s (tolerance 1 %). Explanation: Basic Concepts: Conservation of energy K= 1 m v2 2
Ugr = m g h .
v1
/2
r
v htop
Let : r = 4.5 m , htop = 1.9 m , hbot = 0.6 m .
and
h1/2
hbottom
From the conservation of energy, K1/2 + U1/2 = Ktop + Utop 1 2 m v1/2 + m g h1/2 = 0 + m g htop 2 1 m g h1/2 = m g htop - 2
Solution: We can solve this by using the principle of conservation of energy. We need to know the kinetic and potential energies at two points in time. The girl will be moving the fastest when her kinetic energy is largest which occurs when her potential energy is
1 2 v 4 max
.
homework 06 RAMSEY, TAYLOR Due: Oct 9 2006, 4:00 am From Part 1, 1 2 m vmax = m g [htop - hbot ] . 2 Substitution yields m g h1/2 h1/2 1 = m g htop - m g [htop - hbot ] 4 1 = [3 htop + hbot ] 4 1 = [3 (1.9 m) + (0.6 m)] 4 = 1.575 m . Question 17 part 1 of 2 10 points A single conservative force acting on a particle varies as F = (-A x + B x2 ) ^ , i where A = 6 N/m and B = 52 N/m2 and x is in meters. Find the change in potential energy as the particle moves from x0 = 3.2 m to x1 = 2 m . Correct answer: 410.592 J (tolerance 1 %). Explanation: The potential energy is
x
9
Find the change in kinetic energy of the particle between the same two points. Correct answer: -410.592 J (tolerance 1 %). Explanation: From conservation of energy (conservative force), the change of the kinetic energy is K = -U = -(410.592 J) = -410.592 J .
Question 19 part 1 of 2 10 points Starting from rest at a height equal to the radius of the circular track, a block of mass 16 kg slides down a quarter circular track under the influence of gravity with friction present (of coefficient ). The radius of the track is 34 m. The acceleration of gravity is 9.8 m/s2 . 16 kg
34 m
U (x) = - =
0 A x2
(-A x + B x2 ) d x B x3 . 3 Determine the work done by the conservative forces. Correct answer: 5331.2 J (tolerance 1 %). Explanation: The work done by the conservative force (gravity) is Wgrav = m g R = (16 kg) (9.8 m/s2 ) (34 m) = 5331.2 J .
2
-
If we take U (0) = 0, then the change in the potential energy is U = U (x1 ) - U (x0 ) Ax2 Bx3 Ax2 Bx3 0 1 1 0 - = - - 2 3 2 3 (6 N/m) (2 m)2 (52 N/m2 ) (2 m)3 = - 2 3 2 (52 N/m2 ) (3.2 m)3 (6 N/m) (3.2 m) - - 2 3 = 410.592 J . Question 18 part 2 of 2 10 points
Question 20 part 2 of 2 10 points
homework 06 RAMSEY, TAYLOR Due: Oct 9 2006, 4:00 am If the kinetic energy of the block at the bottom of the track is 3000 J, what is the work done against friction? Correct answer: 2331.2 J (tolerance 1 %). Explanation: W = Wf + K Wf = m g R - K = (16 kg) (9.8 m/s2 ) (34 m) - (3000 J) = 2331.2 J . Question 21 part 1 of 3 10 points A 3 kg block collides with a massless spring of spring constant 90 N/m attached to a wall. The speed of the block was observed to be 1.5 m/s at the moment of collision. The acceleration of gravity is 9.8 m/s2 . How far does the spring compress if the surface on which the mass moves is frictionless? Correct answer: 27.3861 cm (tolerance 1 %). Explanation: When the surface is frictionless, the maximum distance of compression can be found from energy conservation as follows 1 1 m v 2 = k x2 0 2 2 x0 = = m v k k =
10
Correct answer: 0.44712 (tolerance 1 %). Explanation: When there is friction, we should take into account the energy dissipated by friction in using energy conservation 1 1 m v 2 = k x2 + k m g x 2 2 m v 2 - k x2 2 mg x (3 kg) (1.5 m/s)2 = 2 (3 kg) (9.8 m/s2 ) (16.4317 cm) (90 N/m) (16.4317 cm)2 - 2 (3 kg) (9.8 m/s2 ) (16.4317 cm) = 0.44712 . Question 23 part 3 of 3 10 points Given: The coefficient of static friction between the floor and the block is 0.670681. Does the block remain at rest or does it bounce back off the spring once the spring is fully compressed? 1. cannot be determined 2. bounces back 3. stays at rest correct Explanation: Finally, to determine whether the block will bounce back or not, we should compare the maximum static frictional force possible and the force that is exerted by the spring Fspring = k x = 14.7885 N Fmaxf ric = s m g = 19.718 N Fspring < Fmaxf ric , so the block will remain at rest.
3 kg (1.5 m/s) 90 N/m = 27.3861 cm .
Question 22 part 2 of 3 10 points The maximum distance to which the spring was compressed was observed to be 16.4317 cm. What is the kinetic coefficient of friction between the block and the floor?