Remember the important parts from today: centripetal acceleration is ALWAYS directed radially (towards the center of the circle) and has magnitude of v^2/R. Uniform circular motion means that the speed v is constant as the object moves around.
Solutions - Problems from Chapter 5:
1. acceleration is 41.667 m/s^2, or 4.252 g's.
2. 1.519 x 10-2 m
3. speed of Earth around Sun = 2.989 x 104 m/s, so acceleration = 5.954 x 10-3 m/s^2. We haven't talked about finding the magnitude of the force yet - that will come tomorrow!
Wednesday, October 15, 2008
Sunday, October 12, 2008
Unregrettable Regressions
Hi everyone,
I hope your long weekend is going well - I wanted to post some info that will help you perform a linear regression on your data.
Instructions using excel
Instructions using TI-83/84
I hope your long weekend is going well - I wanted to post some info that will help you perform a linear regression on your data.
Instructions using excel
Instructions using TI-83/84
Thursday, October 9, 2008
Practice before Exam 2!
Here is the link to the website for the book. Pick Chapter 4 and browse around the options at the left side of the screen. The physlet problems and the practice problems are great!
Giancoli 5th Edition Website
Giancoli 5th Edition Website
Monday, October 6, 2008
Inertial Reference Frames and the Temple of Doom
Questions:
14.
The bag of groceries exerts a 40 N force(a) directed downwards (b) on the person (c).
16. While the forces are the same, the acceleration resulting from a "tug" may be different in both cases. The friction force, which provides an opposing force to the tension in the rope, largely determines whether either side will stay put.
22. By pressing horizontally, the normal force is increased. Since friction force is proportional to the normal force, the friction force directed up the wall will also increase.
Problems:
8. bullet acceleration = 21,875 m/s^2, force on bullet = 153 N.
17. Using g = 9.8 m/s^2, the maximum upwards acceleration is .557 m/s^2. Using g = 10 m/s^2, the maximum acceleration is .357 m/s^2. These should be solved by setting the tension to the maximum value, and then solving the equations of motion for acceleration.
33. You can approximate the two trains connected to a third that pulls it along by having three blocks in a row with two connecting strings:
The masses of the two left blocks are the same. Write Newton's 2nd for both blocks, and you should get that T1 = ma and that T2 - T1 = ma. This should help!
34. This is the same style as the pendulum problem from class. You should be able to obtain from your FBD (remember NOT to tilt your axes!) that T = mg/cos(theta) and that the acceleration is g*tan(theta).
The only catch is that you don't know acceleration. You DO know that it slows to a stop from 20 m/s in 5 seconds. Hint. Hint.
The answer is 21.8 degrees.
58.
a)
This problem works the same as having one block on the table with a string that holds a hanging block (like problem 81!).
For Cleo: T - Fk = ma and Fn1 - m1g = 0
For Figaro: m2g - T = m2a.
Solving algebraically, a = 0.098 m/s^2.
b)
Using kinematics, v0 = 0, a = 0.098 m/s^2, and delta_x = 0.9 meters.
Solving for delta_t = 4.111 seconds.
66. Force required = 21,000 N. Convert 90 km/h into m/s! displacement = 1.042 meters.
79.
Following the logic we used in class, we can derive that a = g tan(theta) = 4.663 m/s^2. From v = v0 + a*delta_t, with v0 = 0, and delta_t = 18 seconds, v = 83.9 m/s.
14.
The bag of groceries exerts a 40 N force(a) directed downwards (b) on the person (c).
16. While the forces are the same, the acceleration resulting from a "tug" may be different in both cases. The friction force, which provides an opposing force to the tension in the rope, largely determines whether either side will stay put.
22. By pressing horizontally, the normal force is increased. Since friction force is proportional to the normal force, the friction force directed up the wall will also increase.
Problems:
8. bullet acceleration = 21,875 m/s^2, force on bullet = 153 N.
17. Using g = 9.8 m/s^2, the maximum upwards acceleration is .557 m/s^2. Using g = 10 m/s^2, the maximum acceleration is .357 m/s^2. These should be solved by setting the tension to the maximum value, and then solving the equations of motion for acceleration.
33. You can approximate the two trains connected to a third that pulls it along by having three blocks in a row with two connecting strings:
The masses of the two left blocks are the same. Write Newton's 2nd for both blocks, and you should get that T1 = ma and that T2 - T1 = ma. This should help!
34. This is the same style as the pendulum problem from class. You should be able to obtain from your FBD (remember NOT to tilt your axes!) that T = mg/cos(theta) and that the acceleration is g*tan(theta).
The only catch is that you don't know acceleration. You DO know that it slows to a stop from 20 m/s in 5 seconds. Hint. Hint.
The answer is 21.8 degrees.
58.
a)
This problem works the same as having one block on the table with a string that holds a hanging block (like problem 81!).
For Cleo: T - Fk = ma and Fn1 - m1g = 0
For Figaro: m2g - T = m2a.
Solving algebraically, a = 0.098 m/s^2.
b)
Using kinematics, v0 = 0, a = 0.098 m/s^2, and delta_x = 0.9 meters.
Solving for delta_t = 4.111 seconds.
66. Force required = 21,000 N. Convert 90 km/h into m/s! displacement = 1.042 meters.
79.
Following the logic we used in class, we can derive that a = g tan(theta) = 4.663 m/s^2. From v = v0 + a*delta_t, with v0 = 0, and delta_t = 18 seconds, v = 83.9 m/s.
Sunday, October 5, 2008
Complex systems
32.
a) On the FBD of the bucket and girl together, the tension is directed upwards TWICE. The sum of forces in the y-direction becomes 2T - mg = 0 (for constant speed).
Solving, T = 318.5 N
b) The new tension is 350.35 N (1.1* 318.5 N). 2T - mg = ma so a = (700.7 N - 637 N)/65 kg = 0.98 m/s^2.
45.
a) Let F = 730 N and F12 be the force between the two boxes.
Left block:
Fnetx = F - F12 - mu*Fn1 = m1*a
Fnety = Fn1 - m1g = 0
Right block:
Fnetx = F12 - mu*Fn2 = m2*a
Fnety = Fn2 - m2g = 0
Solving (2), Fn1 = m1g and Fn2 = m2g
Substituting:
F - F12 - mu*m1g = m1*a
F12 - mu*m2g = m2*a
Solving this system, a = 2.476 m/s^2, and F12 = 434 N
61.
a)
We basically derived this in class for the 2nd problem. Let positive be defined to be for m2 dropping.
a = (m2g - m1g*sin(theta))/(m1 + m2)
b)
If the system is to accelerate in the positive direction as defined above, then the net force in the numerator of the fraction above must be positive. This will occur if m2g > m1g*sin(theta). The system will accelerate to the right if the inequality goes the other way.
69.
This is similar to problem 46 from the other day, as well as to the do-now problem on Thursday. Here, you know mu, delta_x, and are asked to find v0. Check out your work for Problem 46 for guidelines on solving this.
81.
a) The key to this problem is knowing that the block WAS in equilibrium until the last bit of sand is added - from this, you know that Fs = Fsmax = mu*Fn. The hanging bucket (and sand) is in equilibrium, so you can find that T = m2g.
You should obtain that the mass of the hanging bucket m2 = 12.6 kg. You must subtract the mass of the bucket to get the sand mass = 11.6 kg.
b)
The equations in the x-direction are now equal to m1*ax and m2*ax respectively.
Solving the system of equations, ax = .879 m/s^2.
Problem on page:
acceleration of all three objects is 1.63 m/s^2 in the direction that moves the bucket upwards. The tension is 57.2 N.
a) On the FBD of the bucket and girl together, the tension is directed upwards TWICE. The sum of forces in the y-direction becomes 2T - mg = 0 (for constant speed).
Solving, T = 318.5 N
b) The new tension is 350.35 N (1.1* 318.5 N). 2T - mg = ma so a = (700.7 N - 637 N)/65 kg = 0.98 m/s^2.
45.
a) Let F = 730 N and F12 be the force between the two boxes.
Left block:
Fnetx = F - F12 - mu*Fn1 = m1*a
Fnety = Fn1 - m1g = 0
Right block:
Fnetx = F12 - mu*Fn2 = m2*a
Fnety = Fn2 - m2g = 0
Solving (2), Fn1 = m1g and Fn2 = m2g
Substituting:
F - F12 - mu*m1g = m1*a
F12 - mu*m2g = m2*a
Solving this system, a = 2.476 m/s^2, and F12 = 434 N
61.
a)
We basically derived this in class for the 2nd problem. Let positive be defined to be for m2 dropping.
a = (m2g - m1g*sin(theta))/(m1 + m2)
b)
If the system is to accelerate in the positive direction as defined above, then the net force in the numerator of the fraction above must be positive. This will occur if m2g > m1g*sin(theta). The system will accelerate to the right if the inequality goes the other way.
69.
This is similar to problem 46 from the other day, as well as to the do-now problem on Thursday. Here, you know mu, delta_x, and are asked to find v0. Check out your work for Problem 46 for guidelines on solving this.
81.
a) The key to this problem is knowing that the block WAS in equilibrium until the last bit of sand is added - from this, you know that Fs = Fsmax = mu*Fn. The hanging bucket (and sand) is in equilibrium, so you can find that T = m2g.
You should obtain that the mass of the hanging bucket m2 = 12.6 kg. You must subtract the mass of the bucket to get the sand mass = 11.6 kg.
b)
The equations in the x-direction are now equal to m1*ax and m2*ax respectively.
Solving the system of equations, ax = .879 m/s^2.
Problem on page:
acceleration of all three objects is 1.63 m/s^2 in the direction that moves the bucket upwards. The tension is 57.2 N.
Thursday, October 2, 2008
The "Inclined Plane" Truth - solutions
46. This problem is the same as our do-now, though it's a little hidden. Stopping distance refers to the distance the block travels as it slows to a stop.
b) 47.37 meters (don't forget to convert 95 km/h into m/s!)
c) the acceleration becomes (1/6) the acceleration due to gravity. If a = 1.63 m/s^2, then the new distance is 284.2 meters.
49.
a) If there is no friction, then a = g sin(theta). If we assume a distance d along the slide (it can be anything!) then we can derive that the speed the child has at the bottom of the slide is Vno-friction=(2gd sin(theta))^(1/2).
If there is friction, the problem says the speed at the bottom is HALF this quantity.
With friction, solve for the new acceleration, and solve for the final velocity of the child at the bottom of the slide. The displacement is the SAME as with no friction.
You should get that a = g sin(theta) - mu*g*cos(theta) and that the final velocity is:
(2gd*sin(theta)-2*mu*gD*cos(theta)) ^ (1/2). If you take this expression and set it equal to 0.5*Vno-friction, you should be able to solve for mu.
The answer (phew!) is that mu = 0.75*tan(28) == .399
51.
a) 1.792 seconds
b) The acceleration does NOT depend on the mass, so the time required would be the same.
52.
a) a = g sin(theta) = 3.67 m/s^2
b) v = (2*a*d)^1/2 = 8.17 m/s
53.
a) 1.226 meters
b) 1.634 seconds (This is just like finding the total time a projectile is in the air - find the time required for the block to stop, and then double it for the entire trip.)
54.
b) 47.37 meters (don't forget to convert 95 km/h into m/s!)
c) the acceleration becomes (1/6) the acceleration due to gravity. If a = 1.63 m/s^2, then the new distance is 284.2 meters.
49.
a) If there is no friction, then a = g sin(theta). If we assume a distance d along the slide (it can be anything!) then we can derive that the speed the child has at the bottom of the slide is Vno-friction=(2gd sin(theta))^(1/2).
If there is friction, the problem says the speed at the bottom is HALF this quantity.
With friction, solve for the new acceleration, and solve for the final velocity of the child at the bottom of the slide. The displacement is the SAME as with no friction.
You should get that a = g sin(theta) - mu*g*cos(theta) and that the final velocity is:
(2gd*sin(theta)-2*mu*gD*cos(theta)) ^ (1/2). If you take this expression and set it equal to 0.5*Vno-friction, you should be able to solve for mu.
The answer (phew!) is that mu = 0.75*tan(28) == .399
51.
a) 1.792 seconds
b) The acceleration does NOT depend on the mass, so the time required would be the same.
52.
a) a = g sin(theta) = 3.67 m/s^2
b) v = (2*a*d)^1/2 = 8.17 m/s
53.
a) 1.226 meters
b) 1.634 seconds (This is just like finding the total time a projectile is in the air - find the time required for the block to stop, and then double it for the entire trip.)
54.
Wednesday, October 1, 2008
handout answers
1. a = 3.33 m/s^2 DOWN for the 8 kg block.
2.
a) upwards
c) T = 5000 N
d) M = 625 kg
3.
b) T = 1050 N
c) If up is positive, the acceleration of the helicopter is +5.2 m/s^2, and the package is -9.8 m/s^2. (Many of you had this sign wrong when we were working on it on Friday! Your other work should be correct.)
delta_x for the helicopter = 70.4 m
delta_x for the package = 40.4 m
distance = 70.4 m + 5 m - 40.4 m = 35.0 m
4. a = 0.75 g
5.
a) a = F/6m
b) left tension = F/2, right tension = 5F/6
6.
b) F = mu*mg/(cos(theta) - mu*sin(theta))
7.
a) a = 3.0 m/s^2
c) a = 1.96 m/s^2 (NOT the same as part a!)
8. F = g(m1 + m2)/mu
2.
a) upwards
c) T = 5000 N
d) M = 625 kg
3.
b) T = 1050 N
c) If up is positive, the acceleration of the helicopter is +5.2 m/s^2, and the package is -9.8 m/s^2. (Many of you had this sign wrong when we were working on it on Friday! Your other work should be correct.)
delta_x for the helicopter = 70.4 m
delta_x for the package = 40.4 m
distance = 70.4 m + 5 m - 40.4 m = 35.0 m
4. a = 0.75 g
5.
a) a = F/6m
b) left tension = F/2, right tension = 5F/6
6.
b) F = mu*mg/(cos(theta) - mu*sin(theta))
7.
a) a = 3.0 m/s^2
c) a = 1.96 m/s^2 (NOT the same as part a!)
8. F = g(m1 + m2)/mu
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