Student question: A few questions in the homework refer to a point in which objects move without slippage. I know this means that it is rolling, but I don't know what equations need to be satisfied for this to occur.
Dr. Winters' response: Things slide when there is no friction. Things move without slippage when there is friction. So if a problem says that an object moves or rolls without slipping then it means that you need to consider the force due to friction. And you need to consider the force of friction when you are looking at torques, because the force of friction acts on the outside surface of a wheel or ball, and that is a distance r from the center of mass (so there is torque τ=rF (the angle between friction, which acts tangent to the surface, and the radius r is 90 degrees so sinθ=1). It is probably a torque problem.
Hope that helps. Good luck.
Sunday, November 20, 2011
Saturday, November 19, 2011
HW9 #4
I'm really stuck on this problem... I know how to find the instantaneous angular velocity and the average angular acceleration, but I don't know how to calculate the instantaneous angular acceleration. I tried a few different methods, but none of them are right. For example, I tried alpha = (v^2)/(r^2), since a = alpha*r and in this case, the acceleration is centripetal acceleration. Did anyone figure out how to do this one?
Saturday, November 12, 2011
HW8 #8
Is this question referring to the initial speed of the point or after the time interval given in #5?
HW8 #10
I seem to be stuck on this problem. I converted the rotational velocity so that the units would be rad/s instead of rev/min. I calculated the radius (in m) using the diameter (given in m). I plugged these values into the KE =(1/2)(m)(wr)^2 equation [w = lower case omega], but my answer was incorrect. What am I doing wrong?
Sunday, November 6, 2011
HW7 #7
Student question:
I tried to do #7, but it doesn't seem to work. I have the following values given:
A projectile of mass 5 kg is fired with an initial speed of 198 m/s at an angle of 25◦ with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 3.2 kg and 1.8 kg . The 1.8 kg fragment lands on the ground directly below the point of explosion 2.8 s after the explosion.
The acceleration due to gravity is 9.81 m/s2 .
Find the magnitude of the velocity of the 3.2 kg fragment immediatedly after the explosion.
I know that the 1.8 kg has no horizontal velocity because it falls straight down, so I found the horizontal velocity of the 3.2 kg fragment by doing (5 kg)(198cos25 m/s)/(3.2 kg). I tried to find the vertical velocity by finding the initial vertical velocity of the 1.8 kg. I used X=V0t+1/2at^2 and used conservation of energy in the vertical direction to find the height, or X. I then put this velocity into conservation of momentum in the vertical direction and found the vertical velocity for the 3.2 kg fragment. Finally, I used the pythagorean triangle to find the resulting velocity. I've done this five times and it hasn't worked. Am I doing something wrong or missing some important step? If you could reply as soon as possible, I'd greatly appreciate it. Thanks!
Dr. Winters' response:
Very nice start. This is a fairly complex problem, where you need to look at kinematics and momentum. I like the way you found the horizontal component of the 3.2kg fragment using conservation of momentum in the horizontal direction.
I didn't follow how you found the vertical velocity of the 1.8kg fragment. It sounds as though you first found the vertical velocity and then found the height. I think that's backwards. You can find the height using kinematics (the explosion takes place at the top of the trajectory). Knowing the height and the time to fall, you can find the initial vertical velocity of the 1.8kg fragment.
After that, use conservation of momentum in the vertical direction, as you described, to find the vertical component of velocity of the 3.2kg fragment.
It sounds as though you are doing everything right except attacking the middle calculation backwards. Hope that helps. Good luck.
I tried to do #7, but it doesn't seem to work. I have the following values given:
A projectile of mass 5 kg is fired with an initial speed of 198 m/s at an angle of 25◦ with the horizontal. At the top of its trajectory, the projectile explodes into two fragments of masses 3.2 kg and 1.8 kg . The 1.8 kg fragment lands on the ground directly below the point of explosion 2.8 s after the explosion.
The acceleration due to gravity is 9.81 m/s2 .
Find the magnitude of the velocity of the 3.2 kg fragment immediatedly after the explosion.
I know that the 1.8 kg has no horizontal velocity because it falls straight down, so I found the horizontal velocity of the 3.2 kg fragment by doing (5 kg)(198cos25 m/s)/(3.2 kg). I tried to find the vertical velocity by finding the initial vertical velocity of the 1.8 kg. I used X=V0t+1/2at^2 and used conservation of energy in the vertical direction to find the height, or X. I then put this velocity into conservation of momentum in the vertical direction and found the vertical velocity for the 3.2 kg fragment. Finally, I used the pythagorean triangle to find the resulting velocity. I've done this five times and it hasn't worked. Am I doing something wrong or missing some important step? If you could reply as soon as possible, I'd greatly appreciate it. Thanks!
Dr. Winters' response:
Very nice start. This is a fairly complex problem, where you need to look at kinematics and momentum. I like the way you found the horizontal component of the 3.2kg fragment using conservation of momentum in the horizontal direction.
I didn't follow how you found the vertical velocity of the 1.8kg fragment. It sounds as though you first found the vertical velocity and then found the height. I think that's backwards. You can find the height using kinematics (the explosion takes place at the top of the trajectory). Knowing the height and the time to fall, you can find the initial vertical velocity of the 1.8kg fragment.
After that, use conservation of momentum in the vertical direction, as you described, to find the vertical component of velocity of the 3.2kg fragment.
It sounds as though you are doing everything right except attacking the middle calculation backwards. Hope that helps. Good luck.
Tuesday, November 1, 2011
Adopt-a-Physicist open until Nov 22
We have a unique opportunity to talk to 5 different people who have physics backgrounds. They would like to share their knowledge and opinions. Please take advantage of this offer.
To log on: go to www.adoptaphysicist.org
Log in by scrolling down to our Class: NY: Smithtown AP-C Physics
and our Class PIN (supplied in class).
Read about our physicists and write to them. Write thoughtfully. Find out about their careers, their education, and ask for their suggestions for you. They are happy to answer just about any relevant question, often at length. Keep up a conversation.
Our physicists are:
Dr. Bill Freeman. Manager for a $278M particle accelerator project at Fermilab.
Dr. Bryan Gorman. Mathematical modeling (statistical prediction). One example was for the US Coast Guard for improved search and rescue and drug interdiction.
Dr. Michaela Kleinert. Physics professor at Willamette University in OR. Her interests are in atomic/molecular physics and optics.
Mr. Scott Dodd. Medical Physicist, works in radiation oncology at a cancer treatment center.
Dr. von Foerster. Retired; former university professor and publisher.
To log on: go to www.adoptaphysicist.org
Log in by scrolling down to our Class: NY: Smithtown AP-C Physics
and our Class PIN (supplied in class).
Read about our physicists and write to them. Write thoughtfully. Find out about their careers, their education, and ask for their suggestions for you. They are happy to answer just about any relevant question, often at length. Keep up a conversation.
Our physicists are:
Dr. Bill Freeman. Manager for a $278M particle accelerator project at Fermilab.
Dr. Bryan Gorman. Mathematical modeling (statistical prediction). One example was for the US Coast Guard for improved search and rescue and drug interdiction.
Dr. Michaela Kleinert. Physics professor at Willamette University in OR. Her interests are in atomic/molecular physics and optics.
Mr. Scott Dodd. Medical Physicist, works in radiation oncology at a cancer treatment center.
Dr. von Foerster. Retired; former university professor and publisher.
Sunday, October 23, 2011
HW6 #1
Does anyone know how to solve #1? I have already tried change in energy over time, P=Fv, etc...
HW6 #17
Hey I am having a problem getting #17 to work out on the homework with my numbers. When I tried solving it with other people's numbers the same way, I got the right answer, but mine arent working out. I have an 18 kg child on a playground swing moving with a speed of 4.7 m/s at the lowest point, and the swing is 1 meter long. I need to find the angle the string makes with the vertical. First I tried using conservation of energy with K=U, so .5mv^2=mgh. The m's cancel out and I am trying to solve for h, height above the lowest point. I end up with .5(4.7^2)=9.81h, and when I solve for h, I get 1.12, which doesn't work because it is higher than the length of the string. I also tried solving it by setting cos(theta)=(centripital acceleration)/g, and the acceleration is greater than g so it doesnt work. Please let me know if I am doing something wrong! Thanks!
Sunday, October 16, 2011
HW5 #18
Hey everyone, I just wanted to share a helpful tip for #18. Be careful when converting cubic kilometers to cubic meters. It's not a 1:1,000 ratio because the value is cubed.
Friday, October 14, 2011
HW5 #14
For what length of time should the power be calculated? Is this question asking for the power generated per second/hour/day?
Sunday, October 9, 2011
HW4 #5
Can somebody please help me with this question? I was able to answer the three preceding steps, but I seem to be stuck on this one. I attempted to calculate the horizontal components of T1 and T2 using trigonometry, but this yielded an incorrect solution. I then tried adding these values together, but this was wrong too. Please help!
HW4 #15
If you are having trouble with this problem but you managed to do the other force problems, you might want to take another look at your force diagrams (free body diagrams).
The smaller block hangs over the edge of a table (over a pulley), and is attached to a string. There is a tension T upward, a force due to gravity (Fg = m*g) downward, and the block's resulting acceleration is downward. You use Newton's 2nd law to get an expression for the tension: ∑F=Fg-T=ma so, rearranging and writing Fg=mg, T=mg-ma.
Now look at the larger block. It has a force of gravity downward, a normal force upward, and 2 ropes pulling to the right. That makes it T to the right and another T to the right, so there is 2T pulling to the right (look at the pulley: it makes the string pull twice). With "right" as the positive direction, Newton's 2nd law becomes ∑F=2T=ma. Plug in the expression you had above for T and solve for the acceleration. "The rest is just algebra."
Hope that helps. Good luck.
The smaller block hangs over the edge of a table (over a pulley), and is attached to a string. There is a tension T upward, a force due to gravity (Fg = m*g) downward, and the block's resulting acceleration is downward. You use Newton's 2nd law to get an expression for the tension: ∑F=Fg-T=ma so, rearranging and writing Fg=mg, T=mg-ma.
Now look at the larger block. It has a force of gravity downward, a normal force upward, and 2 ropes pulling to the right. That makes it T to the right and another T to the right, so there is 2T pulling to the right (look at the pulley: it makes the string pull twice). With "right" as the positive direction, Newton's 2nd law becomes ∑F=2T=ma. Plug in the expression you had above for T and solve for the acceleration. "The rest is just algebra."
Hope that helps. Good luck.
HW4 #12
#12 seems to be giving some trouble too. I'll put up this post, but if it doesn't help please ask for more specific suggestions.
Peter Pan is the mass that moves downward vertically. The problem says she starts from vo=0, and you are given the distance and the time, so you can use the x= equation to find her acceleration. That is also the acceleration of the counterweight, which is the block that accelerates upwards on the incline.
Next look at Peter Pan to find the tension in the rope: Fg acts downward, T (the tension in the rope) acts upward, and Peter Pan's acceleration is downward. If you choose downward to be the positive direction (because Peter Pan is accelerating downward) then when you use Newton's 2nd law you get ∑F=Fg-T=ma. (Note that if you had chosen up to be positive then your right hand side would be negative: =-ma.) Relace Fg with m*g to get mg-T=ma, or rearrange to get T=mg-ma. Note that the mass is Peter Pan's mass, which is known, so you can calculate the tension since you already figured out the acceleration.
Now draw a force diagram (free body diagram) for the counterweight on the incline. The force due to gravity is Fg and points downward. The normal force points up-and-to-the-left, perpendicular to the incline. The tension in the rope pulls up-and-to-the-right, along the incline.
Since the counterweight will be accelerating along the incline, make a chart with forces parallel to the incline (it will have acceleration "a"), perpendicular to the incline (will have no acceleration), and "?" (idk) which is in some other direction. The normal force is perpendicular to the incline. The tension in the rope is parallel to the incline. The force due to gravity, Fg, is some other direction, so belongs in the "?" column. Figure out the components of Fg in the parallel and perpendicular directions: Fgsinθ in the parallel direction and Fgcosθ in the perpendicular direction.
Now you can use Newton's 2nd law to look at forces in the direction parallel to the incline (use the direction up-and-to-the right which is the direction of the acceleration as the positive direction): ∑F=T-Fgsinθ=ma And this is NOT equal to zero since there IS an acceleration. Replace Fg with m*g to get T-mgsinθ=ma and solve for the mass, which is the mass of the counterweight which is what you are looking for. "The rest is just algebra."
Hope that helps. Good luck.
Peter Pan is the mass that moves downward vertically. The problem says she starts from vo=0, and you are given the distance and the time, so you can use the x= equation to find her acceleration. That is also the acceleration of the counterweight, which is the block that accelerates upwards on the incline.
Next look at Peter Pan to find the tension in the rope: Fg acts downward, T (the tension in the rope) acts upward, and Peter Pan's acceleration is downward. If you choose downward to be the positive direction (because Peter Pan is accelerating downward) then when you use Newton's 2nd law you get ∑F=Fg-T=ma. (Note that if you had chosen up to be positive then your right hand side would be negative: =-ma.) Relace Fg with m*g to get mg-T=ma, or rearrange to get T=mg-ma. Note that the mass is Peter Pan's mass, which is known, so you can calculate the tension since you already figured out the acceleration.
Now draw a force diagram (free body diagram) for the counterweight on the incline. The force due to gravity is Fg and points downward. The normal force points up-and-to-the-left, perpendicular to the incline. The tension in the rope pulls up-and-to-the-right, along the incline.
Since the counterweight will be accelerating along the incline, make a chart with forces parallel to the incline (it will have acceleration "a"), perpendicular to the incline (will have no acceleration), and "?" (idk) which is in some other direction. The normal force is perpendicular to the incline. The tension in the rope is parallel to the incline. The force due to gravity, Fg, is some other direction, so belongs in the "?" column. Figure out the components of Fg in the parallel and perpendicular directions: Fgsinθ in the parallel direction and Fgcosθ in the perpendicular direction.
Now you can use Newton's 2nd law to look at forces in the direction parallel to the incline (use the direction up-and-to-the right which is the direction of the acceleration as the positive direction): ∑F=T-Fgsinθ=ma And this is NOT equal to zero since there IS an acceleration. Replace Fg with m*g to get T-mgsinθ=ma and solve for the mass, which is the mass of the counterweight which is what you are looking for. "The rest is just algebra."
Hope that helps. Good luck.
HW#4 Q14
I don't know if anyone else is having trouble with this problem, but I'm having trouble figuring out number 14. If anyone could give some helpful advice that would be awesome.
HW4 #3
I am getting questions about how to do #3 and the other tension problems. Here is a start; feel free to add more details if you think there is something missing here, or post a more specific question if this doesn't help quite enough.
A block is suspended from a long string. Since the block hangs from the middle and the angle on the right is the same as the angle on the left, it is reasonable to say that the tension T1 in the string up-and-to-the-left is the same as the tension T2 up-and-to-the-right (there is no reason why one of them would be a bigger force than the other).
The block is hanging from a vertical (up-down) string. Let's call the tension in the vertical string T. If you draw a force diagram for the block itself, you have Fg downward and T upward. From Newton's 2nd law and using up as your positive direction, ∑F=T-Fg=ma=0 and it =0 because the block is not accelerating. So T-Fg=0, and T=Fg=m*g, and you have a numerical answer for the tension T in the vertical string.
Now draw a force diagram (free body diagram) for the knot where all 3 strings attach. You have T downward, T1 up-and-to-the-left, and T2 up-and-to-the-right. Make a chart with columns for forces in the x direction, the y direction, and the "?" (idk) direction. T1 and T2 are in the ? direction, so figure out the x- and y-components of them. In question 2 you figured out the angle from the ceiling, so the horizontal (x) components of T1 is T1cosθ and T2 is T2cosθ, and the vertical (y) components are T1sinθ and T2sinθ.
Now you can use Newton's 2nd law in the vertical direction. If you choose up to be the positive direction again then ∑F=T1sinθ+T2sinθ-Fg=ma=0, and again it is =0 because the knot isn't accelerating. Since T1 is equal to T2 you can call them both T1, and solve for T1. That is the tension in the string.
Hope that helps. Good luck.
A block is suspended from a long string. Since the block hangs from the middle and the angle on the right is the same as the angle on the left, it is reasonable to say that the tension T1 in the string up-and-to-the-left is the same as the tension T2 up-and-to-the-right (there is no reason why one of them would be a bigger force than the other).
The block is hanging from a vertical (up-down) string. Let's call the tension in the vertical string T. If you draw a force diagram for the block itself, you have Fg downward and T upward. From Newton's 2nd law and using up as your positive direction, ∑F=T-Fg=ma=0 and it =0 because the block is not accelerating. So T-Fg=0, and T=Fg=m*g, and you have a numerical answer for the tension T in the vertical string.
Now draw a force diagram (free body diagram) for the knot where all 3 strings attach. You have T downward, T1 up-and-to-the-left, and T2 up-and-to-the-right. Make a chart with columns for forces in the x direction, the y direction, and the "?" (idk) direction. T1 and T2 are in the ? direction, so figure out the x- and y-components of them. In question 2 you figured out the angle from the ceiling, so the horizontal (x) components of T1 is T1cosθ and T2 is T2cosθ, and the vertical (y) components are T1sinθ and T2sinθ.
Now you can use Newton's 2nd law in the vertical direction. If you choose up to be the positive direction again then ∑F=T1sinθ+T2sinθ-Fg=ma=0, and again it is =0 because the knot isn't accelerating. Since T1 is equal to T2 you can call them both T1, and solve for T1. That is the tension in the string.
Hope that helps. Good luck.
Friday, September 23, 2011
HW2 #011
Hey guys, just wanted to share a helpful tip for #011 on this week's UT homework because it's kind of tricky. Don't forget to account for the initial x- and y- components of displacement (it's given to you in question #009) when calculating the answer. Hope this helps :]
Saturday, September 10, 2011
U Texas "unique #"
When you log in to U Texas for the first time, you will want to enroll in a class. Our "unique number" is 2011. There is no homework the first week of class.
Dr. Winters
Dr. Winters
Subscribe to:
Posts (Atom)