Like another comment said, thanks for doing these FRQs. You were the first result when I searched up answers to the FRQs, after I had saw they were released. I also have a question- what difficulty would you rank this set of FRQs relative to other years?
You're welcome. I'd say the kinematics one for #1 was probably the most unusual, because kinematics is not a big topic. But I expected it to be more prevalent with the loss of electricity and circuits and waves. However, I always found the circuit and electricity questions from previous years to be relatively easy for the topic they covered. So in that sense, I think you exchanged a relatively easy question (circuits) for a more difficult standard mechanics question. So I'd say overall it's a little bit harder. The difficulty for the electricity and circuits was more that some teachers were rushed to finish the syllabus, but that probably happened less frequently this time around.
I skimmed the video. When did I make that assumption? Like what time stamp are you referring to? I also think I verified these answers with the official solutions but it's been a while so I could be wrong somewhere. I just didn't see it at first glance
@@AllenTsaoSTEMCoach I am confused about your calculation for net torque in the first question (ai). You used mg directly as the force that provides the torque, so I am assuming that you are saying the torque done by the tension force is equal to the gravity force in magnitude. But that doesn't seem to be correct in terms of the net force of the block, because the acceleration is non-zero.
That's not the correct interpretation. I ignored the tension because that is internal to the system. And the system is the pulley + blocks (and the ropes included). You're thinking of doing the net torque of just the pulley. In that case, yes the torque on the pulley (tensions from the rope) would not be equal to the mass of the blocks. I actually made an entirely separate video where I did do separate free body diagrams for each object (even though the question did not ask you to), and you'll see you basically get the same result, but it's a lot more complicated analytically th-cam.com/video/cHOOAaaXozg/w-d-xo.html
I feel like part (a, i) is a bad question. You solved for the sum of torques BEFORE release. Once accelerating, the tensions, and hence the torques exerted by them would be different than before release.
I believe I solved the torques after release. If it was before release, I would have to include a force from a hand holding the system in place so the net torques would be 0. But after release, the net torque is non zero.
@@AllenTsaoSTEMCoach But once accelerating downward, the tension 1 would be less than m0g, and tension 2 would be greater than 1.5 m0g (as block 2 would be accelerating upward). And since the tensions are what exert torque on the system, it would give a different value to the net torque, wouldn't it? I guess the way they state the problem confuses me. I could buy your result for the "instant" of release, but not after that.
I don't really have an issue with the way you solved the problem -- I think it's the answer they were looking for. I just think it was a poorly stated problem. I think what they should have asked was what total torque do the blocks exert on the pulley before release. I think that is a different situation than the torques after release.
Ah I get what you mean. So the tricky part of this problem is it's the free body diagram of the entire system, not just the pulley. So the tension in the ropes are actually internal forces and don't contribute torque to the entire system. Only the forces of gravity are creating a torque on the entire system. That part is what made this question more difficult overall.
Without a clearly defined axis or orientation, you are free to choose which way is positive. So just like up can be considered positive or down can be considered positive, you can do the same here.
You shouldn't have to put you can if you want to be absolutely sure. The only time they would look for a specific sign is if they gave you an orientation that was supposed to be positive.
I was just wondering about your correction in the description, how would the system inertia decrease after cutting the string if the second pulley still remains on the axle.
Well you've disconnected a block entirely. So the rotational inertia of the pulley + block has been reduced, because the overall mass has been reduced.
They should accept the answer either way. I don't know if there's a hard and fast rule which way is positive. In reality, the vector for angular velocity is not either clockwise or counterclockwise and has a clearly defined direction but that's not covered in AP Physics 1.
@@AllenTsaoSTEMCoach torque direction follows the same convention as angles in math class. Counterclockwise is positive, clockwise is negative. Your graph should be flipped in 5c simply because we know the CCW (+) angular velocity from 0 to Tc. If they asked for angular speed, your graph could be correct, but they specify angular velocity on the y axis.
@@mrostersphysics While that may be a common convention, that is not required in this problem. If they don't indicate a direction, you are free to choose any direction you prefer. Angular speed would not be a proper graph because angular speed should always be positive, so it wouldn't have any portion negative. The formal direction of torque would actually be a vector pointing out of the page. And as with most vectors, you get to decide if that vector direction is positive or negative (similar to if up is positive or down is positive).
If they don't specify which torque direction is positive, it's fine to use either clockwise or counterclockwise as positive. So it would be fine if the graph were positive first then switching to negative.
Part i. You said the tension force causing the torque is equal to the weights of the objects. They are not equal. The objects are accelerating so the forces are NOT balanced. Tension does not equal weight.
You're right, I misspoke. I correctly analyzed the tension in a separate video if someone was interested in how it would look if we analyzed the objects separately.
Actually I can't find in the video where I said that. Can you let me know the approximate time? I wanted to note that correction and the corresponding time on the video for the corrections.
Hello! Why can’t the torque in part a be left as negative? Isn’t torque a vector quantity? How do you know your + and - directions are wrong based off of the negative torque?
Without reading aii, how would you know the direction of the torque of object 1? Initially, I thought they would both be positive since I incorrectly thought that object one goes up and 2 goes down so I got 3.5. Another question: do you have to indicate the positive direction on the AP exam?
You want to look at the force and which direction that force would cause the system to rotate: clockwise or counterclockwise. So if object 1 has gravity pulling down, you ask yourself if that downward force would cause it to rotate clockwise or counterclockwise. You don't *need* to specify a direction but I highly encourage everyone to in order to make sure you are using the correct positive/negative signs in your work.
The slope of the graph (for part c) would be angular acceleration, but you were comparing the torques. T = Iα, but the rotational inertia isn't the same before and after tc, I don't think.
That's a good point. So the I would be smaller so the angular acceleration should be even larger than before. So a slope steeper than 3x of before the string is cut. They likely will only grade that you knew the slope was steeper but not necessarily a specific value since that wouldn't be able to be calculated
Actually now that I reread it, the system is only the pulley for this part. But in my original graph, I was doing the motion of the whole system, and when you cut the block off, there's less rotational inertia because you've removed the block so it's just the pulley on the right block left.
The general shape is correct. What's incorrect is that the slopes do not have to be 3x steeper after the string is cut. You should have the slope steeper, but not by any particular amount
Like another comment said, thanks for doing these FRQs. You were the first result when I searched up answers to the FRQs, after I had saw they were released. I also have a question- what difficulty would you rank this set of FRQs relative to other years?
You're welcome. I'd say the kinematics one for #1 was probably the most unusual, because kinematics is not a big topic. But I expected it to be more prevalent with the loss of electricity and circuits and waves.
However, I always found the circuit and electricity questions from previous years to be relatively easy for the topic they covered. So in that sense, I think you exchanged a relatively easy question (circuits) for a more difficult standard mechanics question. So I'd say overall it's a little bit harder. The difficulty for the electricity and circuits was more that some teachers were rushed to finish the syllabus, but that probably happened less frequently this time around.
dang you uploaded these fast. thanks so much :D
I think this question is totally wrong: how can you regard the tension equal to the weight when there is an acceleration?
I skimmed the video. When did I make that assumption? Like what time stamp are you referring to?
I also think I verified these answers with the official solutions but it's been a while so I could be wrong somewhere. I just didn't see it at first glance
@@AllenTsaoSTEMCoach I am confused about your calculation for net torque in the first question (ai). You used mg directly as the force that provides the torque, so I am assuming that you are saying the torque done by the tension force is equal to the gravity force in magnitude. But that doesn't seem to be correct in terms of the net force of the block, because the acceleration is non-zero.
That's not the correct interpretation. I ignored the tension because that is internal to the system. And the system is the pulley + blocks (and the ropes included).
You're thinking of doing the net torque of just the pulley. In that case, yes the torque on the pulley (tensions from the rope) would not be equal to the mass of the blocks.
I actually made an entirely separate video where I did do separate free body diagrams for each object (even though the question did not ask you to), and you'll see you basically get the same result, but it's a lot more complicated analytically
th-cam.com/video/cHOOAaaXozg/w-d-xo.html
I feel like part (a, i) is a bad question. You solved for the sum of torques BEFORE release. Once accelerating, the tensions, and hence the torques exerted by them would be different than before release.
I believe I solved the torques after release. If it was before release, I would have to include a force from a hand holding the system in place so the net torques would be 0. But after release, the net torque is non zero.
@@AllenTsaoSTEMCoach But once accelerating downward, the tension 1 would be less than m0g, and tension 2 would be greater than 1.5 m0g (as block 2 would be accelerating upward). And since the tensions are what exert torque on the system, it would give a different value to the net torque, wouldn't it? I guess the way they state the problem confuses me. I could buy your result for the "instant" of release, but not after that.
By the way -- thank you for your quick response to my question!
I don't really have an issue with the way you solved the problem -- I think it's the answer they were looking for. I just think it was a poorly stated problem. I think what they should have asked was what total torque do the blocks exert on the pulley before release. I think that is a different situation than the torques after release.
Ah I get what you mean. So the tricky part of this problem is it's the free body diagram of the entire system, not just the pulley. So the tension in the ropes are actually internal forces and don't contribute torque to the entire system. Only the forces of gravity are creating a torque on the entire system. That part is what made this question more difficult overall.
shouldn't angular velocity in the counterclockwise direction be positive? meaning ur graph is flipped?
Without a clearly defined axis or orientation, you are free to choose which way is positive. So just like up can be considered positive or down can be considered positive, you can do the same here.
@@AllenTsaoSTEMCoach so is it necessary for us to clarify that next to the graph on the actual exam
You shouldn't have to put you can if you want to be absolutely sure. The only time they would look for a specific sign is if they gave you an orientation that was supposed to be positive.
how many points do you think I would lose on part A if I calculated net torque correctly but didn't do the absolute value part for magnitude?
You may not lose any. Depends on the grader.
Could you do the other form of the test? There were two sets of questions and this set happened to be the one I didn’t take
Unfortunately I cannot because the College Board does not release those FRQs. Those are considered secure and aren't allowed to be discussed.
@@AllenTsaoSTEMCoach Ah that’s right. Just checked the College Board website.
Thank you for doing these videos
You're very welcome.
For 5 c i did the same graph but reverse over the x axis, cause I had different + and - directions for clockwise and stuff. Is that ok?
Yep
I was just wondering about your correction in the description, how would the system inertia decrease after cutting the string if the second pulley still remains on the axle.
Well you've disconnected a block entirely. So the rotational inertia of the pulley + block has been reduced, because the overall mass has been reduced.
@@AllenTsaoSTEMCoach Thanks for replying! But did they deduct a point for just making a slope of 3 time steeper on the ap exam?
No they didn't
partC is the slope supposed to be larger than three times of that before time C since the moment of inertia also decreases?
Yep you're right. It would be larger than 3x
Have the multiple choice answers been released, or is it just the FRQs for now?
Awesome videos btw
Thanks! The multiple choice won't ever be released, unfortunately. Just the FRQs.
I believe that clockwise is positive by convention.
They should accept the answer either way. I don't know if there's a hard and fast rule which way is positive.
In reality, the vector for angular velocity is not either clockwise or counterclockwise and has a clearly defined direction but that's not covered in AP Physics 1.
@@AllenTsaoSTEMCoach Thanks for the clarification
@@AllenTsaoSTEMCoach torque direction follows the same convention as angles in math class. Counterclockwise is positive, clockwise is negative. Your graph should be flipped in 5c simply because we know the CCW (+) angular velocity from 0 to Tc. If they asked for angular speed, your graph could be correct, but they specify angular velocity on the y axis.
@@mrostersphysics While that may be a common convention, that is not required in this problem. If they don't indicate a direction, you are free to choose any direction you prefer.
Angular speed would not be a proper graph because angular speed should always be positive, so it wouldn't have any portion negative.
The formal direction of torque would actually be a vector pointing out of the page. And as with most vectors, you get to decide if that vector direction is positive or negative (similar to if up is positive or down is positive).
Wouldn’t the interval to 0-Tc be positive since there is a counterclockwise net torque resulting in a positive angular acceleration?
If they don't specify which torque direction is positive, it's fine to use either clockwise or counterclockwise as positive. So it would be fine if the graph were positive first then switching to negative.
Part i. You said the tension force causing the torque is equal to the weights of the objects. They are not equal. The objects are accelerating so the forces are NOT balanced. Tension does not equal weight.
You're right, I misspoke. I correctly analyzed the tension in a separate video if someone was interested in how it would look if we analyzed the objects separately.
Actually I can't find in the video where I said that. Can you let me know the approximate time? I wanted to note that correction and the corresponding time on the video for the corrections.
Hello! Why can’t the torque in part a be left as negative? Isn’t torque a vector quantity? How do you know your + and - directions are wrong based off of the negative torque?
They're asking for the magnitude of the net torque so they want you to say the absolute value of the net torque.
isn't counterclockwise supposed to be the positive direction
It doesn't always have to be.
Without reading aii, how would you know the direction of the torque of object 1? Initially, I thought they would both be positive since I incorrectly thought that object one goes up and 2 goes down so I got 3.5. Another question: do you have to indicate the positive direction on the AP exam?
You want to look at the force and which direction that force would cause the system to rotate: clockwise or counterclockwise. So if object 1 has gravity pulling down, you ask yourself if that downward force would cause it to rotate clockwise or counterclockwise.
You don't *need* to specify a direction but I highly encourage everyone to in order to make sure you are using the correct positive/negative signs in your work.
For the last part, although torque is 3x, hasn't the inertia changed? Which means angular acceleration is not necessarily 3x?
You're correct. I think I noted that in the description.
@@AllenTsaoSTEMCoach Oh sorry I didn't see that! Thank you for confirming
The slope of the graph (for part c) would be angular acceleration, but you were comparing the torques. T = Iα, but the rotational inertia isn't the same before and after tc, I don't think.
That's a good point. So the I would be smaller so the angular acceleration should be even larger than before. So a slope steeper than 3x of before the string is cut.
They likely will only grade that you knew the slope was steeper but not necessarily a specific value since that wouldn't be able to be calculated
@@AllenTsaoSTEMCoach Why would the rotational inertia change after the string is cut?
Actually now that I reread it, the system is only the pulley for this part. But in my original graph, I was doing the motion of the whole system, and when you cut the block off, there's less rotational inertia because you've removed the block so it's just the pulley on the right block left.
@@AllenTsaoSTEMCoach So is your answer in the video correct?
The general shape is correct. What's incorrect is that the slopes do not have to be 3x steeper after the string is cut. You should have the slope steeper, but not by any particular amount
For 5c, I did the same graph but upside down. Do you think I'd get partial credit for that?
The sign doesn't matter, so it would be the same score because they didn't specify which direction is positive.
Thanks