TBR Momentum and Torque Review #8

[circulus vitios]

---

Members do not see this ad.

Rod mass = 0.5 kg
Counterweight mass = 0.01 kg
X = 1 cm with nothing in the pan.

What is the mass of the pan?

A. 0.1 kg
B. 0.5 kg
C. 1.0 kg
D. 5.0 kg

The solution is: (10 cm)(mg) = (1 cm)(0.01 kg)(g) + (10 cm)(0.5 kg)(g)

Why are they ignoring the rod's mass on the left of the fulcrum? Why are they using 10 cm as the radius on the right side of the fulcrum, and why is the full mass of the rod used on the right side of the fulcrum? Shouldn't the mass of the rod be accounted for on both sides of the fulcrum, and shouldn't the mass be a fraction of the overall mass of the rod?

edit: I realize 10 cm is the distance from the fulcrum to the rod's center of mass, but that still makes no sense.

 

[osprey099]

Why are they ignoring the rod's mass on the left of the fulcrum? Why are they using 10 cm as the radius on the right side of the fulcrum, and why is the full mass of the rod used on the right side of the fulcrum? Shouldn't the mass of the rod be accounted for on both sides of the fulcrum, and shouldn't the mass be a fraction of the overall mass of the rod?

edit: I realize 10 cm is the distance from the fulcrum to the rod's center of mass, but that still makes no sense.

You assume the rod's mass is concentrated at the center of mass

 

[circulus vitios]

You assume the rod's mass is concentrated at the center of mass

But the fulcrum isn't at the center of mass. There is an unequal distribution of weight to the left and to the right of where the fulcrum is positioned in my diagram; there is more weight on the right.

 

[milski]

Because it's a massless rod. Sorry, physics tends to handle problems this way - until you're told it is not so, consider everything ideal ball with uniform density and no friction. 😉

 

[Postictal Raiden]

Because it's a massless rod. Sorry, physics tends to handle problems this way - until you're told it is not so, consider everything ideal ball with uniform density and no friction. 😉

Yet its mass is taken into account on the right side of the fulcrum. How can you explain that?

Shouldn't the equation be sat up like this?

10mg + 10*0.5*g = 1*0.01*g + 30mg

 

[milski]

Yet its mass is taken into account on the right side of the fulcrum. How can you explain that?

Shouldn't the equation be sat up like this?

10mg + 10*0.5*g = 1*0.01*g + 30mg

By not reading the question, doh! 🙁 It's completely my mistake. The rod should be part of the equation and you can use its middle point as a center of mass.

Can't see the problem while typing on the phone - I'll come back with the equation.

Ok, read the whole example carefully now. You can replace each rigid body with its weight applied at its center of mass. For the rod, the center of mass is 10 cm to the right and the weight is 0.5g. That leads to the equation in the OP.

Alternatively, you can split the rod in two bars. The left has a center of mass 5cm to the left and weight 0.5/4g. The right center of mass is 15 cm with 3*0.5/4g weight. It should be easy to see that this simplifies to the exactly same equation.

 

[Postictal Raiden]

By not reading the question, doh! 🙁 It's completely my mistake. The rod should be part of the equation and you can use its middle point as a center of mass.

Can't see the problem while typing on the phone - I'll come back with the equation.

Ok, read the whole example carefully now. You can replace each rigid body with its weight applied at its center of mass. For the rod, the center of mass is 10 cm to the right and the weight is 0.5g. That leads to the equation in the OP.

Alternatively, you can split the rod in two bars. The left has a center of mass 5cm to the left and weight 0.5/4g. The right center of mass is 15 cm with 3*0.5/4g weight. It should be easy to see that this simplifies to the exactly same equation.

I'm sorry, but my skull is a little thick when it comes to physics.

I don't understand why the mass of the rod is accounted for on the right side, but not on the left side? Also, why are we only considering half of the rod not the entirety of it? Wouldn't the other 20cm on the right of the center of the mass exert a clockwise torque on the system?

 

[milski]

The mass is accounted at the center of mass of the body. For the rod, that center happens to be on the right side of the fulcrum. If you want to account for each of the parts of the rod on the two sides of the fulcrum, you can do that too. You still need to account for the mass of each at the center of mass of each of these parts. The two ways are equivalent and will lead to equivalent equations.

We are not considering the half of the rod in any of these cases - we are considering its mass and where the center of mass is. For a uniform rod, that's the middle of the rod. You cannot write something like 0.1m * 0.5kg/4 * g for the torque of the left side of the rod - that would imply that all of its mass is concentrated at the its end and the rest of it has no mass at all. Since the density is uniform along the length of the rod, you can consider it to be the same as a point with the same mass at the middle of the rod.

 

[kgpremed11]

Rod mass = 0.5 kg
Counterweight mass = 0.01 kg
X = 1 cm with nothing in the pan.



The solution is: (10 cm)(mg) = (1 cm)(0.01 kg)(g) + (10 cm)(0.5 kg)(g)

Why are they ignoring the rod's mass on the left of the fulcrum? Why are they using 10 cm as the radius on the right side of the fulcrum, and why is the full mass of the rod used on the right side of the fulcrum? Shouldn't the mass of the rod be accounted for on both sides of the fulcrum, and shouldn't the mass be a fraction of the overall mass of the rod?

edit: I realize 10 cm is the distance from the fulcrum to the rod's center of mass, but that still makes no sense.

I just did that problem today and had no idea what they were talking about.

 

[Postictal Raiden]

The mass is accounted at the center of mass of the body. For the rod, that center happens to be on the right side of the fulcrum. If you want to account for each of the parts of the rod on the two sides of the fulcrum, you can do that too. You still need to account for the mass of each at the center of mass of each of these parts. The two ways are equivalent and will lead to equivalent equations.

We are not considering the half of the rod in any of these cases - we are considering its mass and where the center of mass is. For a uniform rod, that's the middle of the rod. You cannot write something like 0.1m * 0.5kg/4 * g for the torque of the left side of the rod - that would imply that all of its mass is concentrated at the its end and the rest of it has no mass at all. Since the density is uniform along the length of the rod, you can consider it to be the same as a point with the same mass at the middle of the rod.

Thank you, this explains it.

I thought that the mass of the rod was distributed throughout the entire length and that I should account for it on both sides of the equation. Knowing that the mass is centered in the middle of the rod helped me understand why the equation was sat up the way it is in the OP.

 

[milski]

Thank you, this explains it.

I thought that the mass of the rod was distributed throughout the entire length and that I should account for it on both sides of the equation. Knowing that the mass is centered in the middle of the rod helped me understand why the equation was sat up the way it is in the OP.

Just keep in mind that the mass being centered there is only an abstraction - it keeps the math simple (and still correct) but is not what the real object is.

 

[ToldYouSo]

The rod's center of mass can be treated as the point where all the mass is; as previously stated it is an abstraction but it makes the calculations much easier and works well. It's analogous to two planets gravitationally interacting where you treat the whole mass of each planet as a point in the middle.

 

[circulus vitios]

TBR's section on torque is really lacking. If you're having trouble understanding how a situation could exist where net force is zero but torque isn't zero, look at these diagrams:

Wheel with force tangential to wheel. Forces are same magnitude and in opposite directions so the net force is zero but there is a clockwise net torque.

Rod with opposite forces of the same magnitude so they cancel out, but they're at different radii from the pivot point. Because they're at different distances from the pivot point, the torques are different and will not cancel out. In this particular case, there will be a clockwise net torque.

 

[ToldYouSo]

I found torque in general to be a little tricky; in both of those examples there is no translational motion so it's not hard to tell that the net force is zero.

 

[4563728]

The mass is accounted at the center of mass of the body. For the rod, that center happens to be on the right side of the fulcrum. If you want to account for each of the parts of the rod on the two sides of the fulcrum, you can do that too. You still need to account for the mass of each at the center of mass of each of these parts. The two ways are equivalent and will lead to equivalent equations.

We are not considering the half of the rod in any of these cases - we are considering its mass and where the center of mass is. For a uniform rod, that's the middle of the rod. You cannot write something like 0.1m * 0.5kg/4 * g for the torque of the left side of the rod - that would imply that all of its mass is concentrated at the its end and the rest of it has no mass at all. Since the density is uniform along the length of the rod, you can consider it to be the same as a point with the same mass at the middle of the rod.

Do we have to consider the mass torque of the rod in all torque problems? In some problems, the rod isnt taken in to consideration, so when do you know when to include it in torque problems?

 

[milski]

Do we have to consider the mass torque of the rod in all torque problems? In some problems, the rod isnt taken in to consideration, so when do you know when to include it in torque problems?

If you don't have to consider the rod, you'll be told to or the data about its mass will be missing from the problem.

 

[Blueprint MCAT Tutor]

Do we have to consider the mass torque of the rod in all torque problems? In some problems, the rod isnt taken in to consideration, so when do you know when to include it in torque problems?

The usual approach for all MCAT problems here is simple: make any and all simplifying assumptions you can (no friction, massless rod, etc.) unless the problem itself brings them up.

Remember, the MCAT is not out to trick you. If it looks like a straightforward problem, then it is. Well-prepped students so often hit trouble b/c they're over-thinking things.

 

[Blueprint MCAT Tutor]

This problem has already been answered, but I just want to add another quick illustration: you could, if you wanted, "split the rod up" into left and right sides and apply the torque on each side. But that would just get you the same answer.

Here an illustration of what it would look like if you "split the rod up"

But again, it's faster to just treat every object in a torque problem as if it "exists" only at its center of mass. It's kind of weird and counter-intuitive, but it works.

 

[4563728]

This problem has already been answered, but I just want to add another quick illustration: you could, if you wanted, "split the rod up" into left and right sides and apply the torque on each side. But that would just get you the same answer.
Here an illustration of what it would look like if you "split the rod up"

But again, it's faster to just treat every object in a torque problem as if it "exists" only at its center of mass. It's kind of weird and counter-intuitive, but it works.

Thanks so much for your help! So in this specific problem, I am assuming we had to involve the torque of the rod because the fulcrum was not placed at the center of mass? I think I understand it for the most part but it was just a little confusing at first.

I had another question about a problem from TBR, from section 5, passage 6 number 39. I don't really understand why the graph is a straight line. If we are graphing the relationship between v^2 vs. wavelength, shouldn't it be a square root graph (specifically, option B)? I really need help with this problem!

 

[BerkReviewTeach]

The relationship is given in the passage as v [proportional to] square root of lambda, so v2 [proportional to] lambda. The graph has v2 on the y-axis and lambda on the x-axis, so the relationship between v2 and lambda is a linear one 9directly proportional). I think you are thinking of the v vs. lambda graph, which would be a square root graph as you are stating. But the graph is not v and lambda, it's v2 and lambda.