Can someone help me with this rotational dynamics problem.

  • #26
So now, 6.4-2Fp=ma

and I know the Torque is (Fp+Fg)R they both act in the same direction, hence why I added them together. I know, Torque also equals Iα where α=a/r and I=.5MR^2 so all together .5MR^2(a/R)=(Fp+Fg)(R) where a=2(Fp+Fg)/M. I want to eliminate Fg so I have only two variables to solve for a, a being one of them. I am eluded to there must be another way of doing this that I am not aware of, in which I can solve for a where I do not have a second friction force. I know on a single cylinder, I have mg pointing down, and the normal force pointing up, I have the mg from the plank pointing down onto the cylinder, I have the force of friction from the ground pointing to the right. and I have the force of friction from the plank pointing to the left. As of now I have been measuring torque from the center of mass of the cylinder. If there is any other way of measuring the torque, I'm not aware of it. Is the torque applied from the bottom of the cylinder would the torque be Fp(2R)=(.5*M2R^2)(a/2R)?
 
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  • #27
tiny-tim
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  • #28
You're right, 2(Fp-Fg)/M=a So 6.4-Fp=M(2(Fp-Fg)/M) I'm missing something huge because I continue to get the wrong answer. I get something that's close but off by a bit.
 
  • #29
Ok so add the moment of inertia of the center of mass to the moment of inertia from the ground? Would the moment of inertia from the ground be .5M2R^2
 
  • #30
tiny-tim
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You're right, 2(Fp-Fg)/M=a So 6.4-Fp=M(2(Fp-Fg)/M) I'm missing something huge because I continue to get the wrong answer. I get something that's close but off by a bit.
(i assume you also used your .5MR^2(a/R)=(Fp+Fg)(R) ?)

you'd better show us your complete calculation :smile:

(i'll guess you're out by a factor of two somewhere because of the two cylinders)
 
  • #31
Alright so would I do (Fp-Fg)(R)=(.5(MR^2)+M(2R^2))a/R for the parallel axis theorem then? T=RxF which is equal to the moment of inertia (using the parallel axis theorem) times a/r. Then solve for a but then if i plug that into my F=ma equation i still have three variables :(
 
  • #33
I believe I figured it out, ok so I took

F-Fp=ma

T=FpR

and T=Iα

For figuring the correct I value, I did the parallel axis theorem, I=I_cm+MD^2

where I_cm=.5MR^2 and MD^2=4MR^2 so my combined I value is

.5MR^2+4MR^2=9/2MR^2=I

The I took my I value and multiplied it by α=a/R to give me torque and I set this equation to my T=FpR

(9/2)MR^2(a/R)=Fp(R)

I solved for Fp and got (9/2)Ma=Fp

I then plugged (9/2)Ma=Fp into my F-Fp=ma equation to get

F-(9/2)Ma=ma

I then plugged in my respective values and solved for a and got a certain value and multiplied it by two (because of the two cylinders, one of which was not accounted for in this calculation) to get my final answer, does this sound right? I used the parallel axis theorem.
 
  • #34
SammyS
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What did you get for the acceleration of the plank ?
 
  • #35
I got a=.877 but there is a flaw some where in my calculating because for all my answers, it say's I'm 10% within the expected value and to check my calculations. What do you think?
 
  • #36
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I got a=.877 but there is a flaw some where in my calculating because for all my answers, it say's I'm 10% within the expected value and to check my calculations. What do you think?
I get an answer between 0.95 and 1.00 m/s2.
 
  • #37
How else can you do this with out having Fg in the equation?
 
  • #38
tiny-tim
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hi ben! :smile:

(just got up :zzz:)
For figuring the correct I value, I did the parallel axis theorem, I=I_cm+MD^2

where I_cm=.5MR^2 and MD^2=4MR^2
(why are you using CAPITALS for distances? everyone else uses little letters :wink:)

no, d (in the parallel axis theorem) is always the distance from the centre of mass (r, not 2r) :smile:
I then plugged in my respective values and solved for a and got a certain value and multiplied it by two (because of the two cylinders, one of which was not accounted for in this calculation) to get my final answer, does this sound right?
i'm not sure what you multiplied by 2 :confused:

i think the simplest way is to treat it as one cylinder with twice the mass

can i now go back over the tactics, since you seemed confused about them earlier?

if you're using the centre of mass, then you don't need the parallel axis theorem, but you do need both Fp and Fg

if you're using the bottom (the centre of rotation, ie the only part that's instananeously stationary), then you do need the parallel axis theorem, but you only need Fp (not Fg)

(and if you need Fg, then it's Fp + Fg for the torque, but Fp - Fg for the F = ma)​

finally, you can only do angular momentum = LP = IPω (and therefore τP = IPα) if P is
i] either the centre of mass
ii] or the (instantaneously stationary) centre of rotation (to be precise: where rotation stays parallel to a principal axis of the body, any point on the instantaneous axis of rotation) …​

from the pf library
dL/dt is easiest to calculate about either the centre of mass (C) or (in a "two-dimensional case" where rotation stays parallel to a principal axis of the body) the centre of rotation (R) … in those cases, it is simply the moment of inertia "times" the angular acceleration:
τnet,c.o.m. = dLc.o.m./dt = Ic.o.m.α
τnet, c.o.r. = dLc.o.r./dt = Ic.o.r.α

Sometimes a more general point P is needed, and then:
LP = Ic.o.m.ω + m PC x vc.o.m.

Where rotation stays parallel to a principal axis, so that L stays parallel to ω, then m RC x vc.o.m. = RC x (ω x RC) = m RC2ω, which, from the parallel axis theorem, is (Ic.o.r. - Ic.o.m.)ω, so Lc.o.r = Ic.o.r.ω

This applies, for example, to a sphere or a cylinder rolling over a step, but not to a cone rolling on a plane, or a wheel rolling on a curved rail. :wink:
 
  • #39
ok, so I decided to do the parallel axis theorem and got the wrong answer.

t=Fpr=(.5(2m)(r^2)+2mr^2)a/r

I have 2m because I am treating it as one cylinder with twice the mass. Once I solved for Fp I plugged that into my F-Fp=Ma and got the wrong answer again.
 
  • #40
tiny-tim
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hi ben! :smile:
t=Fpr=(.5(2m)(r^2)+2mr^2)a/r

I have 2m because I am treating it as one cylinder with twice the mass. Once I solved for Fp I plugged that into my F-Fp=Ma and got the wrong answer again.
show us your complete equations :smile:

(i suspect you used the wrong r either in τ = Fpr or in a = αr)
 
  • #41
A plank with a mass M = 5.60 kg rides on top of two identical, solid, cylindrical rollers that have R = 4.40 cm and m = 2.00 kg. The plank is pulled by a constant horizontal force of magnitude 6.40 N applied to the end of the plank and perpendicular to the axes of the cylinders (which are parallel). The cylinders roll without slipping on a flat surface. There is also no slipping between the cylinders and the plank.

t=Fpr=(.5(2m)(r^2)+2mr^2)a/r

Fp(.044)=(.5(2(2)(.044^2))+2(2)(.044^2))a/.044

.044Fp=.264a

Fp= 6a

F-Fp=Ma

6.4-6a=5.6a

a=.552
 
  • #42
tiny-tim
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t=Fpr=(.5(2m)(r^2)+2mr^2)a/r

Fp(.044)=…
no, the torque about the bottom point is Fp times .088 :wink:
 
  • #43
May I ask your reasoning why you multiplied Fp by 2R? Is that because there are two cylinders?
 
  • #44
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A plank with a mass M = 5.60 kg rides on top of two identical, solid, cylindrical rollers that have R = 4.40 cm and m = 2.00 kg. The plank is pulled by a constant horizontal force of magnitude 6.40 N applied to the end of the plank and perpendicular to the axes of the cylinders (which are parallel). The cylinders roll without slipping on a flat surface. There is also no slipping between the cylinders and the plank.

t=Fpr=(.5(2m)(r^2)+2mr^2)a/r

Fp(.044)=(.5(2(2)(.044^2))+2(2)(.044^2))a/.044

.044Fp=.264a

Fp= 6a

F-Fp=Ma

6.4-6a=5.6a

a=.552
So interpreting the following
Fp(.044)=(.5(2(2)(.044^2))+2(2)(.044^2))a/.044​
I get
[itex]\displaystyle F_p(2R)=\left((1/2)(2m)(2R)^2+(2m)(2R)^2\right)\frac{a}{2R}[/itex]​
You are using 2R instead of R to calculate the moment of inertia.

Otherwise, that all looks good.

But once corrected, I get a different answer than I got before, but it is within 10% of your earlier answer.

Added in Edit:

In fact the answer I got earlier was in error due to a arithmetic mistake.
 
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  • #45
tiny-tim
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May I ask your reasoning why you multiplied Fp by 2R?
because the distance between the force Fp and the torque point is 2R !
 
  • #46
Alright guys this still isn't right. The answers are all off, I plugged in the respective values and continue to get the wrong answer. Solve for Fp with Fp(2r)=(.5(2m)(2r)^2+2m(2r)^2)a/2r then plug into F-Fp=ma and still get the wrong answer.

Still get Fp=6a

F-Fp=ma

a=.551
 
  • #47
SammyS
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Alright guys this still isn't right. The answers are all off, I plugged in the respective values and continue to get the wrong answer. Solve for Fp with Fp(2r)=(.5(2m)(2r)^2+2m(2r)^2)a/2r then plug into F-Fp=ma and still get the wrong answer.

Still get Fp=6a

F-Fp=ma

a=.551
That's still incorrect.


Below is an excerpt from my previous post.
So interpreting the following
Fp(.044)=(.5(2(2)(.044^2))+2(2)(.044^2))a/.044​
I get
[itex]\displaystyle F_p(2R)=\left((1/2)(2m)(2R)^2+(2m)(2R)^2\right)\frac{a}{2R}[/itex]​
You are using 2R instead of R to calculate the moment of inertia.
The quantity, [itex]\displaystyle \ \ F_p(2R)\,,\ \ [/itex] the torque about the point of contact with the ground, is correct.

The quantity, [itex]\displaystyle \ \ \frac{a}{2R}\,,\ \ [/itex] the angular acceleration of the cylinders, is correct.

The quantity [itex]\displaystyle \ \ \left((1/2)(2m)(2R)^2+(2m)(2R)^2\right)\,,\ \ [/itex] the combined moment of inertia for the cylinders, is incorrect. However, the (2m) is correct. The radii of the cylinders doesn't change just because you're using the torque about the point of contact with the ground. Rather than using 2R, you should still be using R in this expression -- in both places that used 2R.
 
  • #48
Thank you Sammy and Tim. You guys are very helpful, thank you for pushing me and allowing me to think. Even though I didn't come up with the answer on my own, I stressed to think about this problem and was incorrect on a few concepts but you guys allowed me to learn more about moments of inertia, and the parallel axis theorem. Both of which I have very little experience in. I'll definitely fine-tune my skills in rotational dynamics and hopefully become better at solving these kinds of problems. Thank you.
 
  • #49
SammyS
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Thank you Sammy and Tim. You guys are very helpful, thank you for pushing me and allowing me to think. Even though I didn't come up with the answer on my own, I stressed to think about this problem and was incorrect on a few concepts but you guys allowed me to learn more about moments of inertia, and the parallel axis theorem. Both of which I have very little experience in. I'll definitely fine-tune my skills in rotational dynamics and hopefully become better at solving these kinds of problems. Thank you.
So, what was your final answer for, a, the acceleration of the plank ?
 
  • #50
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