Struggling with Rotational Dynamics Homework?

[SammyS]

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.

 

[benoconnell22]

How else can you do this without having Fg in the equation?

 

[tiny-tim]

hi ben!

(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 )

no, d (in the parallel axis theorem) is always the distance from the centre of mass (r, not 2r)

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

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 F_p and F_g

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 F_p (not F_g)

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

finally, you can only do angular momentum = L_P = I_Pω (and therefore τ_P = I_Pα) 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:

τ_net,c.o.m. = dL_c.o.m./dt = I_c.o.m.α
τ_{net, c.o.r.} = dL_c.o.r./dt = I_c.o.r.α​

Sometimes a more general point P is needed, and then:

L_P = I_c.o.m.ω + m PC x v_c.o.m.​

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

 

[benoconnell22]

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.

 

[tiny-tim]

hi ben!

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

(i suspect you used the wrong r either in τ = F_pr or in a = αr)

 

[benoconnell22]

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

 

[tiny-tim]

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

Fp(.044)=…

no, the torque about the bottom point is F_p times .088

 

[benoconnell22]

May I ask your reasoning why you multiplied Fp by 2R? Is that because there are two cylinders?

 

[SammyS]

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

$\displaystyle F_p(2R)=\left((1/2)(2m)(2R)^2+(2m)(2R)^2\right)\frac{a}{2R}$​

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.

 

[tiny-tim]

May I ask your reasoning why you multiplied Fp by 2R?

because the distance between the force F_p and the torque point is 2R !

 

[benoconnell22]

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

 

[SammyS]

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

$\displaystyle F_p(2R)=\left((1/2)(2m)(2R)^2+(2m)(2R)^2\right)\frac{a}{2R}$​

You are using 2R instead of R to calculate the moment of inertia.

The quantity, $\displaystyle \ \ F_p(2R)\,,\ \$ the torque about the point of contact with the ground, is correct.

The quantity, $\displaystyle \ \ \frac{a}{2R}\,,\ \$ the angular acceleration of the cylinders, is correct.

The quantity $\displaystyle \ \ \left((1/2)(2m)(2R)^2+(2m)(2R)^2\right)\,,\ \$ 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.

 

[benoconnell22]

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.

 

[SammyS]

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 ?

 

[ehild]

The problem can be solved also considering the rotation of the cylinder about their centre of mass, and calculating the torques of the frictional forces with respect to the CM. See the other thread https://www.physicsforums.com/showthread.php?t=651378

ehild

 

[benoconnell22]

I ended up getting a=.901m/s^2

 

[SammyS]

I ended up getting a=.901m/s^2

[STRIKE]I didn't get that[/STRIKE].

I'll re-check what I got, but I have already gone over my answer a few times.

Added in Edit:

Your result here is good!

 

[benoconnell22]

I'm sorry I meant .934 m/s^2 I was practicing this problem with different quantities and accidentally wrote the answer one of the others.

 

[SammyS]

I'm sorry I meant .934 m/s^2 I was practicing this problem with different quantities and accidentally wrote the answer one of the others.

I didn't get that either.

When you got a = 0.552 m/s² your moment of inertia was 4 times what it should have been, so when you had F_p = 6a, that was also 4 times the size it should have been.

 

[ehild]

See attached picture. The acceleration of the plank is ap. F-2F_p=Ma_p (1)

The motion of the cylinder is translation of the centre of mass and rotation about the CM.
The CM accelerates with a_cm. The friction between cylinder and plank accelerates the CM of the cylinders forward, and accelerates forward rotation; the rotational resistance Fg between ground and cylinder drives translation forwards, but hinders rotation. .
So F_p+F_g=ma_cm. (2)

r(F_p-F_g)=Iβ (β is the angular acceleration)

The cylinders do not slip: βr=0.5 m.
The moment of inertia about the CM is I=0.5 mr².
The torque equation becomes:

F_p-F_g=0.5 ma_cm.(3)

Adding equations (2) and(3):

2F_p=1.5ma_cm (4)


The plank does not slip on the cylinders, so its velocity is the same as the topmost point of the cylinders, which is twice the velocity of the CM:
a_cm=0.5 a_p.

Now you have the equations (1) and (4)

F-2F_p=Ma_p
2Fp=1.5ma_p/2

Add them:

F=Ma_p+0.75ma_p.

ehild

 

[SammyS]

I ended up getting a=.901m/s^2

OK ...

I went over my calculations yet again.

The answer is a = 6.4/7.1 = 0.9014... .

However you came up with this for a previous post (post #51) of yours, this answer looks good.

I apologize for my repeatedly poor arithmetic.