Struggling with Rotational Dynamics Homework?

[benoconnell22]

Homework Statement

I've been trying to attempt this problem for the past day but I don't understand how to answer the questions. It's frustrating. If someone could point me in the right direction that would be nice. My TA did the problem in class but I still don't quite understand it. This is my first time taking a physics course.

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.

(a) Find the initial acceleration of the plank at the moment the rollers are equidistant from the ends of the plank.
magnitude: m/s2
direction:

(b) Find the acceleration of the rollers at this moment.
magnitude: m/s2
direction:

(c) What friction forces are acting at this moment? (Let fp be the frictional force exerted by each roller on the plank, and let fg be the rolling friction exerted by the ground on each roller.)
fp = N
direction:

fg = N
direction:

Homework Equations

Ʃτ=Fd

v=ωr
a=\alphar

The Attempt at a Solution

Not sure where to start. Any help would be well appreciated.

 

[tiny-tim]

welcome to pf!

hi ben! welcome to pf!

(try using the X² button just above the Reply box )

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.

(a) Find the initial acceleration of the plank at the moment the rollers are equidistant from the ends of the plank.

try (a) first …

start by giving things names: call the acceleration of the plank "a", the angular acceleration of the cylinders "α", and the friction force between the plank and each cylinder "C"

write out the F = ma equation for the plank, and the τ = Iα equation for each cylinder, and also the "rolling constraint" equation that relates a to α …

what do you get? ​

 

[benoconnell22]

Ok so I got, F=ma, thus 6.4N=5.6a, a=6.4/5.6≈1.14

and a=αr

1.14=α(.044)

α=25.97

The for torque I got the following equation T=Iα where I=1/2mr^2

plugging in the respective values the equation looks like T=1/2(mr^2)(α)=1/2(2(.044^2))(25.97)=.0502Nm

 

[tiny-tim]

Ok so I got, F=ma, thus 6.4N=5.6a, a=6.4/5.6≈1.14

nooo …

F - C = ma​

(ie you have to use the net force, F - C;

you'll also have to use C to find the angular acceleration)

 

[benoconnell22]

Alright so, I got: 6.4N-C=5.6a

a=6.4-C/5.6

and

α=(6.4-C/5.6)/.044

And Torque: T=Iα= 1/2(2(.044^2))((6.4-C/5.6)/.044)=.00194((6.4-C/5.6)/.044)

My problem is, I don't know where to go from here. I'm trying to find the angular acceleration. So would C=Normal force x mu? Then I can plug that into α? But mu isn't given? Frustrating!

 

[tiny-tim]

And Torque: T=Iα= 1/2(2(.044^2))((6.4-C/5.6)/.044)=.00194((6.4-C/5.6)/.044)

but what is T?

and about what point are you measuring T, and the moment of inertia?

 

[benoconnell22]

Torque is Radius x Force so would it be Radius x 6.4-C? Then set that equal to my Torque function and solve for C? Then I can plug that into my a) equation to find the acceleration?

 

[benoconnell22]

Also, thank you for helping me.

 

[tiny-tim]

Then set that equal to my Torque function and solve for C? Then I can plug that into my a) equation to find the acceleration?

yes , but …

Torque is Radius x Force so would it be Radius x 6.4-C?

i] the 6.4N is not a force on the cylinders, so it does not contribute to the torque
ii] the friction with the ground does contribute to the torque
iii] you don't want to have to find the friction, so about which point should you measure the torque?

(and iv] don't forget there are two cylinders)

 

[benoconnell22]

But the 6.4N is the cause for the rotation of the wheels which is parallel to the cylinders. If there were no 6.4 N they wouldn't rotate. The only other force I can think of that is applied to the cylinder is the downward force of mg on from the plank. Is friction the only force that contributes to the torque? Excuse my ignorance, I'm trying every opportunity I see.

 

[SammyS]

Draw a free body diagram for the plank and for a cylinder. (The free body diagram for the second cylinder is identical to that for the first cylinder.)

 

[benoconnell22]

We have, mg from the plank pointing down, Normal force for the plank pointing up, Force of friction pointing to the left, and 6.4N pointing to the right.

 

[SammyS]

We have, mg from the plank pointing down, Normal force for the plank pointing up, Force of friction pointing to the left, and 6.4N pointing to the right.

That force of friction on the plank if 2f_p, or as called it, 2C.

Now, for a free body diagram for either cylinder...

 

[benoconnell22]

For either cylinder, we have a normal force pointing up, the mg of the plank down onto the cylinder plus the mg of the cylinder itself. The side ways force of the plank's force pointing to the right.

 

[SammyS]

For either cylinder, we have a normal force pointing up, the mg of the plank down onto the cylinder plus the mg of the cylinder itself. The side ways force of the plank's force pointing to the right.

How about f_g, "rolling friction exerted by the ground on each roller".

If there were no friction between the ground & the cylinder, the cylinders would rotate at a different rate than they do with that friction.

 

[benoconnell22]

F_p is to the right on the top of the cylinder and F_g is pointing to the right on the ground. I got that but how do I go about finding the initial acceleration when the rollers are equidistant to the ends of the plank. I'm lost in the abyss .

 

[SammyS]

F_p is to the right on the top of the cylinder and F_g is pointing to the right on the ground. I got that but how do I go about finding the initial acceleration when the rollers are equidistant to the ends of the plank. I'm lost in the abyss .

f_p acting on the cylinder is to the left if the cylinder is being pulled to the right.

The cylinder will exert a force of f_p on the ground which is to the right.

 

[tiny-tim]

hi ben!

(just got up :zzz:)

For either cylinder, we have a normal force pointing up, the mg of the plank down onto the cylinder plus the mg of the cylinder itself. The side ways force of the plank's force pointing to the right.

yes that's right

do you now see how important it is to draw a separate free body diagram for each differently-moving part?

if you don't, you'll get confused as to which force is acting on which body … only directly acting forces count!

F_p is to the right on the top of the cylinder and F_g is pointing to the right on the ground. I got that but how do I go about finding the initial acceleration when the rollers are equidistant to the ends of the plank.

the only significance of the "equidistant" thing is that it means that the normal forces are the same, and so the friction forces are also the same

(it just makes the arithmetic easier! )

ok, you now have a free body diagram which tells you there are only two forces that can give a torque to each cylinder: the plank friction force (F_p), and the ground friction force

you have a completely free choice as to which point you take torques about …

if you only want F_p in the equation, which point should you choose? ​

 

[benoconnell22]

Would I choose the top of the cylinder because it's the area in perpendicular contact with the friction force to the plank?

 

[SammyS]

Would I choose the top of the cylinder because it's the area in perpendicular contact with the friction force to the plank?

The plank exerts zero torque about the top of the cylinder.

The ground exerts zero torque about the bottom of the cylinder.

Now, back to tim's question... Which point should you choose for calculating torque, if you don't want to mess with f_g, the frictional force exerted by the ground?

 

[benoconnell22]

I would choose the point that's in contact with the force of friction by the plank i.e. the top of the cylinder. This does not make any sense to me. So you wouldn't choose the force of friction to the plank? and you don't want to deal with the force of friction due to the ground. Then what other force is there?

 

[SammyS]

I would choose the point that's in contact with the force of friction by the plank i.e. the top of the cylinder. This does not make any sense to me. So you wouldn't choose the force of friction to the plank? and you don't want to deal with the force of friction due to the ground. Then what other force is there?

You can calculate torque about any point you choose.

You're definitely going to need F_p to find the net force on the plank so that you can find ma of the plank.

So, by choosing to calculate torque about the point of contact with the ground, you may be able to solve most of the problem without regard to F_g.

 

[benoconnell22]

Does F=6.4N-Fp=ma look right? Then T=Fp(R) because Fp is the force being applied to the cylinder. and T=Iα (where α=a/R) Then set these two equal to each other. I would choose my point at the center of mass of the cylinder. Where R is the distance from the center of mass of the cylinder and Fp is the force applied to cause the torque.

 

[tiny-tim]

Does F=6.4N-Fp=ma look right?

yes

I would choose my point at the center of mass of the cylinder.

ok, but then you must include the torque from both F_p and F_g

 

[benoconnell22]

Is there another point where I don't have to incorporate both frictional forces? i.e just have Fp?

 

[benoconnell22]

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)?

 

[tiny-tim]

a=2(Fp+Fg)/M.

no, that should be F_p - F_g, shouldn't it?

 

[benoconnell22]

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.

 

[benoconnell22]

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

 

[tiny-tim]

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

(i'll guess you're out by a factor of two somewhere because of the two cylinders)

 

[benoconnell22]

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 :(

 

[benoconnell22]

two*

 

[benoconnell22]

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.

 

[SammyS]

What did you get for the acceleration of the plank ?

 

[benoconnell22]

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?

 

[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