HELP A rolling without slipping problem

AI Thread Summary
A solid cylinder is subjected to a horizontal force while rolling without slipping, prompting a discussion on its acceleration and frictional forces. The acceleration of the center of mass is derived as a_cm = 4F/3M, with friction acting in the direction of the applied force. Participants emphasize the importance of applying Newton's second law for both translational and rotational motion to solve the problem accurately. The frictional force is clarified to oppose slipping, leading to a better understanding of the system dynamics. The conversation highlights the need for careful analysis of forces and torques to derive correct relationships between the variables involved.
Kudo Shinichi
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HELP!A rolling without slipping problem

Homework Statement


A solid cylinder of radius R and mass M has a string wrapped around it and is placed on its side on a horizontal surface. The free end of the string is pulled horizontally with a force F as shown in the figure. As the string unwraps, the cylinder rolls along the surface without slipping.
a) show that the acceleration of the center of mass is given by a_cm=4F/3M
b) what is the magnitude and direction of the frictional force acting on the cylinder?
c) what is the acceleration of the free end of the string?

diagram:
http://tinypic.com/view.php?pic=2qsw4kj&s=4

The Attempt at a Solution


V_cc=0 and it is the velocity at reference frame of center to center
a) accelertaion of CM:
A_tot=sqrt((atan)2+(ac)2)
=sqrt((R*alpha)2)2+(R*(Vcc)2)
because vcc=0 therefore A_tot=R*alpha
4F/3M=4(M*alpha)/3M=4/3*alpha and R=4/3 in this case

b) the frictional force opposite to the applied force, and it is equal to the applied force, which is equal to M*alpha
c) the acceleration of the free end of the string would be same as the acceleration of the cylinder, which is R*alpha

I am not sure whether I did the problem correctly or not, can anyone help me with it? thank you very much.
 
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I can't follow your solution. Why not start with the basics? Draw a free body diagram and apply Newton's 2nd law to both translation and rotation.
 


Doc Al said:
I can't follow your solution. Why not start with the basics? Draw a free body diagram and apply Newton's 2nd law to both translation and rotation.

i tried but how can I related the force to CM?
the angular velocity for CM is omega and alpha is the angular acceleration around the cylinder
What I know now is mass of the cylinder. Even though I also know the acceleration but it is not for center of mass.
 


Kudo Shinichi said:
i tried but how can I related the force to CM?
Using Newton's 2nd law for translational motion: Fnet = macm, where acm is the acceleration of the center of mass.
 


Doc Al said:
Using Newton's 2nd law for translational motion: Fnet = macm, where acm is the acceleration of the center of mass.

Do you mean that acm=F/m, which I could use to should for part a. Sorry but where did 4/3 come from? is it part of the acceleration?
 


Kudo Shinichi said:
Do you mean that acm=F/m, which I could use to should for part c.
Yes, but the force you need here is the net force, not just the applied force "F". That's why I wrote Fnet.
What individual forces act on the cylinder?
Sorry but where did 4/3 come from? is it part of the acceleration?
The 4/3 will appear when you solve for the acceleration. As I said earlier, you'll need to apply Newton's 2nd law twice: for translational motion and for rotational motion.
 


Yes, but the force you need here is the net force, not just the applied force "F". That's why I wrote Fnet.
What individual forces act on the cylinder?

since it is a rotational motion there is \alpha for this equation,
The 4/3 will appear when you solve for the acceleration. As I said earlier, you'll need to apply Newton's 2nd law twice: for translational motion and for rotational motion.[/QUOTE]


net force=F-Ffr=F-\mu*FN=macm
the above is the translational motion, sorry but how can i get rid of \mu

net force=F-Ffr=F-\mu*FN=m*\alpha
there is a new variable appears in this equation,\alpha, which we can write it as r/a

i think what i can do next after i know how to get rid of \mu is set the two equation equal to each other and solve then we will be able to get 3F/4M
is it right?
 


Kudo Shinichi said:
net force=F-Ffr=F-\mu*FN=macm
the above is the translational motion, sorry but how can i get rid of \mu
You don't need to use μ, just leave the friction force as Ffr. In what direction does the friction force act?

The second equation you need is Newton's 2nd law for rotation: \tau = I \alpha
 


Doc Al said:
You don't need to use μ, just leave the friction force as Ffr. In what direction does the friction force act?
frictional force is equal and opposite to the applied force
The second equation you need is Newton's 2nd law for rotation: \tau = I \alpha
\tau = I \alpha
I can write I as 1/2mR2 and \alpha as R/acm
therefore, \tau=(1/2mR2)(R/acm)

or there is another way to describe \tau = I \alpha, which is \tau=radius*force, which is equal to radius*macm
 
  • #10


Kudo Shinichi said:
\tau = I \alpha
I can write I as 1/2mR2 and \alpha as R/acm
therefore, \tau=(1/2mR2)(R/acm)
Careful. \alpha = a/R \ne R/a

or there is another way to describe \tau = I \alpha, which is \tau=radius*force, which is equal to radius*macm
I suggest that you find the net torque due to the forces acting on the cylinder.
 
  • #11


Doc Al said:
Careful. \alpha = a/R \ne R/aI suggest that you find the net torque due to the forces acting on the cylinder.

do you mean use the latter equation to solve(torque=radius*F)?
and by the way, I think that I have inserted my answers for the previous question into the qutation just now...
about the frictional force, it is equal and opposite to the applied force.

If I use the first equation I wrote for torque then torque=(1/2mR2) (a/R)=1/2ma*R=1/2F*R
How do we connect this equation with F-Ffr=macm?
 
  • #12


Kudo Shinichi said:
do you mean use the latter equation to solve(torque=radius*F)?
What torque is exerted by each force acting on the cylinder? What's the net torque?
and by the way, I think that I have inserted my answers for the previous question into the qutation just now...
about the frictional force, it is equal and opposite to the applied force.
No. If friction were equal and opposite to the applied force, then the cylinder would not accelerate.

If I use the first equation I wrote for torque then torque=(1/2mR2) (a/R)=1/2ma*R=1/2F*R
This is OK up until the last step. Note that F is just one of the forces acting on the cylinder.
How do we connect this equation with F-Ffr=macm?
You still need to find the net torque.
 
  • #13


Doc Al said:
What torque is exerted by each force acting on the cylinder? What's the net torque?

No. If friction were equal and opposite to the applied force, then the cylinder would not accelerate.


This is OK up until the last step. Note that F is just one of the forces acting on the cylinder.

You still need to find the net torque.

If there is just one force applied on the system then isn't there just one torque? which means that the net torque is equal the only torque, which is r*F? Sorry, I am a bit confused...because in order to have net torque you need to have more than one force

Or do should I also include the frictional force?
then the net torque would be r*Fsin\theta+r*Ffrsin\theta
Sorry again...how do we find the angle for sin?
 
  • #14


Kudo Shinichi said:
Or do should I also include the frictional force?
Of course you should include the friction force. (For both translation and rotation.)
then the net torque would be r*Fsin\theta+r*Ffrsin\theta
Sorry again...how do we find the angle for sin?
First things first: Which way do the forces act?* Draw the forces on your diagram. Since both forces are horizontal and tangent to the cylinder, the only angles needed are 90 degrees.

*You still haven't figured out which way the friction acts. Note that friction always opposes slipping between surfaces.
 
  • #15


Doc Al said:
Of course you should include the friction force. (For both translation and rotation.)

First things first: Which way do the forces act?* Draw the forces on your diagram. Since both forces are horizontal and tangent to the cylinder, the only angles needed are 90 degrees.

*You still haven't figured out which way the friction acts. Note that friction always opposes slipping between surfaces.

the free body diagram I drew
http://tinypic.com/view.php?pic=i4h2ed&s=4

As you said the friction is indeed opposes to the slipping direction

Or am I wrong on drawing free body diagram?
 
  • #16


Kudo Shinichi said:
As you said the friction is indeed opposes to the slipping direction

Or am I wrong on drawing free body diagram?
You have the direction of the friction wrong. Friction opposes slipping between the bottom of the cylinder and the floor. If there were no friction, the cylinder would spin clockwise thus the bottom of the cylinder would tend to slip to the left against the floor. Friction on the cylinder opposes that slipping, thus it acts to the right.
 
  • #17


Doc Al said:
You have the direction of the friction wrong. Friction opposes slipping between the bottom of the cylinder and the floor. If there were no friction, the cylinder would spin clockwise thus the bottom of the cylinder would tend to slip to the left against the floor. Friction on the cylinder opposes that slipping, thus it acts to the right.

OH!that's why
since the direction of Ffr and applied force are the same
then the translational motion: F+Ffr=macm
the rotational motion:r(F+Ffr)=1/2maR2
F+Ffr=1/2maR
then how do we relate these two motions together?
sub translational motion into the rotational motion: r(macm)=1/2maR2
or set these two equations equal to each other
1/2maR=macm
but then we still can't relate it to 4F/3M
 
  • #18


Kudo Shinichi said:
OH!that's way
since the direction of Ffr and applied force are the same
then the translational motion: F+Ffr=macm
Good.
the rotational motion:r(F+Ffr)=1/2maR2
F+Ffr=1/2maR
Redo this one. Does the torque due to each force act in the same direction? (And be careful with the Rs.)

Once you get the two equations correct, you'll have two equations and two unknowns (Ffr and acm). You can solve them together.
 
  • #19


Doc Al said:
Good.

Redo this one. Does the torque due to each force act in the same direction? (And be careful with the Rs.)

Once you get the two equations correct, you'll have two equations and two unknowns (Ffr and acm). You can solve them together.

Sorry I don't really get what you mean by torque acting on the different direction. If Ffr has the same direction as F and the angles for sine are equal then why do they have different direction
Just a bit ahead of myself, if there are two unknowns then do we use the sub one equation into another or do we set both equations equal to each other?
 
  • #20


Kudo Shinichi said:
Sorry I don't really get what you mean by torque acting on the different direction. If Ffr has the same direction as F and the angles for sine are equal then why do they have different direction
One acts at the top of the cylinder, creating a clockwise torque; the other acts at the bottom, creating a counter-clockwise torque.
Just a bit ahead of myself, if there are two unknowns then do we use the sub one equation into another or do we set both equations equal to each other?
Combine them to eliminate one of the variables. Substituting one into the other would work.
 
  • #21


Doc Al said:
One acts at the top of the cylinder, creating a clockwise torque; the other acts at the bottom, creating a counter-clockwise torque.

Combine them to eliminate one of the variables. Substituting one into the other would work.

R(F-Ffr)=1/2maR2
F-Ffr=1/2maR
this is for the rotational motion

F+Ffr=macm
translational motion

Sorry, I can't find any way to combine these two equations into one...can you help me with it?
 
  • #22


Kudo Shinichi said:
R(F-Ffr)=1/2maR2
F-Ffr=1/2maR
this is for the rotational motion
The right hand side has an extra R. Remember that \alpha = a/R.

F+Ffr=macm
translational motion
Good.

Be careful with notation. Use the same symbol for acceleration in both equations: a = acm.

Sorry, I can't find any way to combine these two equations into one...can you help me with it?
Just add the two equations and see what happens. (After you correct the rotational motion equation.)
 

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