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Rotational Motion and linear accerlation

by scoldham
Tags: rotation, rotational kinematic
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scoldham
#1
Dec18-09, 01:14 PM
P: 54
1. The problem statement, all variables and given/known data

A solid cylinder of weight 50 lb and radius 3.0 inches has a light thin tape wound around it. The tap passes over a light smooth fixed pulley to a 10 lb body hanging vertically in the air. If the plane on which the cylinder moves is inclined 30 degrees to the horizontal, find (a) the linear acceleration of the cylinder down the inclined plane, and (b) the tension of the tabpe, assuming no slippage.

2. Relevant equations

[tex]\sum _\tau = I_c\alpha[/tex]

[tex]I_c = \frac{M_c R_0^2}{2}[/tex]

3. The attempt at a solution

I'm having an issue seeing the relationship linear acceleration and Newton's 2nd law for rotational motion.

Am I correct in thinking about substituting [tex]\alpha = \omega^2 R[/tex] into the 2nd law, then applying appropriate forces?

Edit: See next post
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scoldham
#2
Dec18-09, 04:10 PM
P: 54
Ok... scratch what I said earlier.

[tex]\alpha = \frac{\tau}{I_c} [/tex]

[tex]\alpha = \frac{a_t}{R} [/tex]

So..

[tex]\frac{\tau}{I_c} = \frac{a_t}{R}[/tex]

which means [tex]\tau R = a_t I_c[/tex] (*)

A force diagram tells me

[tex]\tau = m_Ag sin\vartheta - m_Bg[/tex]

The cylinder rolls down the incline [tex]\vartheta[/tex] under the influence of gravity and the weight generated by the blocks mass detracts from that value via the tape.

So, plugging [tex]I_c[/tex] for a solid cylinder and the derived value for [tex]\tau[/tex] into * we get:

[tex]m_Ag sin\vartheta - m_Bg = \frac{m_AR^2a_t}{2} [/tex]

Meaning that [tex]a_t[/tex] (or the linear accleration) would be:

[tex]a_t = 2 (\frac{gsin\vartheta}{R} - \frac{m_Bg}{m_aR^2}) [/tex]

Granted.. I didn't take the units into consideration.. but I am hoping for a more conceptual and general understanding of the problem, units aside.

Am I closer to being on the right track?
Doc Al
#3
Dec18-09, 04:53 PM
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Quote Quote by scoldham View Post
[tex]\alpha = \frac{\tau}{I_c} [/tex]

[tex]\alpha = \frac{a_t}{R} [/tex]
Careful. What you are calling a_t cannot simply be the acceleration of the cylinder down the incline, since the tape is also accelerating up the incline. (Figure out the connection between α and the two linear accelerations.)


A force diagram tells me

[tex]\tau = m_Ag sin\vartheta - m_Bg[/tex]

The cylinder rolls down the incline [tex]\vartheta[/tex] under the influence of gravity and the weight generated by the blocks mass detracts from that value via the tape.
Careful. Since the cylinder and the block have different accelerations, don't try to apply Newton's 2nd law in one step. Instead, write a separate force equation for each, then combine them later.

scoldham
#4
Dec18-09, 05:21 PM
P: 54
Rotational Motion and linear accerlation

Quote Quote by Doc Al View Post
Careful. What you are calling a_t cannot simply be the acceleration of the cylinder down the incline, since the tape is also accelerating up the incline. (Figure out the connection between α and the two linear accelerations.)
The force of the tension in the tape should be subtracted from the force of gravity compelling the cylinder to roll down the incline. So [tex]\tau = m_Ag sin\vartheta - T[/tex] where T is tension. Correct?

I'm calculating [tex]T = m_Bg[/tex] as the m_B is free hanging and pulling on the tape.

Quote Quote by Doc Al View Post
Careful. Since the cylinder and the block have different accelerations, don't try to apply Newton's 2nd law in one step. Instead, write a separate force equation for each, then combine them later.
Wouldn't the linear acceleration of the cylinder and the block (or body as I described in the problem statement) be the same as they are a system (linked by the tape)?
scoldham
#5
Dec18-09, 05:27 PM
P: 54
I've attached a picture of the set up.... just for reference.

Thanks for the assistance.
Attached Thumbnails
rotation_problem.png  
Doc Al
#6
Dec18-09, 05:30 PM
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Quote Quote by scoldham View Post
The force of the tension in the tape should be subtracted from the force of gravity compelling the cylinder to roll down the incline. So [tex]\tau = m_Ag sin\vartheta - T[/tex] where T is tension. Correct?
It's true that the net force on the cylinder is mAgsinθ - T. (I don't know why you are labeling that net force 'tau', since 'tau' usually stands for torque.) Now apply Newton's 2nd law to the cylinder.

I'm calculating [tex]T = m_Bg[/tex] as the m_B is free hanging and pulling on the tape.
No. If the tension equaled the weight of the hanging mass, the mass would not accelerate.

Wouldn't the linear acceleration of the cylinder and the block (or body as I described in the problem statement) be the same as they are a system (linked by the tape)?
No. The cylinder unwinds the tape. (Their accelerations are connected, just not equal.)
Doc Al
#7
Dec18-09, 05:36 PM
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Quote Quote by scoldham View Post
I've attached a picture of the set up.... just for reference.
Interesting. I imagined the orientation of the cylinder and tape reversed. (With the tape being flat against the incline.)
scoldham
#8
Dec18-09, 05:39 PM
P: 54
Well... I have had trouble with Torque vs Force.... So I'm guessing [tex]\alpha = \frac{\tau}{I_c}[/tex] isn't valid? If it is.. how to we calculate tau?

Concerning Tension...

Would [tex]T_{Net} = m_Agsin\vartheta - m_Bg[/tex] ?
scoldham
#9
Dec18-09, 05:41 PM
P: 54
Quote Quote by Doc Al View Post
Careful. What you are calling a_t cannot simply be the acceleration of the cylinder down the incline, since the tape is also accelerating up the incline. (Figure out the connection between α and the two linear accelerations.)



Careful. Since the cylinder and the block have different accelerations, don't try to apply Newton's 2nd law in one step. Instead, write a separate force equation for each, then combine them later.
Quote Quote by Doc Al View Post
Interesting. I imagined the orientation of the cylinder and tape reversed. (With the tape being flat against the incline.)
Does that change your suggestions?
Doc Al
#10
Dec18-09, 06:04 PM
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Quote Quote by scoldham View Post
Well... I have had trouble with Torque vs Force.... So I'm guessing [tex]\alpha = \frac{\tau}{I_c}[/tex] isn't valid? If it is.. how to we calculate tau?
What's the definition of torque?

Concerning Tension...

Would [tex]T_{Net} = m_Agsin\vartheta - m_Bg[/tex] ?
What do you mean by 'net' tension? By setting up equations, you'll solve for the tension.

Quote Quote by scoldham View Post
Does that change your suggestions?
Is that a diagram you created or was it provided with the problem? Where the cylinder touches the incline, is there friction?
scoldham
#11
Dec18-09, 06:13 PM
P: 54
Quote Quote by Doc Al View Post
What's the definition of torque?
Well mathematically [tex]\tau = r \times F[/tex] ... with r being radius and F being force

I suppose that means that torque is force applied radially. So wouldn't that mean that the torque of the tape around the cylinder affect the tension in the tape? Also, does the Force in the equation for torque equal the force of the weight of the block pulling through the tape?

Idea: Edit:see better idea

[tex]\tau = Rm_Agsin\vartheta[/tex]

Therefore,

[tex]R^2m_Agsin\vartheta = \frac{m_AR^2a_t}{2}[/tex]

or

[tex]gsin\vartheta = \frac{a_t}{2}[/tex]

[tex]a_t = \frac{gsin\vartheta}{2}[/tex] right? closer at least?

Better idea:

[tex]\tau = r \times F[/tex]

[tex]\tau = r m_Agsin\vartheta -m_Bg[/tex]

actually.. this is right back where I started...........
Quote Quote by Doc Al View Post
What do you mean by 'net' tension? By setting up equations, you'll solve for the tension.
I'm denoting net tension as the tension in the tape as deduced by the force from the cylinder and the force from the block.

The pulley has negligible mass and is fixed... so wouldn't the tension in either piece of the tape be the same?

Quote Quote by Doc Al View Post
Is that a diagram you created or was it provided with the problem? Where the cylinder touches the incline, is there friction?
The diagram is an mspaint translation of the one provided with the problem. Friction is neglected.
Doc Al
#12
Dec18-09, 06:53 PM
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Quote Quote by scoldham View Post
Well mathematically [tex]\tau = r \times F[/tex] ... with r being radius and F being force
OK, so what forces act on the cylinder? What torque do those forces exert about its center of mass.

I suppose that means that torque is force applied radially.
Only forces with a tangential component exert a torque.
So wouldn't that mean that the torque of the tape around the cylinder affect the tension in the tape?
Not sure what that means.
Also, does the Force in the equation for torque equal the force of the weight of the block pulling through the tape?
No. The force that the tape pulls does not equal the weight of the block.

I'm denoting net tension as the tension in the tape as deduced by the force from the cylinder and the force from the block.
Rather than try to figure out the tension in your head, set up equations for each mass then solve for the tension.

The pulley has negligible mass and is fixed... so wouldn't the tension in either piece of the tape be the same?
Sure.
scoldham
#13
Dec18-09, 06:57 PM
P: 54
So wouldn't that mean that the torque of the tape around the cylinder affect the tension in the tape?
Quote Quote by Doc Al View Post
Not sure what that means.
Would the torque generated by the cylinder apply a force to the tape?
scoldham
#14
Dec18-09, 07:07 PM
P: 54
Quote Quote by Doc Al View Post
Rather than try to figure out the tension in your head, set up equations for each mass then solve for the tension.
I'm not sure I follow what you mean by set up equations for each mass . My best guess leads me back to the same world I've been trapped in for nearly 3 hours....

[tex]\tau = R \times F = Rm_Agsin\vartheta + T[/tex]

[tex]F_B = ma = m_Bg -T[/tex]

.:. [tex]Rm_Agsin\vartheta = -m_Bg[/tex]

There's no unknown in this equation. What is it that I am missing? Should I be solving for a system?

Edit:

IDEA!

[tex]\tau = R \times F = RTsin\vartheta[/tex]

However.... I'm still seeing [tex]F_B = ma = m_Bg -T[/tex] which says [tex]T = -m_Bg[/tex] which you've told me isn't the case.............
Doc Al
#15
Dec18-09, 07:48 PM
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Quote Quote by scoldham View Post
I'm not sure I follow what you mean by set up equations for each mass .
Identify the forces on each mass (the block and the cylinder). Apply Newton's 2nd law to each. You'll get two equations for translational forces/accelerations and a third for the rotational torque/acceleration for the cylinder. Combine those three equations (plus a constraint equation relating the rotational and translational accelerations) and you can solve for the two accelerations and the tension.

Do it step by step.
scoldham
#16
Dec18-09, 08:44 PM
P: 54
Ok... so

Forces for cylinder:

Linear

[tex]m_Agsin\vartheta - T = m_Aa[/tex] =>
[tex]T = m_Agsin\vartheta -m_Aa[/tex]

Radial

[tex]\tau = R \times F = R m_Aasin\vartheta = I_c\alpha[/tex] =>
[tex]\alpha = \frac{R m_Aasing\vartheta}{I_c}[/tex]

Forces for block:

[tex]T- m_Bg = m_Ba[/tex] =>
[tex]T = m_Ba + m_Bg[/tex]

.:.

Solving system for T => [tex]m_Agsin\vartheta -m_Aa = m_Ba + m_Bg[/tex]

So [tex]a = \frac{m_Agsin\vartheta - m_Bg}{m_A + m_B}[/tex]

BUT... linear acceleration from angular:

[tex]a = R\alpha[/tex]

I don't believe that I'm using the right value for Force in the torque equation.... perhaps other problems....

Am I even using the right base equations.... it's starting to seem like maybe that's a no.

I do, however, appriciate the ever so gentle guidance (in all sincerity).... I'll have done every possible algebraic operation to these equations before it's over....

I look forward to seeing what angle I should work next.
Doc Al
#17
Dec19-09, 07:36 AM
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Quote Quote by scoldham View Post
Ok... so

Forces for cylinder:

Linear

[tex]m_Agsin\vartheta - T = m_Aa[/tex] =>
[tex]T = m_Agsin\vartheta -m_Aa[/tex]
OK, but label that acceleration a_A, since it does not equal the acceleration of the block. You've chosen the acceleration such that it will have a positive value if the cylinder goes down the incline; nothing wrong with that.

Radial

[tex]\tau = R \times F = R m_Aasin\vartheta = I_c\alpha[/tex] =>
[tex]\alpha = \frac{R m_Aasing\vartheta}{I_c}[/tex]
No. Of the forces acting on the cylinder, which one exerts a torque about its center?

Forces for block:

[tex]T- m_Bg = m_Ba[/tex] =>
[tex]T = m_Ba + m_Bg[/tex]
Again, give that acceleration a label such as a_B. Here you chose a positive value of 'a' as up. (I always let 'a' stand for the positive magnitude of the acceleration and I always write my equations assuming the most likely direction for the object's acceleration. Here I would assume that the block accelerates down.)

The big problems are:
(1) Not recognizing that the two linear accelerations are different.
(2) Using the wrong force in your torque equation.


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