A few problems I was stuck on


by a_lawson_2k
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a_lawson_2k
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#1
Nov18-06, 02:50 PM
P: 38
I'm brushing up on the physics, and am teaching myself from University Physics w/ Modern Physics 10th Edition (I got 11th off eBay with the solution manual, it should come in about a week). I haven't had too much trouble with the problems up until the section about the dynamics of rotational motion, either way, I got through the exercises (easy problems), but when I get to the regular problems, things start getting tricky. What may be happening here is that I didn't quite pick up on how to set up the problem over the two days I've worked on the chapter. I looked back over it, and it didn't appear I had missed anything, hopefully seeing what I'm doing wrong with these problems may shed light on the situation.

First problem:
Quote Quote by 10-55
A block with mass m=5 kg slides down a surface inclined at 36,9 degrees to the horizontal. The coefficient of kinetic friction is 0.25. A string attached to the block is wrapped around a flywheel on a fixed axis at O. The flywheel has mass 25 kg and moment of inertia of 0.5 kg-m^2 w/ respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0,2 m from the axis.
a) what is the acceleration of of the block down the plane?
b) what is the tension in the string?
I drew the free body diagram, and applied Newton's Second to the (adjusted) x and y axes (offset 36,9 degrees).
[tex]\Sigma F_y = n-mg cos(\theta) = 0[/tex]
[tex]\Sigma F_x = mg sin(\theta)-F_f-T = m*a_y[/tex]
The natural force became mg cos (36.9*) which worked out to be 39,2N
mg sin (36.9*) came out to be 29.4N, so the sum of the forces in the Y direction came out to:
[tex]\Sigma F_x = 19.6-T = m*a_y[/tex]

Here is what really started puzzling me, how do I get the flywheel to come into the equation? It doesn't specify a radius from which I can evaluate the torque, and from there the angular acceleration, so I'm stuck.

Second problem:
Quote Quote by 10-61
A uniform marble, radius r, starts from rest with its center of mass at heigh h above the lowest point of a loop-the-loop track of radius R, a track shaped like the one in figure 7-26 (MS Paint representation provided). It rolls without slipping. Rolling friction and air resistance are negligible.
a) what is the minimum value of h such that the marble will not leave the track at the top of the loop?
--NOTE: radius r is NOT negligible compared to radius R
b) what is the answer to part (a) if the track if lubricated making friction negligible?
I don't know where to start here, sadly. I have no clue how to even approach it, the book didn't provide an example even remotely like this one. I could probably set it up in energy terms and using the sum of the forces in the radial direction as being (mv^2)/r, but I'd have no idea what to do from there. Forgive me if it looks like I'm not trying, it's just I'm not sure how to reconcile the stuff in the problem.

This one I have more of an idea of how to solve:
Quote Quote by 10-69
Uniform solid cylinder, mass M and radius 2R rests on horizontal table top. A string is attached by a yoke to a frictionless axle through the center so the cylinder can rotate about the axle. The string runs over a disk-shaped pulley, mass M, radius R, mounted on a frictionless axle through its center. A block of mass M is suspended from the free end of the string. The string doesn't slip over the pulley surface and the cylinder rolls without slipping on the table top.
a) find the acceleration of the block after the system is released from rest
for the block: [tex]\Sigma F_y = M*a = M*g-T_1[/tex]
for the flywheel: [tex]\Sigma \tau = I*\alpha = (1/2*M*R^2)(a/R)=1/2*M*R*a = (T_2-T_1)*R[/tex]
For the cylinder with the yoke: [tex]\Sigma F_x = T_2 = M*a, a=(2R)\alpha[/tex]

Here is where I'm stuck:
Again for the cylinder with the yoke: [tex]\Sigma \tau = (1/2*M*(2R)^2)(a/(2R)) = a*M*R = ??? [/tex]

I didn't think there would be torque in the system since the axle acted on the center of mass, thus torque would be zero, but otherwise, I can't solve this since there are three unknowns and only two equations.

Thanks in advance for any and all help.
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Doc Al
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Nov18-06, 03:09 PM
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Quote Quote by a_lawson_2k
First problem:


I drew the free body diagram, and applied Newton's Second to the (adjusted) x and y axes (offset 36,9 degrees).
[tex]\Sigma F_y = n-mg cos(\theta) = 0[/tex]
[tex]\Sigma F_x = mg sin(\theta)-F_f-T = m*a_y[/tex]
I assume you meant that to equal [itex]m*a_x[/itex]
The natural force became mg cos (36.9*) which worked out to be 39,2N
mg sin (36.9*) came out to be 29.4N, so the sum of the forces in the Y direction came out to:
[tex]\Sigma F_x = 19.6-T = m*a_y[/tex]

Here is what really started puzzling me, how do I get the flywheel to come into the equation? It doesn't specify a radius from which I can evaluate the torque, and from there the angular acceleration, so I'm stuck.
Sure it gives the radius:
The string pulls without slipping at a perpendicular distance of 0,2 m from the axis.
Doc Al
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#3
Nov18-06, 03:18 PM
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Quote Quote by a_lawson_2k
Second problem:


I don't know where to start here, sadly. I have no clue how to even approach it, the book didn't provide an example even remotely like this one. I could probably set it up in energy terms and using the sum of the forces in the radial direction as being (mv^2)/r, but I'd have no idea what to do from there.
That's exactly what you have to do:
(1) Use Newton's 2nd law to find the minimum speed for maintaining contact at the top of the loop.
(2) Use conservation of energy to find the initial height that will give you the needed speed at the top of the loop. Do not neglect rotational energy.

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Nov18-06, 03:29 PM
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A few problems I was stuck on


Quote Quote by a_lawson_2k
This one I have more of an idea of how to solve:


for the block: [tex]\Sigma F_y = M*a = M*g-T_1[/tex]
for the flywheel: [tex]\Sigma \tau = I*\alpha = (1/2*M*R^2)(a/R)=1/2*M*R*a = (T_2-T_1)*R[/tex]
For the cylinder with the yoke: [tex]\Sigma F_x = T_2 = M*a, a=(2R)\alpha[/tex]

Here is where I'm stuck:
Again for the cylinder with the yoke: [tex]\Sigma \tau = (1/2*M*(2R)^2)(a/(2R)) = a*M*R = ??? [/tex]

I didn't think there would be torque in the system since the axle acted on the center of mass, thus torque would be zero, but otherwise, I can't solve this since there are three unknowns and only two equations.
Comments:
(1) Regarding the block: OK. Note that you have chosen "a" to be down and positive.
(2) Regarding the pulley: Careful with your signs. Use the same convention that you used with the block.
(3) Regarding the cylinder: The string tension is not the only force acting on it. Recall that it rolls without slipping. You'll end up with two equations for the cylinder: one for rotation; one for translation.
a_lawson_2k
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#5
Nov18-06, 06:55 PM
P: 38
Thanks for the response; the first one was solved without any problem.

Regarding the second: I set it up like this:
(height of loop)[tex]\Sigma F_r = m v^2/R = m*g, v=\sqrt{R*g}, v=\omega*R, \omega = \frac{\sqrt{R*g}}{R}[/tex]

From there, I could not reconcile my solution for h with the book's. What I did was set the difference in potential energy (m*g*(h-2R)) equal to the rotational kinetic energy [tex]\frac{1}{2}I\omega^2[/tex]. The problem was presented in my computation of the moment of inertia of the rotating object, namely using the parallel axis theorem, I got the moment of inertia to be [tex]\frac{2}{5}Mr^2+M(R-r)^2[/tex]. The middle term of the square of R-r doesn't cancel out, even without it, my answer is still off. Something tells me I'm missing something, though, is the kinetic energy included with the rotational energy, or is it only the rotational energy?

Also, there is another section of the problem that says 'what is the answer to part a (this part) if the track is well-lubricated, making friction negligible'. Friction was *already* negligible...what's going on here?

As for the cylinder in the third problem, I don't quite follow you. Which equations are you talking about (angular momentum? torque?)
Doc Al
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#6
Nov18-06, 07:18 PM
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Quote Quote by a_lawson_2k
Regarding the second: I set it up like this:
(height of loop)[tex]\Sigma F_r = m v^2/R = m*g, v=\sqrt{R*g}, v=\omega*R, \omega = \frac{\sqrt{R*g}}{R}[/tex]

From there, I could not reconcile my solution for h with the book's. What I did was set the difference in potential energy (m*g*(h-2R)) equal to the rotational kinetic energy [tex]\frac{1}{2}I\omega^2[/tex].
Realize that the ball has both translation and rotational KE.

The problem was presented in my computation of the moment of inertia of the rotating object, namely using the parallel axis theorem, I got the moment of inertia to be [tex]\frac{2}{5}Mr^2+M(R-r)^2[/tex]. The middle term of the square of R-r doesn't cancel out, even without it, my answer is still off.
Why are you using the parallel axis theorem?
Something tells me I'm missing something, though, is the kinetic energy included with the rotational energy, or is it only the rotational energy?
As already said, you're missing the translational KE.

Also, there is another section of the problem that says 'what is the answer to part a (this part) if the track is well-lubricated, making friction negligible'. Friction was *already* negligible...what's going on here?
Careful. Rolling friction was neglible, but not ordinary static friction which is what allows the ball to roll instead of slide. (This is also the key to problem #3.) Without friction, the ball just slides: translational KE, but no rotational KE.

As for the cylinder in the third problem, I don't quite follow you. Which equations are you talking about (angular momentum? torque?)
I'm talking about applying Newton's 2nd law, twice: the cylinder both rotates and translates. (See my tip above about static friction.)
a_lawson_2k
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#7
Nov18-06, 07:29 PM
P: 38
I used the parallel axis theorem because I thought the ball was rotating around the circle, radius R, so the rotational kinetic energy formula could be used. As for the last two, no constant of kinetic friction was given; how do I go about this? As for the cylinder both rotating and translating, the equations for those are energy-related and deal with velocity...since time doesn't come into play, I can't switch velocity over to acceleration.
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Nov18-06, 07:43 PM
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Quote Quote by a_lawson_2k
I used the parallel axis theorem because I thought the ball was rotating around the circle, radius R, so the rotational kinetic energy formula could be used.
No. Better to treat the ball's KE as the sum of:
(1) translational KE of its center of mass
(2) rotational KE about its center of mass

Of course, they are not independent since [itex]\omega[/itex] and v are related.

As for the last two, no constant of kinetic friction was given; how do I go about this?
No coefficient of friction is needed.

As for the cylinder both rotating and translating, the equations for those are energy-related and deal with velocity...since time doesn't come into play, I can't switch velocity over to acceleration.
Newton's 2nd law is what you've been using all along in that problem. You applied it to the rotation of the pulley and the translation of the block. Now you'll apply (twice) to analyze the rotation and translation of the cylinder.

First step: Identify the forces (and the torques) acting on the cylinder.
a_lawson_2k
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#9
Nov18-06, 08:04 PM
P: 38
OK, here is what I came up with:
[tex]Mgh=\frac{1}{2}M(\sqrt{Rg})^2+\frac{1}{2}(\frac{1}{5}Mr^2)(\sqrt{Rg}/r)^2[/tex]
Solving, it did not come out to be the result the book wanted. I used [tex]\omega = v/r[/tex], as the ball never slipped. No idea how to deal with this from here, as usual.

I could identify the forces acting on the cylinder, just not the torques. I know that [tex]\alpha = \frac{a}{2R}[/tex] in this case, I just don't know how to put it into a [tex]\Sigma \Tau[/tex] equation or how it figures into calculating the acceleration from there.
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Nov18-06, 08:24 PM
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Quote Quote by a_lawson_2k
OK, here is what I came up with:
[tex]Mgh=\frac{1}{2}M(\sqrt{Rg})^2+\frac{1}{2}(\frac{1}{5}Mr^2)(\sqrt{Rg}/r)^2[/tex]
Solving, it did not come out to be the result the book wanted. I used [tex]\omega = v/r[/tex], as the ball never slipped. No idea how to deal with this from here, as usual.
That looks pretty complicated. Why not start with this:
[tex]mg(h -2R) = 1/2mv^2 + 1/2I\omega^2[/tex]

Use [itex]\omega = v/r[/itex] and your equation for centripetal force to simplify.


I could identify the forces acting on the cylinder, just not the torques.
What are they and where do they act?
a_lawson_2k
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#11
Nov18-06, 08:43 PM
P: 38
[tex]mg(h-2R)=\frac{7}{10}mv^2[/tex] that's what I get when I plug in the moment of inertia and the [itex]\omega = v/r[/itex]. I'm going in circles here, the book says (23R-17r)/10 and I have no clue how they got there. Again, the book gave no examples even remotely like this, so I'm lost.

From there, the velocity in order to keep the thing moving in the circle was [tex]\sqrt{Rg}[/tex]. Not sure how to figure that in either...

As for the cylinder, I know it's rolling, I don't know to figure out what it's doing to the tension and acceleration of the system. Again, didn't see where to place such forces in either book I consulted.

Frankly, I can't believe I even passed AP Physics, but this would explain the D.
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Nov18-06, 09:06 PM
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Quote Quote by a_lawson_2k
[tex]mg(h-2R)=\frac{7}{10}mv^2[/tex] that's what I get when I plug in the moment of inertia and the [itex]\omega = v/r[/itex]. I'm going in circles here, the book says (23R-17r)/10 and I have no clue how they got there. Again, the book gave no examples even remotely like this, so I'm lost.
Oops... looks like we have to be a bit more careful. The change in height of the ball's center is h -2R +r, not just h - 2R.

From there, the velocity in order to keep the thing moving in the circle was [tex]\sqrt{Rg}[/tex]. Not sure how to figure that in either...
And the radius of the circle that the ball's center traces out is not R, but R -r. So [itex]mv^2/(R-r) = mg[/itex].

As for the cylinder, I know it's rolling, I don't know to figure out what it's doing to the tension and acceleration of the system. Again, didn't see where to place such forces in either book I consulted.
Again: What forces act on the cylinder? (Hint: Only two forces act horizontally, which is the direction of interest.)
a_lawson_2k
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#13
Nov19-06, 11:18 AM
P: 38
Well, I know the tension is acting from the center of mass, thus causing no torque, and the friction going in the opposite direction, where the cylinder meets the table, causes the torque. The sum of the torques would be equal to the moment of inertia times the angular acceleration (the acceleration I'm trying to solve for divided by the radius), I can't seem to find what goes on the other end of the equation.
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Nov19-06, 01:39 PM
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Not sure where you are stuck--we're just applying Newton's 2nd law. For rotation: net Torque = I*alpha; For translation: net Force = m*a.

Call the friction force "f". What torque does it produce (about the cylinder's center of mass)? Set that equal to I*alpha. That's your first equation. (Which you can of course simplify by replacing alpha and I, etc.)

You'll have another equation by setting the net force on the cylinder equal to M*a.
a_lawson_2k
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#15
Nov19-06, 03:49 PM
P: 38
OK, I think I found out what to do
[tex]\SimgaF_x=T_1-F_f=Ma, \Sigma\Tau=I\alpha=(\frac{1}{2}M{2R}^2)(\frac{a}{2R})=MRa=F_f(2R)
F_f=\frac{1}{2}Ma[/tex]
Plugging in, [tex]T_1=Ma+F_f=\frac{3}{2}Ma
\Sigma\Tau=I\alpha=\frac{1}{2}MR^2(\frac{a}{R})=\frac{1}{2}MRa=(T_2-T_1)R
T_1=Ma
\Sigma F_y=Ma=Mg-T_1, a=\frac{g}{2}[/tex]
The book says g/3, I looked over it a few more times and didn't find out what went wrong...

[EDIT] after another look and viewing an example, I found out I had goofed on the signs of the torque; I was able to get the desired result from there. Thanks again for your help and assistance, and hopefully if I get better at this stuff (it's only been a week since I dusted this stuff off), I can help on here as well.
fajoler
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#16
Oct31-10, 12:45 PM
P: 11
Hey for the third problem you were having trouble with, I somehow ended up with g/2 as my answer. Could you tell me where exactly you messed up so I can see where I messed up as well?

I'm not quite sure...
Thanks!


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