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Rotational motion


druS

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OK I'll give this forum a go. It is not actually homework, but study that is related and extra.

A 2.00kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.150m (my note R = 0.075), to a hanging book with a mass 3.00kg. The system is released from rest and the books are observed to move 1.20m in 0.800s.

a) What is the tension in each part of the cord?

b) What is the moment of inertia of the pulley about it's rotational axis.

This is an end of chapter question Q 10.13 in the global edition of University Physics with modern physics by Young and Freedman.

I (think I) can work the basic kinematics on distance and time and capture an acceleration. I haven't managed turning that into the cord tensions. I think I can find an angular acceleration in the pulley which means i should be able to find moment of inertia.

But it has not at all worked.

Who wants to guide a newb through what must be a pretty basic question?

 

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You should draw a free-body diagram, showing the various forces in play, and then set up your Newton's second law equation(s)

Kinematics will give you acceleration, as you have noted. You can get the tensions by combining these.

The moment of inertia is another matter, since you aren't given the pulley's mass. But, first things first.

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So the second law is F = m.a  And I wish I could start there.

Let's start with the lower  book. So (oh good, the cat decided to join this conversation)

x = x o + vo.t + 1/2 at2

I am focused on the lower body, get an acceleration of roughly 5.6m/s2. 

OK How do I translate this to the tensions.

 

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6 hours ago, druS said:

I (think I) can work the basic kinematics on distance and time and capture an acceleration. I haven't managed turning that into the cord tensions. I think I can find an angular acceleration in the pulley which means i should be able to find moment of inertia.

But it has not at all worked.

 

4 hours ago, druS said:

So the second law is F = m.a  And I wish I could start there.

Let's start with the lower  book. So (oh good, the cat decided to join this conversation)

x = x o + vo.t + 1/2 at2

I am focused on the lower body, get an acceleration of roughly 5.6m/s2. 

 

Hello dru, I will move you along a bit, perhaps swansont or someone will help with the next bit.

You are on the right lines, but I don't know where you got that formula from, which may be why you have calculated the wrong value for acceleration.

As an old dog I use


[math]s = ut + \frac{1}{2}f{t^2}[/math]


Where s is distance, u is initial velocity, f is acceleration and t is time.

In this case


[math]u = 0[/math]   ;   [math]s = 1.2[/math]  ; [math]t = 0.8[/math]

all given.

Note this is just stated for 'the system' as all parts of the system remain connected and the rope does not go slack.

Putting the numbers in gives


[math]1.2 = 0 + \frac{1}{2}f\left( {\frac{8}{{10}}} \right)\left( {\frac{8}{{10}}} \right)[/math]


Now I am going to labour this point a bit since it helps the arithmetic.

You do not need a calculator and for school work it is often best to follow this pattern.

get rid of the fractions and decimals thus


[math]2.4 = f\left( {\frac{8}{{10}}} \right)\left( {\frac{8}{{10}}} \right)[/math]

Note the way I have used 8/10 instead of 0.8


[math]240 = f\left( 8 \right)\left( 8 \right)[/math]


[math]240 = 64f[/math]


finally


[math]f = \left( {\frac{{30}}{8}} \right)[/math] metres per second per second.

I would suggest leaving the acceleration as 30/8 for the next part.

 

School examples generally drop out easily like this.

 

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I see you visited the site some hours after I posted but did not reply.

Hopefully you are thinking about it.

 

11 hours ago, studiot said:

Note this is just stated for 'the system' as all parts of the system remain connected and the rope does not go slack.

 

I hope you picked up the significance of this statement.

It means that bothe masses and all parts of the rope move the same distance in the same time resulting in the same acceleration.

So we can simply apply Newton's second law (F = ma) to each mass to obtain the net acceleration force.

For the book mass the rope is pulling in the horizontal direction and its tension is the only force acting on the book so


[math]T = 2\left( {\frac{{30}}{8}} \right) = 7.5N[/math]

For the hanging weight there are two forces acting. The tension in the rope and gravity and these act in opposite directions.

 

Can you form the equation to determine the tension using this information ?

This is where swansont's suggestion comes in handy.

17 hours ago, swansont said:

You should draw a free-body diagram, showing the various forces in play, .................................................But, first things first.

 

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Cheers guys. Waking up in the middle of the night with a brain spark started by this conversation. Yes I've been visiting and cogitating. Studiot, I suspect our formula are equivalent though I like the accuracy in the way you express things.  I'll stick with the formula sheet in our sample exam papers though.

Swansont I think I was jumping to the next step in my head, looking for the moment of inertia. not focusing on one issue at a time had me applying Newtons 3rd law incorrectly. Acceleration works to 3.75m.s-2 (matching Studiot's more accurate expression). And going down is negative.

For the 3kg hanging book I get a tension force above F= m.a = 3.0kg x 3.75m/s2 and weight acting down W = m.a = 3.0 x -9.81

AND what I was missing a resulting force FR = m.a = 3kg . 3.75m/s2

[Sum of] Forces = 0

FR = FT + W (the bit I did not have right!)

3 x (-3.75) = FT + 3.00 x (-9.81)  providing FT = 18.18 N

Then I was wracking my brain with the 2kg book on the (frictionless) table. The only difference is that there is no relevant weight force (normal force and weight cancel to zero in the y direction). (What I had been doing incorrectly for the 3kg book) In the x direction the resultant force simple equals the tension force and we get 7.5N.

Which fortunately matches the text answers.

OK next step?

I'm presuming that the torque applied by these tension forces will not equate to the actual torque observed and the difference will be related to the moment of inertia of the pulley. (but just for now it is 3:20 in the morning and I'm going back to sleep a happy student).

 

Edited by druS
more explanation required
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2 hours ago, druS said:

I'm presuming that the torque applied by these tension forces will not equate to the actual torque observed and the difference will be related to the moment of inertia of the pulley. (but just for now it is 3:20 in the morning and I'm going back to sleep a happy student). 

The linear acceleration of the point of contact of the rope and pulley is the same as you have already found.

 

Can you relate this to the rotating properties of the pulley (radius, moment of inertia etc) ?

 

2 hours ago, druS said:

Studiot, I suspect our formula are equivalent though I like the accuracy in the way you express things.  I'll stick with the formula sheet in our sample exam papers though.

I don't like using x as the basic distance variable since you may need to resolve the motions into x and y coordinates or actually work along the line of motion.

Further xo has no proper meaning in terms of distance travelled and will give misleading answers in cases where resolution is needed (eg because the motion is not along the x axis).

Not only is it wildly unconventional, I think it's just the authors puffing themselsves up.

Edited by studiot
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Thanks again. Yes this final step didnt stress me too much.

accel = alpha x R;

T = F x R

Sigma T = I x Alpha

Which ultimately gave me 0.16 kg.m/s which is text consistent.

Are there easy and simple introductions to LaTex? Probably dont have time to test it mid trimester but it is goinng to have to happen sooner or later.

 

Thank Studiot.

 

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13 hours ago, druS said:

Thanks again. Yes this final step didnt stress me too much.

accel = alpha x R;

T = F x R

Sigma T = I x Alpha

Which ultimately gave me 0.16 kg.m/s which is text consistent.

Are there easy and simple introductions to LaTex? Probably dont have time to test it mid trimester but it is goinng to have to happen sooner or later.

 

Thank Studiot.

 

So you now have the satisfaction of having puzzled most of the question out for yourself.

:)

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