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Angular Momentum of a counterweight

  • Thread starter Sheneron
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[SOLVED] Angular Momentum

1. Homework Statement
A 4.60 kg counterweight is attached to a light cord, which is wound around a spool (refer to Fig. 10.20). The spool is a uniform solid cylinder of radius 8.00 cm and mass 1.30 kg.
(a) What is the net torque on the system about the point O?
Magnitude
(b) When the counterweight has a speed v, the pulley has an angular speed = v/R. Determine the total angular momentum of the system about O.
(c) Using the fact that = dL/dt and your result from (b), calculate the acceleration of the counterweight.
2. Homework Equations
[tex] L = r X P[/tex]
[tex] \tau = r X F[/tex]
[tex] L = I\omega[/tex]

3. The Attempt at a Solution
I was able to solve the first part:
[tex] \tau = rF [/tex]
[tex] \tau = (0.08)(4.6)(9.8) = 3.61 Nm[/tex]

I am not sure how to solve part B though, here is what I tried but it was wrong:

[tex] L_{net} = rp + I\omega [/tex]
[tex] L_{net} = r X mv + I*\frac{v}{r} [/tex]
[tex] L_{net} = v (r X m + \frac{I}{r})[/tex]

I factored out the v because it doesn't give you the velocity, and the answer space looks like this:

________ X v kg *m^2/s
 

Answers and Replies

Doc Al
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I am not sure how to solve part B though, here is what I tried but it was wrong:

[tex] L_{net} = rp + I\omega [/tex]
[tex] L_{net} = r X mv + I*\frac{v}{r} [/tex]
[tex] L_{net} = v (r X m + \frac{I}{r})[/tex]
Looks OK to me--keep going.

Hint: What's the rotational inertia of a solid cylinder?
 
360
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Rotational inertia of a solid cylinder is 0.5MR^2

With my equation there, at the rXm, do I have to do the cross product?
 
Doc Al
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Rotational inertia of a solid cylinder is 0.5MR^2
Good.

With my equation there, at the rXm, do I have to do the cross product?
You can only take cross products of vectors, so r X m doesn't make sense. (I was wondering if you meant that X as a cross product or just as multiplication.) The time to do the cross product is before you factor out the speed.

In the definition of angular momentum of a particle, [itex]\vec{L} = \vec{r}\times m\vec{v}[/itex], [itex]\vec{r}[/itex] is the position vector of the particle. The cross product turns out to be easy to evaluate in terms of r (the radius). So what is the angular momentum of the particle?
 
360
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I don't know. I am a little unclear about how to do cross products.
 
360
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Can someone tell me how I would go about cross producting [itex] \vec{L} = \vec{r}\times m\vec{v}
[/itex]
 
360
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1. Homework Statement
A 4.60 kg counterweight is attached to a light cord, which is wound around a spool (refer to Fig. 10.20). The spool is a uniform solid cylinder of radius 8.00 cm and mass 1.30 kg.
(a) What is the net torque on the system about the point O?
Magnitude
(b) When the counterweight has a speed v, the pulley has an angular speed = v/R. Determine the total angular momentum of the system about O.
(c) Using the fact that = dL/dt and your result from (b), calculate the acceleration of the counterweight.


2. Homework Equations
[tex] L = r X P [/tex]
[tex] \tau = r X F [/tex]
[tex] L = I\omega [/tex]
3. The Attempt at a Solution
I was able to solve the first part:
[tex] \tau = rF [/tex]
[tex] \tau = (0.08)(4.6)(9.8) = 3.61 Nm [/tex]

I am not sure how to solve part B though,

[tex] L_{net} = rp + I\omega [/tex]
[tex] L_{net} = (r)mv + I*\frac{v}{r} [/tex]

[tex] L_{net} = v (rm + \frac{I}{r}) [/tex]
[tex]L_{net} = v [ (.08)(4.6) + \frac{1}{2}(1.30)(.08)]} [/tex]
[tex]L_{net} = v(0.42)[/tex]

I factored out the v because it doesn't give you the velocity, and the answer space looks like this:

________ X v kg *m^2/s

Is that how I would go about part B? Thanks
 
Last edited by a moderator:
Doc Al
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Looks good to me.
 
360
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Can you help me with part c?
 
Doc Al
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Can you help me with part c?
What's the full statement of part c? (Your initial post has something missing.) You'll need to take the derivative of L (that's what dL/dt means).
 
360
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sorry, yes I know i have to take the derivative...
Here is the post:
Using the fact that t = dL/dt and your result from (b), calculate the acceleration of the counterweight.

so would it be: 3.61 = dL/dt and dL/dt= 0.42dv/dt and dv/dt is acceleration so would it be a= 3.61*0.42
 
Doc Al
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haha yeah i just realized that.

Thank ye
 
Re: [SOLVED] Angular Momentum

just wondering what units the final answer is in - if the counterweight is just accelerating downwards, do we still write the answer in rads/sec^2 ?
 
Re: [SOLVED] Angular Momentum

or would it be in N*s/kg?

(N*m)/(kg*m^2/s)
 
Doc Al
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Re: [SOLVED] Angular Momentum

just wondering what units the final answer is in - if the counterweight is just accelerating downwards, do we still write the answer in rads/sec^2 ?
The final answer for what question? rads/sec^2 are units for angular acceleration; the counterweight has linear acceleration, which has units of m/s^2.
 
Re: [SOLVED] Angular Momentum

The final answer for what question? rads/sec^2 are units for angular acceleration; the counterweight has linear acceleration, which has units of m/s^2.
thankyou! :) i was thinking angular acceleration *facepalm* :)
 

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