Physics Conceptual Questions - Mixed

OlderDan

Quote by cristo

I disagree... Consider the postion vector of the COM. [tex] M \bold{R} =\sum_i m_i \bold{r}_i \[/tex] Differentiating twice yields [tex] M \bold{\ddot{R}} =\sum_i m_i \ddot{r_i} = \sum_i\sum_{j\neq i} \bold{F}_{ij} + \sum_i \bold{F}_i [/tex] Now, since [tex]\bold{F}_{ij}+\bold{F}_{ji}=0. [/tex] the first term on the rhs vanishes, hence [tex] M \bold{\ddot{R}} = \sum_i\bold{F}_i [/tex] So, in order for the COM to have constant velocity, the sum of all external forces acting on the ith particle must be zero.

That is not the correct interpretation of your equation. Want to restate it?

OlderDan

Quote by threewingedfury

ok so for the friction problem - T cos theta?

KE + PE problem - only when the work done by the resultant external force is zero?

Earths CoM - a little less than halfway between Earth's surface and the outer boundary of the atmosphere?

particles - the forces exerted by the particles on each other sum to zero?

Theres so many answers on here that I got confused

You created much of the confusion by posting two sets of two unrealted problems in the same thread. It is far better to keep problems separate.

Your suggested answers to all the problems are wrong.

cristo

Quote by OlderDan

That is not the correct interpretation of your equation. Want to restate it?

If the sum of the external forces acting on the system equals zero, then the COM will move with a constant velocity.

threewingedfury

The right answer (IMHO): the horizontal component of the applied force is T cos theta. The block is moving at constant velocity so the resultant horizontal force is zero. Therefore, the friction force is also T cos theta (in the opposite direction).

The answer mu(mg-Tsin theta) is also right - except it doesn't appear in the list of choices.

Your suggested answers to all the problems are wrong.

So the T cos theta isnt right?

OlderDan

Quote by cristo

If the sum of the external forces acting on the system equals zero, then the COM will move with a constant velocity.

That sounds better. An example of this would be the air in a balloon. There are many external forces acting on the individual molecules of air as they coillide with the balloon surface, but their sum is often zero.

OlderDan

Quote by threewingedfury

So the T cos theta isnt right?

T cosθ is one correct answer, and it is the only correct answer in the list. The friction force is also the coefficient of kinetic friction times the normal force, its magnitude is

μ(mg - T sinθ)

which is not one of the choices.

threewingedfury

so for the CG of the Earth's atmosphere - I dont really understand exactly how to figure this out. I figured it would be closer to the Earths surface, but is it halfway between the surface and the boundary of the atmosphere?

cristo

Alephzero gives a good hint in post #12

Quote by AlephZero

Re the "CG of the earth atmosphere" question, the question is about the whole atmosphere, not just the bit of it which is above your own head. Think about a sphere with a thin layer of material (atmosphere) covering the whole surface.

threewingedfury

but I dont understand how the Earth and the atmosphere play roles in the question - I think of the Earth weighing so much, but it cant be the center of the Earth - Im all confused

cristo

What makes you think it can't be the centre of the earth? As Alephzero says, imagine the atmosphere as a thin layer of material coating the earth. Now, where do you think the centre of gravity of this object lies?

threewingedfury

well I figured because it said the center of mass of the atmosphere that it didnt really include the earth

cristo

It doesn't include the earth. Ok, imagine it wasnt there, so you had a thin hollow spherical "shell" of uniform density. Where would you expect the centre of gravity of this shell to be?

threewingedfury

in the center?

cristo

Yup, so the answer to your question is close to the centre of the earth.

vanesch

Quote by cristo

I disagree... Consider the postion vector of the COM. [tex] M \bold{R} =\sum_i m_i \bold{r}_i \[/tex] Differentiating twice yields [tex] M \bold{\ddot{R}} =\sum_i m_i \ddot{r_i} = \sum_i\sum_{j\neq i} \bold{F}_{ij} + \sum_i \bold{F}_i [/tex] Now, since [tex]\bold{F}_{ij}+\bold{F}_{ji}=0. [/tex] the first term on the rhs vanishes, hence [tex] M \bold{\ddot{R}} = \sum_i\bold{F}_i [/tex] So, in order for the COM to have constant velocity, the sum of all external forces acting on the ith particle must be zero.

You should read more carefully. I was talking about the SECOND question, and nobody is talking about a uniform motion of the COG, but rather about the conservation of KE + PE.

This question:

The sum of the kinetic and potential energies of a system of objects is conserved:
a. only when no external force acts on the objects
b. only when the objects move along closed paths
c. only when the work done by the resultant external force is zero
d. always
e. none of the above

Is c correct? If theres no work done by the external force then I figured the energy would be conserved.

Here, I maintain that the correct answer is e, for the reasons I mentionned before.

vanesch

Quote by threewingedfury

So the T cos theta isnt right?

It IS the right answer. In fact, given that:
1) the motion of the block is uniform
2) the forces acting on it are
- weight (vertical)
- binding force of the surface (vertical)
- friction force (horizontal)
- known force T under angle T

we must have that the total force is 0 (because uniform motion), and hence that the horizontal and vertical components are 0.
The vertical component is given by the sum of the weight, the vertical component of T and the binding force, but we don't care.

The horizontal component is given by the sum of the friction force and the horizontal component of T, which equals T cos theta.
So this means that the friction force must equal T cos theta (in the other direction).

OlderDan

Quote by vanesch

You should read more carefully. I was talking about the SECOND question, and nobody is talking about a uniform motion of the COG, but rather about the conservation of KE + PE.

This question:

Here, I maintain that the correct answer is e, for the reasons I mentionned before.

I probably should not be speaking for cristo, but I don't think he was disagreeing with you. I think he was responding to threewingedfury, but since he did not quote the post to which he was replying, the sequence suggested otherwise.

So to all who read here, PLEASE avoid creating such a jumpled thread by keeping problems in separate threads, and please identify the posts to which you are responding if it is not the one directly above yours.

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threewingedfury	so for the CG of the Earth's atmosphere - I dont really understand exactly how to figure this out. I figured it would be closer to the Earths surface, but is it halfway between the surface and the boundary of the atmosphere?

cristo Mentor	What makes you think it can't be the centre of the earth? As Alephzero says, imagine the atmosphere as a thin layer of material coating the earth. Now, where do you think the centre of gravity of this object lies?