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Resolution of vectors for problems related to mechanics

  • #1

Homework Statement


Refer the given image. [Prob 2.9]

Homework Equations


F=ma

The Attempt at a Solution


I drew the normal vector perpendicular to the surface of the cone and resolved it as
##Nsin\theta=mg##
##Ncos\theta=\frac {mv_{0}^2} {r}## where ##v_{0}## and ##r## are the speed and radius of the trajectory respectively.
Solving for ##r## we get,
##r=\frac {v_{0}^2} {g}tan\theta##

My doubt is however is on the matter of resolving the required vectors. Is it necessary to resolve the normal reaction into components or is it correct to resolve ##mg## along the normal and along the lateral surface of the cone? Similar situations arise when solving problems related to pendulums. I get confused whether tension vector must be resolved or mg.
 

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  • #2
Orodruin
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It does not matter. It is just a question of which basis you work in.

The point if you resolve the normal and the lateral is that the acceleration also needs to be resolved into those components.
 
  • #3
stockzahn
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The Attempt at a Solution


I drew the normal vector perpendicular to the surface of the cone and resolved it as
##Nsin\theta=mg##
##Ncos\theta=\frac {mv_{0}^2} {r}## where ##v_{0}## and ##r## are the speed and radius of the trajectory respectively.
Solving for ##r## we get,
##r=\frac {v_{0}^2} {g}cos\theta##
Unfortunately there is something wrong with the math - I think the approach is correct.

The Attempt at a Solution


My doubt is however is on the matter of resolving the required vectors. Is it necessary to resolve the normal reaction into components or is it correct to resolve mg along the normal and along the lateral surface of the cone? Similar situations arise when solving problems related to pendulums. I get confused whether tension vector must be resolved or mg.
In general it doesn't matter, but often one of the reference frames is more advantageous (for the calculation and the illustration of the results).
 
  • #4
Orodruin
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Unfortunately there is something wrong with the math - I think the approach is correct.
I agree. The algebra done to get from the resolved components to the final result is not correct.
 
  • #5
ehild
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Homework Statement


Refer the given image. [Prob 2.9]

Homework Equations


F=ma

The Attempt at a Solution


I drew the normal vector perpendicular to the surface of the cone and resolved it as
##Nsin\theta=mg##
##Ncos\theta=\frac {mv_{0}^2} {r}## where ##v_{0}## and ##r## are the speed and radius of the trajectory respectively.
Correct

Solving for ##r## we get,
##r=\frac {v_{0}^2} {g}cos\theta##
No, it is not right!
 
  • #6
I'm apologize for my ignorance. I've made the required change.
I agree. The algebra done to get from the resolved components to the final result is not correct.
 
  • #7
What's wrong in this approach ? (refer the image provided)
 

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  • #8
Also if any of the vectors can be resolved then when resolving acceleration and mg in the direction perpendicular to the surface and along it then we get...
(refer image).
However the body doesn't move along the surface but through this approach a net force appears. Where am I going wrong?
 

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  • #9
stockzahn
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What's wrong in this approach ? (refer the image provided)
The approach is correct:

##Nsin\theta=mg##
##Ncos\theta=\frac {mv_{0}^2} {r}## where ##v_{0}## and ##r## are the speed and radius of the trajectory respectively.
The final result (you showed in your original first post) was not though (like ehild already pointed out):

Solving for ##r## we get,
##r=\frac {v_{0}^2} {g}cos\theta##
You seemed to change the ##cos## into a ##tan## afterwards and it now is right.
 
  • #10
stockzahn
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Also if any of the vectors can be resolved then when resolving acceleration and mg in the direction perpendicular to the surface and along it then we get...
(refer image).
However the body doesn't move along the surface but through this approach a net force appears. Where am I going wrong?
In your calculation shown in the pic, the signs are not correct. The component of gravitational force points downwards, the component of the centrigugal force upwards. If you change the sign from plus to minus (and set the result zero) you will obtain the same correlation as in your first post.
 
  • #11
In your calculation shown in the pic, the signs are not correct. The component of gravitational force points downwards, the component of the centrigugal force upwards. If you change the sign from plus to minus (and set the result zero) you will obtain the same correlation as in your first post.
Thanks but another doubt arises in my mind. When am I supposed to consider centripetal acceleration (towards the centre of trajectory) and centrifugal acceleration (away from centre) ?
 
  • #12
The approach is correct:
Let's say the problem was to find the magnitude of normal reaction acting on the body. Clearly both the values of normal can't be the same (refer 2nd image). Then what's the right value of the normal reaction ?
 
  • #13
stockzahn
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Thanks but another doubt arises in my mind. When am I supposed to consider centripetal acceleration (towards the centre of trajectory) and centrifugal acceleration (away from centre) ?
There is an elementary difference between the centripetal force, which is a "real" force, and the centrifugal force, which is an inertial, and therefore, a pseudo force. You can compare it with the application of Newton's second law:

$$m a = F_{net}$$

The net force ##F_{net}## corresponds to the real force acting on the ball by the wall (and the gravity) - the centripetal force. The LHS corresponds to the centrifugal force, which is the the resulting change of kinematic state due to the centripetal force, but since they are equal you can take advantage of the formula for the centrifugal force for movements on circular trajectories.
 
  • #14
stockzahn
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Let's say the problem was to find the magnitude of normal reaction acting on the body. Clearly both the values of normal can't be the same (refer 2nd image). Then what's the right value of the normal reaction ?
I'm not sure if I misunderstand, do you doubt that ##N=\frac{mg}{sin\theta}=\frac{mv^2}{r\,cos\theta}## (if the height of the ball is constant)?
 
  • #15
I'm not sure if I misunderstand, do you doubt that ##N=\frac{mg}{sin\theta}=\frac{mv^2}{r\,cos\theta}## (if the height of the ball is constant)?
Let me start from scratch.
Case 1: I resolve normal reaction vector as one along y axis as ##Nsin\theta=mg##
Case 2: I resolve ##mg## along the surface of cone and perpendicular to the surface as ##mgsin\theta=N##
Now by your previous statement both the methods are correct however we end up with 2 different values of ##N##. This is what I'm confused about.
 
  • #16
Orodruin
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Let me start from scratch.
Case 1: I resolve normal reaction vector as one along y axis as ##Nsin\theta=mg##
Case 2: I resolve ##mg## along the surface of cone and perpendicular to the surface as ##mgsin\theta=N##
Now by your previous statement both the methods are correct however we end up with 2 different values of ##N##. This is what I'm confused about.
Your case 2 is wrong. You forgot that the total component perpendicular to the surface needs to be non-zero because the acceleration has a component that is perpendicular to the surface. The correct equation is
$$
N - mg\sin(\theta) = \frac{mv_0^2}{r} \cos(\theta).
$$
 
  • #17
stockzahn
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Let me start from scratch.
Case 1: I resolve normal reaction vector as one along y axis as ##Nsin\theta=mg##
Case 2: I resolve ##mg## along the surface of cone and perpendicular to the surface as ##mgsin\theta=N##
Now by your previous statement both the methods are correct however we end up with 2 different values of ##N##. This is what I'm confused about.
 

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