- #1
Sanchayan Dutta
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My Approach is shown in the picture.
Which method is correct for this problem involving motion of
- Thread starter Sanchayan Dutta
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AI Thread Summary
The discussion centers on the analysis of electric forces acting on a particle Q in motion, particularly at the highest and lowest points of its trajectory. Participants agree that the two electric forces on Q are in the same direction when Q is at the top, but there is confusion regarding the diagram's accuracy and the conditions described. The tension in the string is noted to be zero at the top, with contributions from centripetal force, weight, and electric forces needing to be considered. There is debate about the minimum tension occurring when Q is at the bottom, with suggestions to clarify the method used. Overall, the execution of the methods discussed requires refinement to accurately represent the physical situation.
My Approach is shown in the picture.
- #2
nasu
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It may be that none of the answers are right.
For the first method, are the two electric forces on Q in the same direction, when Q is on top?
For the first method, are the two electric forces on Q in the same direction, when Q is on top?
- #3
Sanchayan Dutta
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yes they are in same direction.nasu said:
- #4
nasu
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Do you think that your diagram represents the situation described in the problem?
I mean "particle B at the highest point".
I mean "particle B at the highest point".
- #5
J Hann
- 264
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The tension in the string is zero at the top for the given conditions.
Add the contributions to the tension (centripetal force, weight of the mass, and the two electrical forces
such that the tension is zero.
It might be easier to think of the acceleration at the top (m v^2 / R = weight + the two electrical forces)
Add the contributions to the tension (centripetal force, weight of the mass, and the two electrical forces
such that the tension is zero.
It might be easier to think of the acceleration at the top (m v^2 / R = weight + the two electrical forces)
- #6
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They are not. Maybe, as nasu wonders, you are taking Q at the lowest point instead of the highest.Sanchayan Dutta said:
In your method 2, you write that the tension is minimum when Q is at the bottom. What makes you think that?
Seems to me that is exactly Sanchayan's method 1, but the execution was lacking.J Hann said:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19.
For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Let's declare that for the cylinder,
mass = M = 10 kg
Radius = R = 4 m
For the wall and the floor,
Friction coeff = ##\mu## = 0.5
For the hanging mass,
mass = m = 11 kg
First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on.
Force on the hanging mass
$$mg - T = ma$$
Force(Cylinder) on y
$$N_f + f_w - Mg = 0$$
Force(Cylinder) on x
$$T + f_f - N_w = Ma$$
There's also...
This problem is two parts. The first is to determine what effects are being provided by each of the elements - 1) the window panes; 2) the asphalt surface. My answer to that is
The second part of the problem is exactly why you get this affect.
And one more spoiler:
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