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When calculating time period of simple pendulum (an approximation of SHM at small amplitudes) we take gravitational force greater than tension in string and resolve gravitational force in two orthogonal components and equate one of them with tension. While in case of conical pendulum we take tension greater than gravitational force and resolve it into two orthogonal components and equate one of them with gravitational force on metallic bob. Similarly in banked road problem we take normal reaction greater than gravitational force.

Why in one case tension is more and in other gravitation is more?

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When you analyze motion problems using Newton's Laws you draw a free body diagram for each body you want to analyze. Once you have done this you choose a coordinate system in which you will write Newton's 2nd law equations. I believe you are having a fundamental uncertainty about how to choose that coordinate system.

For the simple pendulum, you see the the bob has zero radial acceleration. The only acceleration is angular. Based on that, it is convenient to choose a coordinate system which is tangential and transverse to the instantaneous velocity. Other coordinate systems are possible, but aren't as convenient. With this choice, you would resolve the weight vector into a component parallel to the string, which happens to also be parallel to the tension. Then the Newton's 2nd Law equations will be $$ T-mg\cos \theta = ma_{radial} = 0.$$ $$-mg\sin \theta = ma_{angular} = m\ell\alpha$$ I have chosen $\theta$ to be measured with respect to the gravitational field direction. Because the string length is not changing, the radial velocity is constantly zero and the radial acceleration is zero.

For the conical pendulum, the bob is moving in a circular path at a constant vertical position. The bob is accelerating centripetally, in the horizontal direction. You have the same force vectors, but because of the type of motion it is more convenient to use a different coordinate system to analyze it. The best system to use is toward the center of the circle (horizontally) and perpendicular to that (vertically). In that case, the tension force should be broken into components (again, because of the coordinate system you have chosen to use out of convenience).

Both systems have the same two force vectors: tension along the string and gravity (mg) vertically. The difference in how to resolve which vector depends on which coordinate system you choose to use.

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In the case of the simple pendulum the resultant force is directed along the path of motion which is orthogonal to the direction of the tension. In the case of the conical pendulum the resultant force is directed to the middle of the circle, which is perpendicular to the direction of gravity. When you draw the forces you will see which force, tension or gravity, has to be split into components

Other way of looking at it, in the simple pendulum the Bob moves perpendicular to the string, so force components along the string should cancel each other out, at least in the turning points. In the conical pendulum the Bob moves in a horizontal circle, so force components in the vertical direction should cancel each other.

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  • $\begingroup$ Resultant is result of gravitational force and tension and their orientations so I still do not get answer.I want in actual why one is greater or one is smaller.I do understand your answer but it doesn't solve my question. $\endgroup$ – Vishvajeet Patil May 11 '14 at 11:54
  • $\begingroup$ In simple pendulum resultant force is a component from gravity, as it is perpendicular to the tension. In conical pendulum the resultant force is a component of tension as it is perpendicular to gravity. $\endgroup$ – KvdLingen May 11 '14 at 11:59
  • $\begingroup$ But why so? is my actual question $\endgroup$ – Vishvajeet Patil May 11 '14 at 12:01
  • $\begingroup$ See my edit which explains that $\endgroup$ – KvdLingen May 11 '14 at 12:03
  • $\begingroup$ Why bob moves perpendicular?I am just saying is anything here related to kinetic energy that we give to initiate the motion. $\endgroup$ – Vishvajeet Patil May 11 '14 at 12:05

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