[Physics] Tension in string and gravitational force on bob relative strength

forcesfree-body-diagramnewtonian-gravitynewtonian-mechanicsstring

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?

Best Answer

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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