let a mass be suspended from a string (fixed at one end to the ceiling)........?

Question

the lower end of the string is made to describe a circular path while the string makes an angle θ with the vertical. The body undergoes uniform circular motion. At any point of time, the tension(T) is opposite to the cos component of mg, such that we can write--- T=mg cosθ
But it is seen that the cos component of T is opposite to mg, such that--- T cosθ = mg
But how is that possible??? The use of these two equations in a related problem yields two different answers.
Which one is correct and why????

Answer 1

dear, the problem arises due to the co-ordinate axis chosen by you. axes should be chosen so force or its resolved component lies along dieection of motion and in perpendicular direction. we will choose axis along vertical and horizontal direction. so Tcosθ = mg and Tsinθ provides necessary centripetal force.
thanks.

Answer 2

The mass is not accelerating in the y direction. That means that the weight, mg must be equal to the vertical component of the tension, Tcos(theta).

So, mg = T cos(theta) is true.

You also tried to resolve the weight into components that are parallel and perpendicular to the Tension in the string. This is less useful and if the tension were truly equal to the component of the weight that is parallel to the tension, then there would be no acceleration toward the center of the circle.

Tsin(theta) = mv^2/R , is how the tension and centripetal acceleration are related. The two equations in this answer can be used to describe the scenario.

Answer 3

Short version: the tension in the string and the weight does not balance; they are not even in the same direction. Weight is vertical, while tension is tilted. They must be added vectorially to result in the centripetal force of the circular motion.

Long version:
Make a drawing of the body attached to the string at a certain moment in time. The tension in the string can act on the body only one way - towards the suspension point. The gravitation acts straight down. And there is one more force to consider: the centripetal force for the circular motion; this can act only in one direction - horizontal, towards the center. You forget this when you balance the forces along the string. More detailed explanation below.

There are two natural directions for the coordinate system: with one axis vertical (along gravity) and along the string.

The vertical coordinate system appears more natural, since we have two forces along the axes. Now look at the balance of forces in the two relevant directions for the vertical system:
1. Horizontal: The horizontal component of the tension is not balanced by anything, and gives an accelerated motion. This is actually the centripetal force of the circular motion, thus
Fc = T sin(theta)
2. Vertical: There is no change in the vertical position of the body, so weight mg is balanced by the vertical component of T, so you have T cos(theta) = mg, so T = mg /cos(theta)
If you solve these two equations for the centripetal force you get
Fc = mg tan(theta)


In the system of coordinates with one axis along the string, you must take into account the centripetal force as resultant:
1. Perpendicular to the string:
mg sin(theta) = Fc cos(theta)
Note that this gives directly the centripetal force as above.
2. Along the string:
T - mg cos(theta) = Fc sin(theta)
so T is NOT balanced by mg cos(theta). But you can replace Fc from the first equation and solve for T. You'll have the nice surprise to obtain the same result for T as in the other system of coordinates.

Answer 4

The symmetry of the situation is broken by the requirement that there be a force to balance the centripetal acceleration. So the vertical component (cos theta) of the tension balances gravity, and the horizontal component (sin theta) balances the acceleration a = v^2/r which is required to maintain the circular motion.

Answer 5

Dear Angshuman ,
Your problem is quit obvious , becouse you have not made the free body diagram properly . I also used to do the same thing . but then my tutor (Mr. Salil Vikram ) cleared my concepts about the topic .
You will find that the cos component of tension is completely cancelled by mg but the tension is not completely cancelled by cos component of mg . Rather it is cancelled by [mg*cos0 + (mv^2/r)*sin0].
Try this equation and you will get the right answer.
-The Abhinav

Answer 6

T = mg cosine theta because tension is always a vector product and it depends on the scalar m and vectors g and cosine theta.

Answer 7

note that "0-" refers to theta

mg & T are opposite to each other in directions.
so they cannot be equal vectorially.
please check.