Resolving Vector W: Tips & Tricks for Uniform Circular Motion

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1. A particle is performing uniform circular motion
ucm1.jpg


the necessary centripetal force is provided by T2 -mgcos(theta) ****** no problem :)

2 Now the problem
ucm2.jpg


how to resolve vector W in this case.

i'm confused :( how to resolve the vector. (confused in picking the anlge between the vector and the component)

pls help me by providing some hints pls..
 
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Extend the line along W until it intersects the horizontal axis. You have just created a right triangle. One right side is W, one right side is the segment of the horizontal axis and the hypotenuse is T1. Can you find the component of W along T1?
 
kuruman said:
Extend the line along W until it intersects the horizontal axis. You have just created a right triangle. One right side is W, one right side is the segment of the horizontal axis and the hypotenuse is T1. Can you find the component of W along T1?

my aim is to find a component of W which balances tension (T1) of the string and another component which tries to decrease the velocity of the object in circular motion in first quadrant.
 
I understand your aim. As you know, to describe the components of a vector, you need two perpendicular axes. What do you think these axes should be in this case?
 
ucmpro.jpg


or

ucmpro1.jpg


but in both cases the angle between the vector w and its x componet is not congurent to (theta) {i think so}
 
sphyics said:
... but in both cases the angle between the vector w and its x componet is not congurent to (theta) {i think so}

Your diagrams are fine, but you did not tell which way your axes point. In what direction is x?
 
In that case, the weight has zero x-component and -mg for the y-component. What would happen if you chose the x-axis along T1 and the y-axis perpendicular to it, up and to the left? Note that this is a very convenient coordinate system because the centripetal acceleration and tension are along x. Of course you will have to find the x and y components for the weight in the system of axes.