Resolving Vector W: Tips & Tricks for Uniform Circular Motion

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A particle in uniform circular motion requires centripetal force, which is expressed as T2 - mgcos(theta). The main challenge discussed is resolving vector W, particularly in identifying the angle between W and its components. To aid in this, it is suggested to extend W to create a right triangle with the horizontal axis, where T1 serves as the hypotenuse. The conversation emphasizes the need for a proper coordinate system, with axes aligned to T1 and perpendicular to it, to accurately find the components of W. Understanding these components is crucial for balancing tension and managing the object's velocity in 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?
 
ucmpro1-1.jpg
 
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.
 
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