Circular Motion: Net Force, Speed & Direction at Top - M,L,g

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SUMMARY

A mass M attached to a string of length L moves in a vertical circular path. At the top of the loop, the tension in the string is twice the weight of the mass, leading to a net force equation of F_net = T + mg = mv²/L. The magnitude of the net force is calculated as F_net = 3mg, and the speed of the mass at the top is determined to be v = √(3gL). The direction of the net force is downward, towards the center of the circular path.

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A mass M attached to a string of length L moves in a circle in a vertical plane.AT the top of the circular path, the tension in the string is twice the weight of the ball.
--find magnitude and direct of Net force and speed of the ball at the top.Express aswer in term of M,L,g

net force= T-mg = mv^2/L
2mg-mg = mv^2L
F=Square root gL

F=mv^2/L
Square root gL=mv^2L
Square root of( (L*square root Lg)/m )
 
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They want the magnitude and direction of the net force in terms in terms of m, g and L... so v shouldn't be in your answer...

When the ball is at the top of the loop T and mg are in the same direction:

Fnet = T + mg = mv^2/L

so what is Fnet in terms of m and g... what is the direction of Fnet. What is v in terms of m, g and L...
 
Fnet = T + mg
2mg+mg=3mg , the direction for Fnet is toward the center? not sure

3mg=mv^2/L
L3mg=mv^2 m cancel
v= Square root of 3Lg, the direction for velocity is horizontal and tangentially to the circle, to the right
 
yes, v=sqrt(3Lg) looks right to me. Yes, direction of Fnet is towards the center... in other words downwards since the mass is at the top of the circular path.
 

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