How Is Tension Calculated in a Pendulum String at 45 Degrees?

[carney]

The mass of the ball is m, as given below in kg. It is released from rest. What is the tension in the string (in N) when the ball has fallen through 45o as shown.

Hint: First find the velocity in terms of L and then apply Newton's 2nd law in normal and tangential directions. If you do it correctly, L should disappear from your equation.


m[kg] = 1.58;

 

[thrill3rnit3]

Where's your work?

 

[topgun08]

The below information may not be correct...

Perhaps start with drawing a Free Body Diagram at the degree angle. Their is T tension, the force downward mg, and the force balancing tension, mgsin(theta).

Summing the forces to find the centripetal Force one gets:
Centripetal Force = T - mgsin(theta)
---Apllying Newton's 2nd Law ---
ma = T - mgsin(theta)

in centripetal motion a = (v^2)/r where r is L.

Thus
mv^2/L = T - mgsin(theta)
T = mv^2/L + mgsin(theta)

The only issue is that we have two variables (T and L) and only one equation. So we need another equation. We turn to using Energy.

Original PE = New PE + KE
mgL = mg(.5L) + .5mv^2
L = v^2/g

plugging back into original equation one solves for T

I think this is right, but maybe wait for a more advanced member to comment.

 

[sfsy1]

Original PE = New PE + KE
mgL = mg(.5L) + .5mv^2
L = v^2/g

The change in GPE is (sin45)Lgm, not 0.5Lgm.