Tension in cord at 30.0 degrees

[Kennedy111]

Homework Statement

A 0.160 kg ball attached to a light cord is swung in a vertical circle with a radius of 70.0 cm. At the top of the swing, the speed of the ball is 3.26 m/s. The center of the circle is 1.50 m above the floor.

Determine the magnitude of the tension in the cord when the cord is 30.0 degrees below the horizontal.

m= 0.160 kg
r = 70.0 cm or 0.700 m
v = 3.26 m/s
v (30.0 degrees) = 5.59 m/s
d = 1.50 m

Homework Equations

T = mv^2/r + mgsin(degree)

The Attempt at a Solution

T = mv^2/r = mgsin(degree)
= (0.160 kg)(5.59m/s)^2 / 0.700 m + (0.160 kg)(9.81 m/s^2)(Sin30)
= 7.922765714


May need to double check the velocity I used here...

Thank you

 

[TSny]

The Attempt at a Solution

T = mv^2/r = mgsin(degree)
= (0.160 kg)(5.59m/s)^2 / 0.700 m + (0.160 kg)(9.81 m/s^2)(Sin30)
= 7.922765714


May need to double check the velocity I used here...

Thank you

Look's ok to me. How many significant figures should you retain in the answer?

 

[azizlwl]

To me it is conservation of energy that determine the speed at other point compare to top of the loop.