# 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

Homework Helper
Gold Member

## 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.