# Circular Motion

A kid swings from a rope 8.5 m long on a stationary pole on a playground. His body mass is 70 kg and he is moving 6.0 m/s when he passes through the lowest part of his path.

What is the tension in the rope at the lowest part of his path?

When he reaches the highest point, what angle does the rope make with the vertical?

https://brainmass.com/physics/circular-motion/circular-motion-and-tension-in-a-rope-38319

## SOLUTION This solution is **FREE** courtesy of BrainMass!

At the lowest part the forces acting are

Gravitational force downwards

Tension in the rope upwards

The net force = Tension - mg provides the centripetal force = m v^2/r for the kid to move in a circular path

m= 70 kg

g= 9.81 m/s^2

v= 6.0 m/s

r= 8.5 m

T-mg = mv^2/r

or

T=mg + mv^2/r = 983 N =70x9.81+70x6^2/8.5

tension in the rope at the lowest part of his path= 983 N

To calculate the maximum height (h)to which the kid reaches

we use the equation v^2= 2 gh

v= 6.0 m/s

g= 9.81 m/s^2

Therefore h= 1.83 m =6^2/ (2 x9.81)

Cos theta = (r-h)/r (theta is the angle with the vertical)

r= 8.5 m

h= 1.83 m

Substituting the values:

Cos theta = (8.5-1.83)/8.5= 0.7847

or theta = 38 degrees

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