- #1

barryj

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## Homework Statement

What is the tension of the string at these places.

See figure. Vertical circular motion. m = 1kg, assume g = 10 m/s^2, velocity = 10 m/sec

a) what is the tension in the string when the mass is 30 deg below the horizon.

b)what is the tension at the top of the loop

c) what is the tension at the bottom of the loop[/B]

## Homework Equations

centripetal force F = mv^2/r = 1(10)^2/10 = 10N

weight = mg = (1)(10) = 10 N[/B]

## The Attempt at a Solution

I broke up the weight into two parts, one along the direction of the string, the other tangential to the circle. The component of the weight along the direction of the string is mgcos(60 deg) = 1(10)(.866) = 8.66N. So I figure the tension in the string must be the sum of the centripital force plus the component of the weight in the direction of the string or T = 10 + 8.66 = 18.66.

The tension at the top is 0 N and at the bottom 20 N.[/B]