- #1

- 95

- 2

## Homework Statement

A stuntman whose mass is 70 kg swings from the end of a 4.0-m-long rope along the arc of a vertical circle. Assuming he starts from rest when the rope is horizontal, find the tensions on the rope that are required to make him follow his circular path at a height of 1.5 m above the bottom of the circular arc.

## Homework Equations

## a_c = mv^2 / r ##

## F = ma ##

## The Attempt at a Solution

From conservation of energy, ## 4mg = 1/2 m v^2 + 1.5mg \Rightarrow v = 7 m/s ## is the velocity at that point. To calculate the angle the center and current position makes at the horizontal, we notise this is just a right triangle with hypotenuse ## 4 ## and opposite leg ## 2.5 ##. It's easy to see that the angle is about ## 38.68 ^{\circ} ##. Let the tension from the point to the center be T. IT's easy to see that the y-component of the centripetal acceleration is the y component of the tension subtracting mg. Therefore, ## \frac{70 \cdot 7^2}{4} \sin 38.68 = T \sin 38.68 - 70 \cdot 9.8 \Rightarrow T = 1955 N##. But it's a contradictory to other link here: https://www.physicsforums.com/threads/a-rotational-motion-problem.147022/

Anyone can shine light on this?

Thanks,

minimario :)