Calculating Tension in Rotational Motion: Stuntman on a Vertical Circle

AI Thread Summary
The discussion focuses on calculating the tension in a rope as a stuntman swings in a vertical circle. At the start of his motion, when the rope is horizontal, the tension is zero since he begins from rest, resulting in no centripetal force. At the bottom of the arc, the tension must counteract both the gravitational force and provide the necessary centripetal force, leading to the equation F = mv^2/r + mg. The speed at the bottom is determined using conservation of mechanical energy, resulting in a speed of approximately 8.9 m/s. The final tension in the rope is calculated to be about 2.07 x 10^3 N.
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Please help...rotational motion

Homework Statement


A stuntman swings from the end of 4 m long rope along the arc of a vertical circle. If his mass is 70kg, find the tension in the rope required to make him follow his circular path, assuming he starts from rest when the rope is horizontal:
a) at the beginning of his motion

b) at the bottom of his arc


Homework Equations



centripetal acceleration = v^2/r

F = mv^2/r

The Attempt at a Solution


For part a) at the beginning of his motion, is the tension equal to zero? The tension of the rope is pointing to the center and therefore the equation is F = mv^2/r. However, v is equal to 0 because he starts from rest. is that right?

For part b) at the bottom of the arc, tension in the rope and mg causes acceleration and therefore the equation is F - mg = mv^2/r. But what's v?

Can someone help me please?
 
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mizzy said:
For part a) at the beginning of his motion, is the tension equal to zero? The tension of the rope is pointing to the center and therefore the equation is F = mv^2/r. However, v is equal to 0 because he starts from rest. is that right?
Right.

For part b) at the bottom of the arc, tension in the rope and mg causes acceleration and therefore the equation is F - mg = mv^2/r. But what's v?
You'll need to figure out the speed at the bottom. (What's conserved?)
 


Doc Al said:
Right.


You'll need to figure out the speed at the bottom. (What's conserved?)

Kinetic energy
 


mizzy said:
Kinetic energy
Total mechanical energy is conserved as man swings, not just kinetic.
 


Doc Al said:
Total mechanical energy is conserved as man swings, not just kinetic.

Ok. In that case:

KEmiddle + PEmiddle = KEbottom +PEbottom
1/2mv^2 +mgh = 1/2mv^2 +mgh
0 + mgr = 1/2mv^2 + 0
mgr = 1/2mv^2
(70)(9.8)(4) = 1/2(70)v^2

therefore, v = square root off 2gr = 8.9m/s

F1 - mg = mv^2/r
F1 = mv^2/r +mg
= (70)(8.9)^2/4 + (70)(9.8)
= 2.07 x 10^3 N

Can you confirm if this is right? THANKS
 


Looks good! (Don't round off until the end.)
 


Thanks for your help! =)
 

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