Calculating Tension in Rotational Motion: Stuntman on a Vertical Circle

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Homework Help Overview

The discussion revolves around calculating the tension in a rope during the rotational motion of a stuntman swinging in a vertical circle. The problem involves analyzing forces at different points in the motion, specifically at the start and at the bottom of the swing.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the tension in the rope at the beginning of the motion, questioning whether it is zero due to the initial velocity being zero. They also discuss the forces acting on the stuntman at the bottom of the arc, particularly the relationship between tension, gravitational force, and centripetal acceleration.

Discussion Status

Some participants have provided guidance on the conservation of mechanical energy, suggesting that total mechanical energy is conserved as the stuntman swings. There is an ongoing exploration of how to calculate the speed at the bottom of the swing and how it relates to the forces acting on the stuntman.

Contextual Notes

Participants are working under the assumption that the stuntman starts from rest and are considering the effects of gravitational potential energy and kinetic energy in their calculations. There is a focus on understanding the dynamics of the motion without reaching a final consensus on the calculations presented.

mizzy
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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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