Man swing from on a rope (velocity)

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Homework Help Overview

The problem involves a man swinging from a cliff using a rope, where he releases the rope at a specific angle and travels through the air to land on the ground. The context includes concepts from mechanics and energy conservation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of velocity at the release point, the relevance of mass in the equations, and the geometry involved in determining angles. There are attempts to apply energy conservation principles and questions about the setup of the problem.

Discussion Status

The discussion is active, with participants exploring different equations and concepts related to the problem. Some guidance has been offered regarding the use of energy conservation and the relationship between angles in the scenario, but no consensus has been reached on the final approach or calculations.

Contextual Notes

Participants are considering the implications of the rope's suspension and whether it exerts a force on the man, which could affect energy calculations. There is also a focus on the angles involved in the swing and their geometric relationships.

AdsBerger
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Homework Statement


A man standing on a 15 m high cliff tries to swing on a 12 m rope and let's go at 30° of the horizontal (travelling 150°). He let's go and travels through the air, to then land on the ground.

What is his distance from his landing spot to the base of the cliff?


Homework Equations





The Attempt at a Solution



I am able to figure out the equation to find the height from the ground of which he let's go.
Not able to figure out the velocity when he let's go.
There is no mass for him, so those equations are irrelevant. If he is starting at a horizontal angle, then I am unsure of what equations to use to find his velocity.
 

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The man has mass m. The actual value of m is irrelevant, you will see. What equations you are considering irrelevant?

The angle of the velocity is not 30°with respect to the horizontal. The man swings along a circle and the velocity is tangent to the circle, so perpendicular to the radius.ehild
 
Last edited by a moderator:
I am thinking i will have to use the equation v = √{2gL[1-cos(a)]}
v = √{2(9.8)(12) [1-cos(a)]}
What would the angle be though?
 
To find the velocity ##V## one may use the energy conservation law:
$$
m \frac {v^2} {2} = m g Δh.
$$

So mass ##m## will cancel and the difference of heights ##Δh## you may find out using geometry laws.
 
Consider:
1. can you use energy conservation (k.e. + p.e. = constant) or does the rope exert a force on the man to add to or subtract from his total energy?

2. Once you know v at the release point you are all set. Right?
 
AdsBerger said:
I am thinking i will have to use the equation v = √{2gL[1-cos(a)]}
v = √{2(9.8)(12) [1-cos(a)]}
What would the angle be though?

Where is that equation from?

As for the angle of velocity look at the picture. The velocity is normal to the radius. What is the angle beta?

ehild
 

Attachments

  • manrope.JPG
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The angle beta would be 50. i calculated his velocity to be 10.85 m/s
 
AdsBerger said:
The angle beta would be 50. i calculated his velocity to be 10.85 m/s

Why 50? ? It is a right-angled triangle (the yellow one) and one angle is 30°, beta is the other angle...

ehild
 

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