Man swing from on a rope (velocity)

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SUMMARY

The discussion centers on calculating the landing distance of a man swinging from a 15 m high cliff using a 12 m rope. The man releases the rope at a 30° angle to the horizontal, resulting in a total travel angle of 150°. The key equations utilized include the velocity formula v = √{2gL[1-cos(a)]} and the conservation of energy principle, leading to a calculated release velocity of 10.85 m/s at an angle of 50°. The analysis emphasizes the importance of understanding the geometry of the swing and the forces involved.

PREREQUISITES
  • Understanding of basic physics concepts such as energy conservation and projectile motion.
  • Familiarity with trigonometric functions and geometry related to right-angled triangles.
  • Knowledge of gravitational acceleration (g = 9.8 m/s²) and its application in motion equations.
  • Ability to manipulate and solve equations involving angular measurements and velocities.
NEXT STEPS
  • Explore the principles of projectile motion to understand the trajectory of the man after release.
  • Study the conservation of mechanical energy in swinging systems to deepen understanding of energy transformations.
  • Learn about the dynamics of circular motion and how tangential velocity is derived from angular motion.
  • Investigate the effects of different release angles on the landing distance in similar scenarios.
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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
 

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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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