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March 29, 2024, 08:54:05 am

Author Topic: Physics Question - Acrobat Swinging On Rope  (Read 1263 times)  Share 

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Jefferson

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Physics Question - Acrobat Swinging On Rope
« on: January 21, 2020, 01:05:47 pm »
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Hi everyone,
Please help me with the following question:

An 83 kg acrobat plans to swing from one side of the circus arena to the other from a 21 m rope. If the maximum the performer's arm can exert on the rope is 1450 N downwards, what is the maximum speed in km/h that can be attained throughout the stunt without losing grip?

Answers were not provided :(.
Thank you so much.
« Last Edit: January 21, 2020, 01:07:25 pm by Jefferson »

Coolmate

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Re: Physics Question - Acrobat Swinging On Rope
« Reply #1 on: January 21, 2020, 03:08:04 pm »
+5
Hi everyone,
Please help me with the following question:

An 83 kg acrobat plans to swing from one side of the circus arena to the other from a 21 m rope. If the maximum the performer's arm can exert on the rope is 1450 N downwards, what is the maximum speed in km/h that can be attained throughout the stunt without losing grip?

Answers were not provided :(.
Thank you so much.

Hey Jefferson! ;D

I did the question this way and I am not sure if it is the right way of doing the question though, but I think this is how you are meant to do the question:

$${F_c}={\frac{m{v^2}}{r}}$$ $${v^2}={\frac{{F_c}r}{m}}$$ $${v^2}={\frac{1450 * 21}{83}}$$ $${v^2}={366.8674699 \div 3.6}$$ $$\sqrt{v^2}=\sqrt{101.9076305}$$ $$v={36km/h}$$

I hope this helps! ;D
Coolmate 8)
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Jefferson

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Re: Physics Question - Acrobat Swinging On Rope
« Reply #2 on: January 21, 2020, 05:05:38 pm »
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Hi Coolmate,
Thanks for replying.

How did you know that this is an application of centripetal force? (doesn't look like uniform circular motion. From what I'm visualising in this scenario, he would speed up on the descent and slow down while ascending due to gravity).
Also, is Fc really 1450N? (they say it's a downward force). Does gravity/tension not come into play?

Thanks again.

Bri MT

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Re: Physics Question - Acrobat Swinging On Rope
« Reply #3 on: January 21, 2020, 06:15:29 pm »
+5
Not Coolmate but here's some explanations:

You are absolutely right that gravity and tension are important in this situation & that the acrobat would be speeding up as they go downwards before slowing down as the rope swings up. What you are missing here is understanding of centripetal motion.

As the acrobat swings across they trace out part of a circle. Gravity pushes the acrobat down while tension pulls the acrobat towards the center of the circle. This results in the acrobat having the greatest speed when they are at the bottom of the circle they are tracing out as they swing across (at that point they've done all the going down they're going to do and are just about to go up).

So now we know that the point we are interested in is the bottom of the arc. Since this is part of a circular motion we can use the relationship between centripetal force and speed which Coolmate has supplied.

First,  we need to find the centripetal force.  Remember that this is the net force the acrobat is experiencing and is directed to the centre of the circle.

The acrobat's arm can pull down on the rope with a force of 1450 N. This means that the rope can pull up on the acrobat's arm with a force of 1450 N (note: remember Newton's 3rd law). We are interested in the forces acting on the acrobat at max speed (tension & gravity) and this is how we get the centripetal force. It's important you understand centripetal force isn't unrelated to these; centripetal force is always due to other forces.

An important note with centripetal motion questions is to be aware of what direction forces are acting in. E.g. at the bottom of a circle weight is subtracted to find centripetal force whereas at the top of a circle weight is added.

Hope this helps!


Made some slight edits, please feel free to reply with working or further questions :)
« Last Edit: January 21, 2020, 06:40:43 pm by Bri MT »

Jefferson

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Re: Physics Question - Acrobat Swinging On Rope
« Reply #4 on: January 21, 2020, 07:50:07 pm »
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Not Coolmate but here's some explanations:

You are absolutely right that gravity and tension are important in this situation & that the acrobat would be speeding up as they go downwards before slowing down as the rope swings up. What you are missing here is understanding of centripetal motion.

As the acrobat swings across they trace out part of a circle. Gravity pushes the acrobat down while tension pulls the acrobat towards the center of the circle. This results in the acrobat having the greatest speed when they are at the bottom of the circle they are tracing out as they swing across (at that point they've done all the going down they're going to do and are just about to go up).

So now we know that the point we are interested in is the bottom of the arc. Since this is part of a circular motion we can use the relationship between centripetal force and speed which Coolmate has supplied.

First,  we need to find the centripetal force.  Remember that this is the net force the acrobat is experiencing and is directed to the centre of the circle.

The acrobat's arm can pull down on the rope with a force of 1450 N. This means that the rope can pull up on the acrobat's arm with a force of 1450 N (note: remember Newton's 3rd law). We are interested in the forces acting on the acrobat at max speed (tension & gravity) and this is how we get the centripetal force. It's important you understand centripetal force isn't unrelated to these; centripetal force is always due to other forces.

An important note with centripetal motion questions is to be aware of what direction forces are acting in. E.g. at the bottom of a circle weight is subtracted to find centripetal force whereas at the top of a circle weight is added.

Hope this helps!


Made some slight edits, please feel free to reply with working or further questions :)

Hi Bri MT,
Thanks for the reply and new edit,

Is the net force (centripetal force) at the bottom of the movement given by
Fc = 1450 N - (83kg * 9.8 ms-2)
                      = 636.6 N ?

I find this a little peculiar, as the performer's arm exerts a downward force (same as gravity).
So instead, should the centripetal force be equal to the tension force, i.e.
Fc = 1450 + (83kg * 9.8 ms-2)
                      = 2263.4 N ?

Both scenarios seem odd to me. Could you please clarify further?
Thank you!

Bri MT

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Re: Physics Question - Acrobat Swinging On Rope
« Reply #5 on: January 21, 2020, 08:35:24 pm »
+1
No worries!

It's the first option:

Centripetal force = tension - mg

Or in other words, tension = centripetal force + mg

What's happening here is that their arm is inded pulling down on the rope due to gravity & this contributes to the tension in the rope. This is already part of the 1450 N. Option 1 balances out the component of the upwards tension that's opposing their weight with their weight.

If this is confusing, consider the case where they are at rest hanging from the rope. You would not be able to get F=0 by adding T and mg together; one of the forces needs to be negative