Child on swing (rotational motion, forces) question

[anewera]

Homework Statement

A high school buddy exerts a horizontal force on a swing that is suspended by a rope that is 5.0 meters long, holding at an angle of 30 degrees with the vertical. The child in the swing has a mass of 30kg and dimensions that are negligible compared to the length of the swing. The mass of the rope and the seat are negligible.
a) What are the forces acting on the child?
b) Calculate the tension in the rope, in Newtons.
c) Calculate the horizontal force exerted by the high school buddy in N

The buddy now releases the swing from rest
d) Calculate the tension in the rope just after the release (the swing is instantaneously at rest) in N.
e) Calculate the tension in the rope as the swing passes through its lowest point in N.

Homework Equations

T = (m * v^2 / r) + mg
F = ma

The Attempt at a Solution

a) Well there's no friction and since it's stationary, I don't think there should be a centripetal force but there's Weight...and a Normal force?

b)I was thinking T = mg cos(theta) but then I realized it's not just the rope that is supporting all the weight of the child. The buddy is keeping the swing from moving forward...so that throws me off.

c) I really don't know how to do this, mg would only make sense if there was just a downward vector so now I'm confused.

d) So now that the child isn't holding it I'm guessing the tension is mg cos (theta).

e) T = (m * v^2 / r) + mg, plug in and solve should work?

Thanks in advance...more than half the other class failed the test for rotational motion and I really want to know how to do these harder questions...

 

[tiny-tim]

Welcome to PF!

Hi anewera ! Welcome to PF!

(try using the X² tag just above the Reply box )

a) Well there's no friction and since it's stationary, I don't think there should be a centripetal force but there's Weight...and a Normal force?

This is badly-worded question.

Yes, on the wording given, your answer is correct … the only two forces on the child are its weight, and the reaction force from the seat.

But I suspect your teacher wants answers as if the question was about the forces on the seat.

b)I was thinking T = mg cos(theta) but then I realized it's not just the rope that is supporting all the weight of the child. The buddy is keeping the swing from moving forward...so that throws me off.
c) I really don't know how to do this, mg would only make sense if there was just a downward vector so now I'm confused.

Either do a vector triangle, or do components in a suitable direction.

d) So now that the child isn't holding it I'm guessing the tension is mg cos (theta).

Do F = ma, but this time a is not zero, is it?

 

[anewera]

Hi, thanks for replying(, and just in time too, it's almost midnight over here).

AP Exams apparently usually have some questions with bad wording (that's where my teacher got this question). "Why isn't this not true?" haha.

Well I pretty much solved for the horizontal force and then the tension with the little nudge you gave me in the right direction.

I thought about part e and use conservation of energy, and then used the tension formula for rotational motion and it turned out to be really easy.

For part d though, you said to use F=ma but since the swing has just been released and instantaneously is at rest, when they ask for the tension, shouldn't it just be mgcos(theta)?...since the rope is all that's supporting the child now...

 

[tiny-tim]

Hi anewera!

(have a theta: θ )

For part d though, you said to use F=ma but since the swing has just been released and instantaneously is at rest, when they ask for the tension, shouldn't it just be mgcos(theta)?...since the rope is all that's supporting the child now...

Yes, i wasn't thinking it through to the end …

a is non-zero, but it's entirely tangential, so of course it won't affect the tension.

Yes, with nobody pushing, the radial acceleration is zero (because the angular velocity is still zero), and so the radial components add to zero, giving T = mgcosθ.