Find speed with tension, mass given. thanks

[MissPenguins]

Homework Statement

009 (part 1 of 2) 10.0 points
A 33 kg child sits in a swing supported by two chains, each 3.7 m long. If the tension in each chain at the lowest point is 259 N, find the child’s speed at the lowest point. (Neglect the mass of the seat.) Answer in unit m/s.
b. Find the force of the seat on the child at the
lowest point. Answer in unit N

Homework Equations

F = mv²/r

The Attempt at a Solution

Rearrange equation, and I got sqrt(Fr/m) = sqrt(259N * 3.7m)/33 kg) = 5.3888...
I got it wrong. What did I do wrong? Thanks.

 

[dr_k]

Homework Statement

009 (part 1 of 2) 10.0 points
A 33 kg child sits in a swing supported by two chains, each 3.7 m long. If the tension in each chain at the lowest point is 259 N, find the child’s speed at the lowest point. (Neglect the mass of the seat.) Answer in unit m/s.
b. Find the force of the seat on the child at the
lowest point. Answer in unit N

Homework Equations

F = mv²/r

The Attempt at a Solution

Rearrange equation, and I got sqrt(Fr/m) = sqrt(259N * 3.7m)/33 kg) = 5.3888...
I got it wrong. What did I do wrong? Thanks.

Good Evening Miss Penguins,

You're correct that this is a circular motion problem, but you neglected some of the forces. First you need to draw a picture, and determine the free body diagram for the isolated child.

The isolated child experiences a gravitational attraction, or weight, and a contact force from the seat, the normal force N. Here's a picture to get you started:

http://img109.imageshack.us/img109/6937/swingw.jpg

Hint: The normal force N has a magnitude of 2 times the tension, since the seat is massless.

 

[Delphi51]

I don't know if this is the same as already suggested, but my thought is that the forces add up to the centripetal force:
Fc = 2T - mg

 

[MissPenguins]

Good Evening Miss Penguins,

You're correct that this is a circular motion problem, but you neglected some of the forces. First you need to draw a picture, and determine the free body diagram for the isolated child.

The isolated child experiences a gravitational attraction, or weight, and a contact force from the seat, the normal force N. Here's a picture to get you started:

http://img109.imageshack.us/img109/6937/swingw.jpg

Hint: The normal force N has a magnitude of 2 times the tension, since the seat is massless.

so I used the wrong equation for it? Thanks.

 

[MissPenguins]

I don't know if this is the same as already suggested, but my thought is that the forces add up to the centripetal force:
Fc = 2T - mg

I don't know, but for the first answer, they are asking for speed though, and the equation u suggested doesn't have speed.

 

[Delphi51]

You will of course replace Fc with mv²/R as you did in the first go.
The difference is that mg is involved.

 

[MissPenguins]

You will of course replace Fc with mv²/R as you did in the first go.
The difference is that mg is involved.

I rearranged the equation and ended up with v = sqrt [(2T-mg)r]/m = sqrt [(2(259 N) - (33kg)(9.8m/s²))(3.7)] / (33 kg) = 21.818787...
Does that look reasonable?

 

[dr_k]

I rearranged the equation and ended up with v = sqrt [(2T-mg)r]/m = sqrt [(2(259 N) - (33kg)(9.8m/s²))(3.7)] / (33 kg) = 21.818787...
Does that look reasonable?

That m should be inside the square root, I believe.

 

[MissPenguins]

That m should be inside the square root, I believe.

Yeah it is inside. I submitted the answer, and it is wrong. :(

 

[dr_k]

Yeah it is inside. I submitted the answer, and it is wrong. :(

It doesn't look that way, according to the equation you wrote.

Can you prove that the magnitude of the Normal force is equal to twice the tension?

 

[MissPenguins]

It doesn't look that way, according to the equation you wrote.

Can you prove that the magnitude of the Normal force is equal to twice the tension?

What do you mean by it doesn't look that way? lol...well, I put it in the calculator that way. So v = sqrt of [((2T-mg)r) / m] = sqrt of [((2(259 N)-(33kg)(9.8m/s²)(3.7m) / 33 kg] = 21.8187878787 WRONG. :(

 

[MissPenguins]

What do you mean by it doesn't look that way? lol...well, I put it in the calculator that way. So v = sqrt of [((2T-mg)r) / m] = sqrt of [((2(259 N)-(33kg)(9.8m/s²)(3.7m) / 33 kg] = 21.8187878787 WRONG. :(

Alright, I was able to get a right thanks to a friend. I use ((259*2) - (33)(9.8)) = 194.6, and then v = sqrt of (Fr/m) => sqrt of ((194.6)(3.7)/33 = 4.671058539.



Then in part 2, it says find the force of the seat on the child at the lowest point. I have no clue how to do it. :( Please help, thanks.

 

[Delphi51]

Two ways to look at part 2.
First, there is no mass other than the child so the tension forces are entirely applied to the child via the seat.
Second, apply "sum of Forces = ma" to the child. The only forces are gravity and the seat force.

 

[MissPenguins]

Two ways to look at part 2.
First, there is no mass other than the child so the tension forces are entirely applied to the child via the seat.
Second, apply "sum of Forces = ma" to the child. The only forces are gravity and the seat force.

Nvm, I got it, thanks a lot anyway. :)