Grade 12 Circular Motion Question

[mrcheeses]

Homework Statement

http://i829.photobucket.com/albums/zz219/fuzzy1233/p.png

Homework Equations

Fnet=ma=mv^2/r
Fg=mg

The Attempt at a Solution

Since sides length 3,4 and 5m are given and is in a right angle, I can solve for the angle between 5 and 3 (53.1 degrees) and between 5 and 4 (37 degrees). Then with the angle and a side (3 or 4), I solve for the x component, r, which is found to be 2.41m. Since the 4m rope is given to be 22N, I use trig and solve for the x component of force and found it to be 13.23 N. I then use Fnet=mv^2/r

Fx?=mv^2/r
13.23N=(1.3kg)v^2/2.41
v= 4.9 m/s

edit: I got 5.8 m/s this time and I believe that could be correct. For the total Fx, I added the two x components of the tensions that I solved using trig.
I am not sure if this is right and need verification. Thanks!

 

[MrWarlock616]

Hey how did you get the x-component of force??...I need help

 

[mrcheeses]

Umm I just did it again and got 5.8 m/s.

I split it into the 2 triangles like I did to find the radius, and solved for the two x components of the two tensions and added them together to get the total x component force.

 

[MrWarlock616]

Wait..so the 3 m length is also a string?

Let's see if I have this right...the angle between 5 and 3 would be arcsin(0.8)=53.1° ...and the angle between 5 and 4 would be arcsin(0.6)=36.86°...
So the radius would be 3*sin(53.1°)=0.799 m... and then what.. ? The tension in the 4 m string is equal to mgcos(53.1°) ?? Am I right?

 

[cepheid]

Homework Statement

http://i829.photobucket.com/albums/zz219/fuzzy1233/p.png

Homework Equations

Fnet=ma=mv^2/r
Fg=mg

The Attempt at a Solution

Since sides length 3,4 and 5m are given and is in a right angle, I can solve for the angle between 5 and 3 (53.1 degrees) and between 5 and 4 (37 degrees). Then with the angle and a side (3 or 4), I solve for the x component, r, which is found to be 2.41m. Since the 4m rope is given to be 22N, I use trig and solve for the x component of force and found it to be 13.23 N. I then use Fnet=mv^2/r

Fx?=mv^2/r
13.23N=(1.3kg)v^2/2.41
v= 4.9 m/s

edit: I got 5.8 m/s this time and I believe that could be correct. For the total Fx, I added the two x components of the tensions that I solved using trig.
I am not sure if this is right and need verification. Thanks!

You can save yourself a whole bunch of unnecessary calculator work in this problem. For this problem, you never have to explicitly calculate the angle in degrees. Consider the angle θ between the 4 m rope and the vertical.

From the 3-4-5 triangle it's clear that sinθ = 3/5 = 0.6.

Now if you break up this triangle into two smaller right triangles by drawing the radius of the circle in, then you get that the radius of the circle is given by

r = (4 m)sinθ = (4 m)*0.6 = 2.4 m

*Exactly 2.4 m*, not 2.41 m, which you were just getting due to calculator rounding error. No calculator required here. Not much point in computing the angle using arcsin and then just taking the sine of it again.

Similarly, the horizontal component of the tension is given by (22 N)sinθ = (22 N)*0.6 = 13.2 N

You are certainly correct that you need to compute the horizontal tension in the *top* rope as well. The net horizontal force will be equal to the centripetal force, which will be equal to the sum of the two horizontal tensions from each rope.

To get the tension in the top (3 m) rope, use the fact that the net *vertical* force is 0, i.e. the vertical components of the two tensions have to balance each other.

 

[MrWarlock616]

cepheid thank you so much for opening my stupid eyes...I understood it much better now ty :)

 

[mrcheeses]

Thanks Cepheid. I think I got it :)

 

[mrcheeses]

I got about 13.2 N for both Tensions, but If that is right, then Fy is not equal to zero, which I think it should be?

Edit: nevermind I got it. forgot to account for force of gravity. Thanks :)