How Do You Calculate Tension and Speed in Circular Motion?

[krej]

Homework Statement

A 240 g block on a 58.0 cm-long string swings in a circle on a horizontal, frictionless table at 85.0 rpm.

A. What is the speed of the block?
B. What is the tension in the string?

Homework Equations

Look at the link in #3.

The Attempt at a Solution

I'm not sure how to do this problem. Here is my work so far that shows what I've done: http://img200.imageshack.us/img200/9189/phys001.jpg

And it seems like no matter what I try to solve for, I need a theta to do anything. Is there a way to solve for theta that I'm not seeing? Or is there a way to solve the problem without needing theta?

 

[mg0stisha]

58.0 cm shouldn't be the hypotenuse in your diagram. Try looking at it a little harder.

EDIT: is that a diagram you're given or one you drew?

 

[krej]

I drew this diagram, the only thing given was the question.

Where else would the 58.0 cm go? The only other place I could see it going is at the radius, but it says the strings length is 58.0cm, which isn't necessarily the radius(it could be making really small circles with like a 5cm radius, but just have a long string).

I was also wondering what the 85rpm is used for. Would I be able to replace the acceleration with it? a = v^2/r = omega. Then omega is the angular velocity, which is what the 85.o rpm is, right?

 

[mg0stisha]

The 58.0 cm is the radius, it says that it swings in a circular path on a horizontal frictionless table (implying that it is actually rotating on the table). That should help.

And yes, Omega (85.0 rpm) is an angular velocity (notice the units).

Try again now with this information. Hopefully it'll work out!

 

[krej]

I still can't get it. :( I don't really know what I'm doing wrong either. Maybe my algebra is just off?

Right now I'm trying to find the speed, what I have is two main equations.

\SigmaFr= ma = T*cos(\vartheta)
\SigmaFz= 0 = T*sin(\vartheta) - mg

From the second equation, I found the tension which is T = (m*g)/(sin(\vartheta)

So I put that in the first equation and started solving for v.
(mgcos(\vartheta))/sin(\vartheta) = mv^2/r
r*g*cot(\vartheta) = v^2

Since I don't know cot(\vartheta), I tried to do some trig to find it with what I have, but I'm not sure if this would be right.

cos(\vartheta) = r/T
cos(\vartheta) = r/(mg/sin(\vartheta))
cos(\vartheta) = (r*sin(\vartheta))/m*g
m*g*cos(\vartheta) = r*sin(\vartheta)
cot(\vartheta) = r/m*g

Then I took that cot(\vartheta) and tried plugging it back into my previous equation with it.

r*g*cot(\vartheta) = v^2
cot(\vartheta) = r/m*g

r*g*(r/m*g) = v^2
r^2/m = v^2
sqrt(r^2/m) = v
sqrt(0.58^2 / .24) = 1.184 = v

Then I went to the site to enter the answer, and it said that was wrong. Would you mind helping me figure out what I did wrong? Hopefully I explained everything I did with enough detail for you to understand.

 

[mg0stisha]

I think you might be able to get the speed just using the rpm's and length of the string. Use more basic thinking, I think you're hearing hoofbeats and thinking zebra :) let me know how it goes!

 

[krej]

Oh wow, is it:
omega = v^2/r
omega*r = v^2
sqrt(omega*r) = v
?

But what would the units on that be? It would be rpm*meters, but the answer needs to be in m/s. How would you convert that?

 

[mg0stisha]

think circumference.

 

[krej]

Shoot, the time is up for the assignment. I had to have it done by 11pm, but I still couldn't think of how exactly to do it. :(

Oh well, thank you for all of your help! Even though I didn't get the answer, it still helped me learn a bit more.

 

[mg0stisha]

Sorry! If you would've told me sooner i could've helped more!

How I did it...

Speed: Find the circumference, multiply by 85 and divide by 60 to get the average velocity in m/s

Tension: F_c=T+F_g, F_c=mv²/r (like you said), solve for T!

Sorry it was too late.

 

[krej]

Ah ok, thanks. Now it makes more sense.

And it's all right, I still ending up getting about an 80% on the homework, which is good enough for me right now since I was forced to do it all at the very last minute.

Thanks again. :)

 

[mg0stisha]

No problem!