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# physics question

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physics
Nov 9, 2014

#5
+27342
+13

I agree that the answers are correct.

The first comes from:

$$m\frac{v^2}{r}=T\sin{\theta}$$ and $$mg=T\cos{\theta}$$ where T is tension in the string.

and the second from

$$Period=\frac{2\pi r}{v}$$

In both cases we have:

$$r=L\sin{\theta}$$

.

Nov 10, 2014

#1
+95179
+5

Hi 315,

This is very difficult to see/ read

I can't answer anyway but I think Alan will have a problem as well.  :)

Nov 9, 2014
#2
+394
+13

Hi Ms. Melody,

My eyes are young and work well for reading fine print and old text books.:)

The second answer can be reduced.

I was reading a similar question in an ancient text book that uses look up tables and slide rules.

The formula is just a variation on the circumference of a circle.

T= (2pi *L*sin(theta) )/Sqr(L*g*sin(theta)*tan(theta))

Rearranges to

T= 2pi *sqrt(L2*sin2(theta))/(g*L *sin(theta)*tan(theta))

Reduces to

T= 2pi *sqrt(L*cos(theta)/g)

Though mathematically the same, these ancient text books often simplified the equations as much as possible to make the use of a slide rule easer.

As far as I know, I’m the only student in my school who uses a slide rule. They never need batteries and it makes a great back scratcher. Hahaha

_7UP_

Nov 9, 2014
#3
+95179
+5

Where do you buy a slide rule SevenUp.  Do you have good antiques stores around your place?   LOL

Nov 9, 2014
#4
+394
+13

Actually, I have three. One is a gift from my grandfather. It’s made from bamboo, and he used it when he was in college and my father did too.

Another is a gift from my cousin who goes to an Amish school in Pennsylvania, where electricity –even from a battery --is not used, so no calculators. He carved this himself, and it is more a work of art than practical.

The third one, my older brother found on eBay. It’s plastic and precise. That is the one I carry to school. It's a little longer than the others so it makes a better back scratcher. :)

_7UP_

Nov 10, 2014
#5
+27342
+13

I agree that the answers are correct.

The first comes from:

$$m\frac{v^2}{r}=T\sin{\theta}$$ and $$mg=T\cos{\theta}$$ where T is tension in the string.

and the second from

$$Period=\frac{2\pi r}{v}$$

In both cases we have:

$$r=L\sin{\theta}$$

.

Alan Nov 10, 2014
#6
+1832
0

thank you all

Nov 18, 2014