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# Physics: Circular motion finding angle theta?

A mass m = 6.300 kg is suspended from a string of length L = 1.330 m. It revolves in a horizontal circle. The tangential speed of the mass is 3.205 m/s. What is the angle theta between the string and the vertical (in degrees)?

Looks something like this:

Theta is the angle on the left side of the vertical line, opposite of the ball.

--------------------

......../|\.........

......./.|.\........

....../..|..\.......

...../...|...\......

...O..............

I'm on hour 2 and I cannot get the damn answer for this.

I keep getting that Tsin(theta) = ma_c and Tcos(theta) = mg

This gives me 38.24 deg as the angle. But its wrong. I've stared for a while now and can't get it.

Any help?

### 2 Answers

- D gLv 78 years agoFavorite Answer
these equations are not simple

your equations are correct as far as you go

let me call theta = B

let me call the radius of the circle made by the mass at the bottom of the string ' r '

sin B = r / L

r = L*sin B

a_c = v^2 / r

also

sin B = m*a_c / T

cos B = mg / T

tan B = sin B / cos B = a_c / g

all the other parts cancel

tan B = v^2 / r*g

substituting in for r

tan B = v^2 / (g*L*sin B)

sin B* tan B = v^2 / ( g* L)

(sin B)^2/ cos B = v^2 / (g * L)

I am having trouble seeing the simplifying equation for that angle..

to find the period or something is easy given the angle but the reverse is harder

Ill try to work on it more email me if you have some information regarding equations and simplifications you used maybe the hint will jar me into the correct path...

- favelaLv 44 years ago
the only way i understand a thank you to try this's to get the radial & tangential accelerations in polar coordinates, of their maximum known sort (whilst the two r & @ are changing), (@=theta). that's not person-friendly. you're able to do it via two times differentiating the region vector(the two value & unit vector). the result's; a_r = d^2r/dt^2 - r(d@/dt)^2 a_t = 2(dr/dt)(d@/dt) + rd^2@/dt^2 Now out of your given info. d@/dt = at = SqRt(2@a) d^2@/dt^2 = a dr/dt = advert@/dt = Aat = ASqRt(2@a) d^2r/dt^2 = advert^2@/dt^2 = Aa placed those derivatives into the eq for a_r and question (a) falls out. you are able to desire to be waiting to do a matching element for (b) via equating a_r to a_t and putting each and every thing in terms of @ and constants and then fixing for @. between the wierdest issues i've got ever seen.