# Homework Help: Centripetal forces

1. Mar 17, 2014

### mpittma1

1. The problem statement, all variables and given/known data

A 1.5 kg mass tied to the roof rotates with constant speed in a horizontal
circle. The string makes an angle of 30 deg to the vertical.

a) determine the velocity , the centripetal acceleration and the centripetal
force on the mass.

b) determine the tension in the string.

c) say the mass is given a push so that now the mass rotates with a constant
velocity of 9.4 m/s . Determine the angle the string makes with the vertical
, the centripetal force on the mass and the tension in the string.

2. Relevant equations

∑Fy = TcosΘ - mg = 0

3. The attempt at a solution

To solve part a:

First I found my R by,

R = LsinΘ = 1.5sin30° = .75m

Second my T by,

T= mg/cosΘ = 1.5*9.8/cos30° = 16.97 N

Now I find my velocity by,

V= sqrt((R*T*sinΘ)/m)) = sqrt((.75*16.97*sin30°)/1.5) = 2.06 m/s

the centripetal acceleration,

v2/R = a

a = TsinΘ/m = 16.97*sin30°/1.5 = 5.66 m/s2

for part b I already found my tension to be 17N.

Part C is where I am having trouble, how can i find the angle Θ with the new constant velocity if im not given a radius or tension?

2. Mar 17, 2014

### TSny

Hello, mpittma1. Welcome to PF!

Note that you have 3 "unkowns": θ, T, and R. But you also have 3 equations: your two "Relevant equations" and your equation for R in terms of L and sinθ.

Can you see how to do the algebra to get to an equation where the only unknown is θ?

3. Mar 17, 2014

### mpittma1

Hey TSny,

So my three equations are:

R = LsinΘ (1)

TcosΘ - mg =0 (2)

&

TsinΘ = mv2/R (3)

I see how you could plug equation 1 into equation 2 and come up with

TsinΘ = mv2/LsinΘ which is closer because I have my m, v, and L but Im still not seeing how to get rid of the T by using equation 2.

4. Mar 17, 2014

### TSny

Use (2) to find T in terms of θ. [EDIT: Or divide (3) by (2) after rearranging (2)]

5. Mar 17, 2014

### mpittma1

So I am coming up with:

TanΘ = v2/LsinΘg

after dividing eqn 3 by eqn 2 and plugging in eqn 1 for R.

After doing some more algebra I come up with tanΘsinΘ = v2/Lg

something doesnt seem right there though....... how could i solve for tanΘsinΘ?

6. Mar 17, 2014

### TSny

Looks good so far! Can you express the left side of the equation solely in terms of sinθ?

7. Mar 17, 2014

### mpittma1

well I could easily express the left side as sin2Θ/cosΘ

im not seeing how to express it solely as sinΘ though

8. Mar 17, 2014

### TSny

Can you think of a trig identity that relates sinΘ and cosθ?

9. Mar 17, 2014

### mpittma1

so heres what im doing:

sin2Θ/cosΘ → using a half angle formula (1-cos2Θ)/2 * 1/cosΘ →

(1-cos2Θ)/2cosΘ

my eqn thus far is (1-cos2Θ)/cosΘ = 2v2/Lg

how could you simplify the left side further?

Last edited: Mar 17, 2014
10. Mar 17, 2014

### TSny

I'm not sure the half-angle formula will help. Can you make use of sin2Θ + cos2θ = 1?
[Edit: I said earlier to express everything in terms of sinθ, which will work. But it might be a little easier to express sin2Θ/cosΘ in terms of cosθ.]

Last edited: Mar 17, 2014
11. Mar 17, 2014

### mpittma1

I was trying to find a way to get a +cos2Θ but its just not there

12. Mar 17, 2014

### TSny

So, use the trig identity to express sin2θ in terms of cos2θ.

13. Mar 17, 2014

### mpittma1

(1-cos2Θ)/cosΘ

or secΘ-cosΘ so im almost there

14. Mar 18, 2014

### TSny

Good

I don't recommend writing it this way. Use (1-cos2Θ)/cosΘ for tanθsinθ in your equation from post #5 and see if you can proceed from there.

15. Mar 18, 2014

### mpittma1

After plugging in my values for v L and g i get:

(1-cos2Θ)/cosΘ = 6.011

any suggestions any solving for Θ?

16. Mar 18, 2014

### TSny

Let x = cosθ and write your equation in terms of x. Can you solve for x?

17. Mar 18, 2014

### mpittma1

ok so what i did:

let x = cosΘ

so, 1-x2/x = 6.011

1-x2 = 6.011x

0 = x2 + 6.011x -1

then after solving for the roots of x

x = -6.173 and .162

plugging x = .162 into x = cosΘ

gives me Θ = 80.7° which is the correct answer!!!

thank you for all your help!

18. Mar 18, 2014

Good work!