Homework Help: Centripetal Motion along the horizontal with an angle

1. Oct 11, 2012

mpmor1

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

A 0.20 kg stone attached to a 0.8 m long string is rotated in a horizontal plane. The string makes an angle of 25° with the horizontal. Determine the speed of the stone.

2. Relevant equations

V= √(Fr/m)

3. The attempt at a solution

find the vertical component, call it y, then say F= y/sin(θ)

2. Oct 11, 2012

Staff: Mentor

What forces act on the stone? (Hint: Two forces act.) Consider the vertical and horizontal force components and apply Newton's 2nd law to each.

3. Oct 11, 2012

mpmor1

if the total force is broken into its horizontal and vertical components, can its force of weight (1.962N) be treated as the vertical component?

4. Oct 11, 2012

Staff: Mentor

What do you mean by 'total force'? Realize that in the formula in your first post, that "F" has to be the centripetal force.

I think that's a confusing way to go about it. Better to think of centripetal acceleration = v2/r.

Did you draw yourself a free body diagram? What forces act? The weight is one of the forces.

5. Oct 11, 2012

mpmor1

The two forces acting on the stone are its weight, pointing straight down, and the unknown centripetal force which can be seen as the tension in the string

6. Oct 11, 2012

Staff: Mentor

Almost.

The two forces acting on the stone are its weight and the tension in the string. The horizontal component of those forces must produce the centripetal acceleration. (Centripetal force is just the name we give to the net force in the centripetal direction. It's not a separate force. It never appears on a free body diagram.)

7. Oct 11, 2012

mpmor1

So if the horizontal component is the centripetal Force, that being 4.2075N, then this can simply be plugged into the equation V= sqrt(Fr/m)?

8. Oct 11, 2012

Staff: Mentor

How did you solve for that component? (But you're right.)
Yep. (But there's a slightly easier way.)

Last edited: Oct 11, 2012
9. Oct 11, 2012

mpmor1

If the mass of the stone is .2kg then the weight is 1.962N; this would make the horizontal component equal to that divided by the tangent of the angle, 25.

10. Oct 11, 2012

Staff: Mentor

I was editing my response while you were writing your post.

11. Oct 11, 2012

mpmor1

That doesn't yield the correct answer online

12. Oct 11, 2012

Staff: Mentor

What did you use for r?

13. Oct 11, 2012

mpmor1

.8m, the length of the string

14. Oct 11, 2012

Staff: Mentor

There's your problem. r stands for the radius of the circular path, which is not the length of the string. Use a little trig to figure out r.

15. Oct 11, 2012

mpmor1

thanks

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