Centripetal Motion along the horizontal with an angle

[mpmor1]

Homework Statement

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.

Homework Equations

V= √(Fr/m)

The Attempt at a Solution

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

 

[Doc Al]

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.

 

[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?

 

[Doc Al]

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?

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.

 

[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

 

[Doc Al]

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

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.)

 

[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)?

 

[Doc Al]

So if the horizontal component is the centripetal Force, that being 4.2075N,

How did you solve for that component? (But you're right.)

then this can simply be plugged into the equation V= sqrt(Fr/m)?

Yep. (But there's a slightly easier way.)

 

[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.

 

[Doc Al]

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

 

[mpmor1]

That doesn't yield the correct answer online

 

[Doc Al]

That doesn't yield the correct answer online

What did you use for r?

 

[mpmor1]

.8m, the length of the string

 

[Doc Al]

.8m, the length of the string

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.

 

[mpmor1]

thanks