# Centripetal Force / Tension / Speed

Centripetal Force / Tension / Speed (HELP!)

## Homework Statement

A 100 g bead is free to slide along an 80 cm long piece of string ABC. The ends of the string are attached to a vertical pole at A and C, which are 40 cm apart. When the pole is rotated about its axis, AB becomes horizontal.
a. Find the tension in the string.
b. Find the speed of the bead at B.

## Homework Equations

Fr= mv2 / r

v= 2$$\pi$$r / T

v= $$\sqrt{}m /r$$ ?

Ft = Fn - mg???

## The Attempt at a Solution

m = 100g = 0.1m
length =80cm

I drew a free-body diagram with just the beed. Normal force going up, mg going down and tension going towards the pole on line A and C.

I think b) is:

v= $$\sqrt{}$$ rg
= $$\sqrt{}$$ 0.4m * 9.80m/s2
= 1.979 m/s

I usually know where to go from here but im really stuck with the Tension.
Any ideas? :)

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Last edited:

Doc Al
Mentor
m = 100g = 0.1m
length =80cm
No, the radius of the bead's motion is not 40 cm. The distance AC is 40 cm; use that fact, the length of the string, and some trig to figure out the radius and angles.

I drew a free-body diagram with just the beed. Normal force going up, mg going down and tension going towards the pole on line A and C.
There's no "normal force" since the bead is not resting on a surface. There's just the tension force from the string (which acts twice: once toward A and once toward C) and the weight.

Apply Newton's 2nd law for both the horizontal and vertical directions.

Trig:

I know I have a 90 degree angle and 40cm between A & C. But I cant seem to find the right formula. Are you talking about the c2=a2+b2?

Or the sin0=opp/hyp? Im just having trouble putting the hypotenuse on the other side of the equation. Does it equal to:
sin0 = 40cm / hyp
hyp= 40cm/sin-1

Tension:

I always seem add the normal force automatically. I think thats a bad habit. So if its in the air or accelerating vertically, it does not have a normal force?

Should I continue with this formula:

For B-A: Ft sin0 = mv2/r
For B-C: Ft cos0 = mv2/r

Do I use v from my first message: $$\sqrt{}$$ rg = 1.979 m/s

Doc Al
Mentor
Trig:

I know I have a 90 degree angle and 40cm between A & C. But I cant seem to find the right formula. Are you talking about the c2=a2+b2?
You'll need that formula plus the fact that the length of the string (which comprises two of the sides) is given. That will allow you to find all three sides of the triangle.

Tension:

I always seem add the normal force automatically. I think thats a bad habit. So if its in the air or accelerating vertically, it does not have a normal force?
A normal force is exerted between two objects (or an object and a surface). The only thing touching the bead is the string.

Should I continue with this formula:

For B-A: Ft sin0 = mv2/r
For B-C: Ft cos0 = mv2/r
The direction of the centripetal acceleration is horizontal (to the left in your diagram).

Do I use v from my first message: $$\sqrt{}$$ rg = 1.979 m/s
No, that is incorrect. First get the sides of the triangle, then set up your force equations. The tension and the speed will be the unknowns in your equations, which you'll solve for.

You'll need that formula plus the fact that the length of the string (which comprises two of the sides) is given. That will allow you to find all three sides of the triangle.

c2 = a2+b2
c (hyp) = 80-b
sin0 = 40 / 80-b sin0 * 80 - 40 = b
cos0 = b / 80-b cos0 * 80 /2 = b

cos0 * 80 /2 = sin0 * 80 - 40
cos0 * 80 = sin0 * 80 - 40 * 2
cos0 / sin0 = 80-40 *2 / 80
tan0 = 1
0 = tan-1 1
0 = 45 degrees?

I think im having a lot of trouble with the trig. I reviewed trig material but I cant seem to get it! Any resources I could look at?

The direction of the centripetal acceleration is horizontal (to the left in your diagram).

Tcb sin0 -ma
T = Tcb cos0 ?

Last edited:
Doc Al
Mentor
c2 = a2+b2
c (hyp) = 80-b
OK. Note that "a" is the vertical side of the triangle, thus a = 40 cm.

You have two equations with two unknowns. Solve for b & c. (Note that "b", the horizontal side of the triangle, will be the radius you'll use in your formula for centripetal acceleration.)