Rotational motion pendulum problem

[joe426]

Homework Statement

Homework Equations

ƩF = ma

The Attempt at a Solution

I know how to break this problem up into the x and y direction

Y Direction:
Ft cosθ - Fg = 0

X Direction:
Ft sinθ - Fg = mv² / r

The hint says to find the inward radial component. I'm guessing this means to find r, because I think that is the only thing missing for me to solve this. To find r I think I would use some combination of the length of the string and theta, but don't exactly know.

Thanks for any help!

 

[TSny]

The "inward radial component of the string's tension" is just the horizontal component of the the tension force (i.e., what you called the X component of Ft).

You've got -Fg acting in both the Y-direction and X-direction?

You should be able to express r in terms of d and theta using trig on an appropriate right triangle.

You'll have two unknowns, Ft and v. But you'll also have two equations. Try eliminating Ft and solving for v.

 

[frogjg2003]

When it was talking about the vertical and inward radial components, it was talking about finding Ft_y and Ft_x, which is what you have already done. You know r because you know d and θ. Just solve the two equations for v, and you're all set.

 

[joe426]

There is no -Fg in the x direction, silly me.
r = Ft sinθ?

Y Direction:
Ft cosθ - Fg = 0
Ft = Fg / cosθ ??

X Direction:
Ft sinθ = mv² / r
(Fg / cosθ)sinθ = mv² / (Fg / cosθ)sinθ

? lol I am so confused

 

[TSny]

There is no -Fg in the x direction, silly me.
r = Ft sinθ?

Get an expression for r in terms of d and theta, not in terms of Ft and theta. Consider the right triangle where the string is the hypotenuse and the radius r is the side opposite theta.

Y Direction:
Ft cosθ - Fg = 0
Ft = Fg / cosθ ??

yes

X Direction:
Ft sinθ = mv² / r
(Fg / cosθ)sinθ = mv² / (Fg / cosθ)sinθ

For r, use the expression derived from the right triangle mentioned above.

 

[joe426]

Get an expression for r in terms of d and theta, not in terms of Ft and theta. Consider the right triangle where the string is the hypotenuse and the radius r is the side opposite theta.

yes


For r, use the expression derived from the right triangle mentioned above.

Ohhh. r = dsinθ. Whick makes...

((mg / cosθ)sinθdsinθ) / m = v²

Then I would just take the square root of both sides.

Thanks for all the help again guys!