Finding the tension in a circular pendulum without radius or angle.

[Parallax]

Homework Statement

Some kid is playing with a yoyo of mass m. The yoyo string is let out to length L, and is spun in a horizontal circle at a constant rate of ω. The yoyo string makes an angle of θ with the horizontal

m = 39 grams = 0.039 kilgrams
L = 46cm = 0.46m
ω = 3 rads/sec

Calculate the tension in the string in Newtons.

Homework Equations

Tx = Tcosθ = mω²r = mv²/r

Ty = Tsinθ = mg = 0.3822N

r = L cosθ

h = L sinθ

The Attempt at a Solution

The vertical component of the tension was easy, the only force acting in this direction is gravity with a force of m*g Newtons.

The horizontal component is more confusing... Since the height, radius and length of string (hypotenuse) form a right triangle, the lengths of sides should correspond to the ratios on them.

But I don't seem to know the vertical length, just the force's magnitude.
I'm trying to solve for the hypotenuse's force's magnitude, but only know the length.
And I don't seem to know anything at all about the horizontal length/radius of the circle has been formed or its force.

So using Fx^2 + Fy^2 = T^2 or x^2 + y^2 = L^2 is out. The angular velocity seems hard to use without knowing the radius.

Whats a step that takes me to finding the tension, radius, or θ?

 

[Tanya Sharma]

Welcome to PF!

Tx = Tcosθ = mω²r

r = L cosθ

This should be sufficient to get tension in the string in terms of given parameters.

 

[Parallax]

Welcome to PF!



This should be sufficient to get tension in the string in terms of given parameters.

Do you mean

mω²Lcosθ ?

That still leaves me with an unknown of T and an unknown of θ

 

[Tanya Sharma]

Do you mean

mω²Lcosθ ?

That still leaves me with an unknown of T and an unknown of θ

You do not require θ .

Tcosθ = mω²Lcosθ .

 

[Parallax]

You do not require θ .

Tcosθ = mω²Lcosθ .

So T = mω²L...

It seems to work, tough I'm still not sure how ω corresponds to 1/s^2. I'm so used to translating it linear units via the radius.

And that the force along the hypotenuse is less than its y component... Blowing my mind.

 

[Tanya Sharma]

So T = mω²L...

Yes

It seems to work, tough I'm still not sure how ω corresponds to 1/s^2. I'm so used to translating it linear units via the radius.

Sorry...I do not understand your question.The net force in the radial direction is Tcosθ and this force is equal to the centripetal force given by mω²r .Here the radius is Lcosθ .

Could you reexpress your doubt ?

And that the force along the hypotenuse is less than its y component... Blowing my mind.

No..The force along the hypotenuse is the tension T .The component of which in the vertical direction i.e the y-component is Tsinθ which is less than T .

 

[Parallax]

But the mass * gravity comes to weight of 0.3822 Newtons, and the tension is only 0.1615 Newtons.

 

[Tanya Sharma]

But the mass * gravity comes to weight of 0.3822 Newtons, and the tension is only 0.1615 Newtons.

You haven't taken the normal force from the ground into account. The force equation in the y-direction will be Tsinθ + N = mg .

 

[Parallax]

You haven't taken the normal force from the ground into account. The force equation in the y-direction will be Tsinθ + N = mg .

It never even occurred to me that the yoyo was in contact with the grounds. Thanks!

 

[Matterwave]

This is a horrible problem in my opinion. The problem never specified that there was contact with a ground, but the numbers they give will not work unless there is some additional force holding the yoyo up. There is, in general, a minimum angular velocity necessary for the yoyo to be suspended by the tension force of the string alone.