Problem using Newton's 2nd Law F= m x A

[Bob K99]

Homework Statement

A ball whirling on the end of a string, length L
Maximum tension in string before it breaks: T sub break
Find an equation for the maximum speed of the ball , V sub max, in terms of the ball's mass, M, the breaking
tension T sub break, and the gravitational acceleration g

Ignore the angle theta the string makes with the horizontal plane

Homework Equations

F= mass times acceleration
Acceleration = m x v squared / r (radius)

The Attempt at a Solution

I used algebra and the Pythagorean theorem to get
v = the square root of L x ( T squared - M squared x G squared ) / M
but the correct answer is V = the square root of L x ( T squared - M squared x G squared) / M x T
I cannot figure out how the T sub break showed up in denominator
[/B]

 

[gneill]

Hi Bob K99, Welcome to Physics Forums.

Can you show your work in detail? We can't tell where things might have gone wrong without seeing what you did.

Note that you can use the x₂ and x² icons in the edit window menu bar to produce subscripts and superscripts for formulas.

 

[Ray Vickson]

Homework Statement

A ball whirling on the end of a string, length L
Maximum tension in string before it breaks: T sub break
Find an equation for the maximum speed of the ball , V sub max, in terms of the ball's mass, M, the breaking
tension T sub break, and the gravitational acceleration g

Ignore the angle theta the string makes with the horizontal plane

Homework Equations

F= mass times acceleration
Acceleration = m x v squared / r (radius)

The Attempt at a Solution

I used algebra and the Pythagorean theorem to get
v = the square root of L x ( T squared - M squared x G squared ) / M
but the correct answer is V = the square root of L x ( T squared - M squared x G squared) / M x T
I cannot figure out how the T sub break showed up in denominator [/B]

As "gneill" has suggested, you can use the "x²" and "x₂" buttons (in the grey ribbon at the top of the input panel) to produce x^anything or x_anything. However, you can also do it in plain text by writing x^{anything} or x_{anything}, which can be shortened to x^a and x_a if your "anything" is a single letter or digit "a".

 

[Bob K99]

Okay I will try to use the menu bar.

I had to make some assumptions in my solution and my high school algebra and trig are a bit rusty so bear with me

I formed a right triangle with the ball and string. The hypotenuse is T ( tension of string ) , the vertical leg of the right triangle is the weight of
the ball or mass times gravity. The horizontal leg of triangle is centripetal force F
So F² = T² - m² x g²F² =( mass X velocity² / r )² Equation 1

It is at this point that my math starts to break down
Ideally to get the posted answer I should get the equation
F = m x v² / r = T² - (m x g)² substitute Length of string L for R
V² = L ( T² - (m x g )² / M
V = square root of L ( T² - M² X G² / M Equation 2

However I ignored the fact that F² = ( m X v² /r )² in equation 1

Also the correct answer has M (mass) x T (tension of string at break) in the denominator equation 2

Correct answer is V = square root of L x ( T² - M² x G²) divided
by M x T

Possibly I should have broken the problem down to 2 equations
1) Equation for forces on ball Using F & T

2.) Equation for Distance Using length L and radius r

Then substitute a variable to get 1 equation

 

[gneill]

Okay. Note that L and R are not the same thing, so directly substituting one for the other is invalid.

Here's a diagram that depicts the situation:

The tension and centripetal force vectors have had their directions reversed so that they project past the end of the ball for convenience; It's the angles and magnitudes that are important to us. You should be able to use similar triangles to find a relationship between r and L that will allow you to replace r in your equations.

 

[Bob K99]

Okay. Note that L and R are not the same thing, so directly substituting one for the other is invalid.

Here's a diagram that depicts the situation:

View attachment 96711

The tension and centripetal force vectors have had their directions reversed so that they project past the end of the ball for convenience; It's the angles and magnitudes that are important to us. You should be able to use similar triangles to find a relationship between r and L that will allow you to replace r in your equations.

Yes, you are right about r and L not being equal I will try your suggestion and keep working on it.
Thanks

 

[Bob K99]

Thanks to gneill I think I figured the problem out Since we have similar triangles the sides are proportional ( see his diagram above)

T² = (mg)² + (mv²)² / r² Law of right triangles

T² - m² g² = m² v⁴ / r² Eq 2

Similar triangles r/ mv² /r =L /T this simplifies to R² /mv² = L/T

Substituting for r² in eq 2 will give L ( T² - m² g²) / MT = v²

V = square root of L ( T² - m² g² ) / M T

Voila