Ball on a string with initial velocity

AI Thread Summary
A ball with a mass of 1.72 kg is attached to a peg with a 1.54 m string and released from a vertical position with an initial speed of 4.81 m/s. The problem involves calculating the ball's speed when the string makes a 21.7° angle above the positive X-axis using energy conservation principles. The initial attempts to solve the equation contained errors, particularly in the use of sine and the absence of a necessary factor of 2. After correcting these mistakes, the final calculation yielded a speed of approximately 6.49 m/s, which is the correct answer. The discussion highlights the importance of careful attention to detail in physics calculations.
joe_coolish
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Homework Statement



A ball with mass m = 1.72 kg is attached to a peg at the origin using a length of massless string of L = 1.54 m. It is released from the top of a vertical circle with speed v0 = 4.81 m/s. What is the ball’s speed (m/s) when the string makes an angle θ = 21.7° above the positive X axis?

8_problem_ballstring.jpg


Homework Equations



Kf+Uf = Ki+Ui
K=0.5mv^2
U=mgy

The Attempt at a Solution



I started by using Kf+Uf = Ki+Ui => .5mVi^2 + lmg = .5mVf^2 + lsinθmg
and then I solved for Vf:

Vf=SQRT(Vi^2 + lg - lsinθg)

that came out to be 5.71, but the correct answer is 6.49. Any help would be wonderful!
 
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joe_coolish said:
I started by using Kf+Uf = Ki+Ui => .5mVi^2 + lmg = .5mVf^2 + lcosθmg

Hi joe_coolish, welcome to PF! :smile:

You've used the cosine, while you should have used the sine.
 
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Thank you for your reply and for the welcome!

I changed the formula to Vf=SQRT(Vi^2 + lg(1 - sinθ))
which gave me this: SQRT(4.81^2+(1.54*9.8)*(1-SIN(RADIANS(21.7)))) = 5.713832509

(I like to use EXCEL for these things)

Which was my original answer. Sorry, I must have copied it down wrong :) I'll edit my original post.

But still, that is the wrong answer... Am I missing something?
 
joe_coolish said:
Thank you for your reply and for the welcome!

I changed the formula to Vf=SQRT(Vi^2 + lg(1 - sinθ))
which gave me this: SQRT(4.81^2+(1.54*9.8)*(1-SIN(RADIANS(21.7)))) = 5.713832509

(I like to use EXCEL for these things)

Which was my original answer. Sorry, I must have copied it down wrong :) I'll edit my original post.

But still, that is the wrong answer... Am I missing something?

In your original equation you have a factor 0.5.
In your final equation there should be a corresponding factor 2 somewhere...

(And yes, Excel is the perfect tool for things like this. :smile:)
 
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I like Serena said:
In your original equation you have a factor 0.5.
In your final equation there should be a corresponding factor 2 somewhere...

Ah! I need to lay off the late nighters...

Thank you!

SQRT(4.81^2+(2*1.54*9.8)*(1-(SIN(RADIANS(21.7))))) = 6.493047349


It's always that dang 2!
 
joe_coolish said:
Ah! I need to lay off the late nighters...

Thank you!

SQRT(4.81^2+(2*1.54*9.8)*(1-(SIN(RADIANS(21.7))))) = 6.493047349

Yep! :smile:

joe_coolish said:
It's always that dang 2!

Uhh, sometimes it's a dang minus sign! :)
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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