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
Messages
9
Reaction score
0

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!
 
Last edited:
Physics news on Phys.org
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.
 
Last edited:
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:)
 
Last edited:
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! :)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top