physics--energy of conservation question

[paki20]

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Q: A ball is suspended from a 5 m long cable that makes a 30 degree angle with the vertical. The ball is released and swings down. What is the ball's speed at the lowest point?

A: 3.6 m/s using law of conservation of energy


This should be a simple question but for some reason I am not getting 3.6 m/s. I know the answer depends on the difference in heights but I just don't know how to figure out the original height. Thanks for any help!

 

[Dr Gerrard]

The 30 degrees from the vertical tells you the height.

Length of string = hypotenuse = (adjacent)/cos(30)

adjacent = 5 * (.86) = 4.3

height is equal to 5 - 4.3 = .7

.7 * m * g = (1/2) * m * v^2

1.4 = v^2

v = 1.16---- I am not sure what I did wrong

EDIT - I had initially done sin(30), but changed that to cos and now am even more off.

EDIT Again --- All right, so even if you assume that the initial height is 5, you still only come up with square root of 10 as your answer, which is less than the "correct" answer. I don't know what the problem is.

 

[paki20]

The 30 degrees from the vertical tells you the height.

Length of string = hypotenuse = (adjacent)/cos(30)

adjacent = 5 * (.86) = 4.3

height is equal to 5 - 4.3 = .7

.7 * m * g = (1/2) * m * v^2

1.4 = v^2

v = 1.16---- I am not sure what I did wrong

EDIT - I had initially done sin(30), but changed that to cos and now am even more off.

EDIT Again --- All right, so even if you assume that the initial height is 5, you still only come up with square root of 10 as your answer, which is less than the "correct" answer. I don't know what the problem is.

ok FINALLY figured it out...its tricky..i'll post my work in a second

 

[prophecygirl]

I got 0.67m for the height (it took me a lot longer than Dr. Gerrard though! 🙂), but the velocity worked out to 3.66 m/s. I ended up with the square root of 13.4 -- I bet you just put your decimal place in the wrong place by accident. The square root of 14 should give a similar answer.

Hope that helps!

EDIT: I used 10 m/s squared for g -- I bet it would have come out exactly to 3.6 if I had used 9.8 m/s squared instead.

 

[paki20]

Need to use law of conservation of energy which is delta K = delta PE

Ki+PEi = Kf + PEf

so mgy1 = 1/2mvf^2 + mgyf

solve for vf:

square root of (2*g*(yi-yf)) = vf

plug in the values:
yi = 5*cos(30)
yf = 5 m

and you get vf = 3.6 m/sec



EDIT: LOL hold on something doesn't look right about this...let me double check it...the height is higher when the cable is at 30 degrees than when the cable is vertical (5m) so yi has to be more than yf....the DIFFERENCE in heights is what matters which is .7 m...which is why we came to the same answer..just want to make sure i have the CONCEPTUAL part correct

 

[paki20]

The 30 degrees from the vertical tells you the height.

Length of string = hypotenuse = (adjacent)/cos(30)

adjacent = 5 * (.86) = 4.3

height is equal to 5 - 4.3 = .7

.7 * m * g = (1/2) * m * v^2

1.4 = v^2

v = 1.16---- I am not sure what I did wrong

EDIT - I had initially done sin(30), but changed that to cos and now am even more off.

EDIT Again --- All right, so even if you assume that the initial height is 5, you still only come up with square root of 10 as your answer, which is less than the "correct" answer. I don't know what the problem is.

yeah your work is right and you do get 3.6 m/second using your work..i think you just miscalculated because you forgot to multiply by "g = 10m/s^2)

 

[paki20]

so at its lowest point is there some KE and PE or does all the initial PE get converted to KE??

 

[ssiding]

So the cables is 30 degrees above the horizontal thus wouldn't the heigt of he mass be sin of 30 degrees X 5 = 2.5. So we could use the kinemativs equation vf^2 = vi^2 + 2ad. The object is released from rest so inital velocity is zero, so final velocity should be around 7m/s. Can you guys tell me what's wrong with the procedure, because oviously it's getting the wrong answer.🙁
Thanks

 

[ssiding]

ok so its 30 degrees from the vertical, then inorder to get the adjacent side you would do cos of 30 X 5m = 4.33. I don;t understand how Dr gerrad knew that you should substract the 4.33 from 5 to get the cord height of .7??? Is this a rule from highschool trig that I should remember, I can;t understand how you knew to do that. 😕


Because when you have the height of .7 then you can plug the numbers into a kinematic equation of vf^2 = vi^2 + 2ad. Since the object is released from rest initial velocity is zero thus (2)(10m/s^2)(.7m)= vf^2 which gets the right answers.🙂

 

[prophecygirl]

I think there are probably several ways to figure it out, but I drew a triangle with the 5 m cable (in the initial position -- 30 degrees with respect to the vertical) as the hypotenuse, a horizontal side along the ceiling and a vertical side straight down. Because we know the angle between the cable and the vertical is 30 degrees, the angle between the cable and the ceiling must be 60 degrees (together those two angles form a right angle so they must add up to 90 degrees). The vertical side of the triangle (i.e., the distance between the ceiling and the ball's initial position) is sin 60 times 5 m = 4.33 m. (Sin 60 has the same value as cos 30.)

When the ball is released and swings down, the cable is hanging straight down so the distance between the ceiling and the ball is 5m. The difference between the initial position of the ball (4.33 m) and the final position (5 m) is 0.67 m (rounded to 0.7 m).

Hope that helps -- it's tough to explain without drawing a picture! 🙂