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Sphere describing a circle, find tension

  • #1
A sphere of 3 kg describe a circle with a horizontal speed of 1,2 m/s. Knowing that L = 800 mm, determine (a) the angle theta (b) the tension in the wire

Answer:
(a) 24,2°
(b) 32,3 N


I find 2 equations

ZFn = m . an
T - Pcos(theta) = m . v^2/p

ZFt = m . at
Psin(theta) = m . at

As calculated at? (tangential acceleration)
 

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  • #2
tiny-tim
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… I find 2 equations

ZFn = m . an
T - Pcos(theta) = m . v^2/p

ZFt = m . at
Psin(theta) = m . at

As calculated at? (tangential acceleration)
Hi Apprentice123! :smile:

(I can't see the diagram yet, but I assume this is a string of length 800 mm with a sphere on the end rotating at a steady speed)

i] there is no tangential acceleration … the acceleration is purely radial (centripetal)

ii] you must balance the centripetal acceleration the weight and the tension, by taking components both horizontally and vertically …

try again! :smile:
 
  • #3


Hi Apprentice123! :smile:

(I can't see the diagram yet, but I assume this is a string of length 800 mm with a sphere on the end rotating at a steady speed)

i] there is no tangential acceleration … the acceleration is purely radial (centripetal)

ii] you must balance the centripetal acceleration the weight and the tension, by taking components both horizontally and vertically …

try again! :smile:
The two formulas that I found are:

ZFn = m . an
T - Pcos(theta) = m . v^2/p

ZFt = m . at
Psin(theta) = m . dv/dt
 
  • #4
tiny-tim
Science Advisor
Homework Helper
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ZFt = m . at
Psin(theta) = m . dv/dt
Apprentice123, there are no tangential forces, and no acceleration!!! :rolleyes:

v is constant!

You need vertical components, not tangential ones.

What is the vertical acceleration, and what are the vertical forces? :smile:
ZFn = m . an
T - Pcos(theta) = m . v^2/p
(btw, I've been wondering for ages … in what language does P stand for weight? :confused: :wink:)

(oh, and have a theta for copying: θ :wink:)


T needs a θ also, and you need to write p in terms of L.
 
  • #5


Apprentice123, there are no tangential forces, and no acceleration!!! :rolleyes:

v is constant!

You need vertical components, not tangential ones.

What is the vertical acceleration, and what are the vertical forces? :smile:


(btw, I've been wondering for ages … in what language does P stand for weight? :confused: :wink:)

(oh, and have a theta for copying: θ :wink:)


T needs a θ also, and you need to write p in terms of L.

Correct ?

ZFn = m.an
T - P = m . v^2/L
T = 34,83 N

ZFt = m.at
at = 0
Tcos(theta) - P = 0
theta = 30,9°
 
  • #6
LowlyPion
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Correct ?

... = m . v^2/L
The centripetal acceleration is m*v2/r.

But r in this case is not L. It is the radius about the point that it rotates, and that point is not where the string is tied.
 
  • #7
tiny-tim
Science Advisor
Homework Helper
25,832
251
Hi Apprentice123! :wink:

Now I see where you got this from …
ZFn = m . an
T - Pcos(theta) = m . v^2/p
Yes, that would be correct if the centripetal acceleration were along the string (the P has cosθ, and the T has no θ).

But, as LowlyPion :smile: points out, it's horizontal.

Centripetal acceleration has nothing to do with physics, it's just geometry

never mind why it's going in a circle (in this case, because of the string) …

all that matters is the shape of the circle, and the speed …

it's a horizontal circle, so the centripetal acceleration is also horizontal. :smile:
 
  • #8


Hi Apprentice123! :wink:

Now I see where you got this from …


Yes, that would be correct if the centripetal acceleration were along the string (the P has cosθ, and the T has no θ).

But, as LowlyPion :smile: points out, it's horizontal.

Centripetal acceleration has nothing to do with physics, it's just geometry

never mind why it's going in a circle (in this case, because of the string) …

all that matters is the shape of the circle, and the speed …

it's a horizontal circle, so the centripetal acceleration is also horizontal. :smile:
Ok. Thanks
 

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