Sphere describing a circle, find tension

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Homework Help Overview

The problem involves a sphere of mass 3 kg moving in a horizontal circular path with a speed of 1.2 m/s, attached to a wire of length 800 mm. The discussion centers around determining the angle theta and the tension in the wire, while addressing the forces acting on the sphere.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the equations of motion relevant to the sphere's circular path, including centripetal acceleration and the forces acting on the sphere. There is a focus on the need to balance vertical and horizontal components of forces, as well as questioning the presence of tangential acceleration.

Discussion Status

Several participants have provided guidance on the correct interpretation of forces and acceleration in the context of circular motion. There is an ongoing exploration of the relationships between the variables involved, with some participants suggesting reconsideration of the assumptions made regarding tangential forces and the geometry of the situation.

Contextual Notes

Participants note the importance of understanding the distinction between radial and tangential components of acceleration, as well as the need to clarify the definitions of variables used in the equations. There is also mention of the diagram not being visible to all participants, which may affect their understanding of the problem setup.

Apprentice123
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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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Apprentice123 said:
… 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:
 


tiny-tim said:
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
 
Apprentice123 said:
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.
 


tiny-tim said:
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°
 


Apprentice123 said:
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.
 
Hi Apprentice123! :wink:

Now I see where you got this from …
Apprentice123 said:
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:
 


tiny-tim said:
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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