Sphere describing a circle, find tension

[Apprentice123]

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)

 

[tiny-tim]

… 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!

(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!

 

[Apprentice123]

Hi Apprentice123!

(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!

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

 

[tiny-tim]

ZFt = m . at
Psin(theta) = m . dv/dt

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

v is constant!

You need vertical components, not tangential ones.

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

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? )

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

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

 

[Apprentice123]

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

v is constant!

You need vertical components, not tangential ones.

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


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

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

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°

 

[LowlyPion]

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.

 

[tiny-tim]

Hi Apprentice123!

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 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.

 

[Apprentice123]

Hi Apprentice123!

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 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.

Ok. Thanks