Question

A small sphere tied to the string of length 0.8m is describing a vertical circle so that the maximum and minimum tensions in the string are in the ratio 3:1. The fixed end of the string is at a height of 5.8m above ground.

a.Find the velocity of the sphere at the lowest position.
b.If the string suddenly breaks at the lowest position,when and where will the sphere hit the ground?
(Take g=10m/s²)​

Answer 1

assume a circle of radius r which is formed vertically along the positive y axis in a 2d imagined plane.

of course the point at which the tension in the string would be max is the lowest position and the top position would have the minimum tensions.

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given:

radius of circle = length of string = 0.8m

maximum and minimum tensions in the string are in the ratio = 3:1

to find:

(a) velocity of the sphere at the least position.

(b) range of the parabolic path covered when string breaks.

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solution:

according to the law of conservation of energy,

$\boxed {\red {\tt loss \: in \: ke\:=\:gain \: in \: pe }}$

===> $\tt \dfrac {mv^2}{2} - \dfrac {mu^2}{2}=mg×2r$

===> $\tt \cancel {m}(v^2-u^2)=\cancel {m}(2×2×gr)$

===> $\tt v^2-u^2=4gr$............. (1)

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use Newton's second law to find second equation.

☆☆ $\boxed {\tt f=ma}$

$\tt t^1-mg=ma^1$

using radial accelerations,

===> $\tt t^1-mg=\dfrac {m (v^2)}{r }$

===> $\tt t^2-mg=\dfrac {mu^2}{r}$

given:

tension ratios are 3:1

$\tt \dfrac {v^2+mgr}{r } = 3 (\dfrac {u^2-mgr}{r })$

$\tt 3u^2=v^2+4rg$

$\tt 3 (v^2-4gr)=v^2+4gr$

$\tt 2v^2 = 16gr$

$\tt v^2=80×0.8$

$\boxed {\tt v = 8m/s }$

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question 2:

the particle when broken off with an initial velocity of 8m/s,

it will cover a Projectile motion with a range=R.

We know that,

===> $\boxed {\green {\tt range = u^x×t }}$

the vertical range is given to us as 5.8m. we'll subtract th radius so as to get vertical displacement.

vertical displacement = 5m

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to find range we'll need time of flight.

$\tt time = (\dfrac {2×5}{10})^{1/2}$

====> time = 1sec

range = 8×1

= 8m

thus the answer of your question is 8m.

hope this helps!!

Answer 2

$\huge{\colorbox{pink}{ Answer }}$

$Tensions \: at \: the \: higest \: point \\ T' = \frac{mv²}{r} - mg \: cos \: 180\\ T' = \frac{mv²}{r} - mg \: \\ {T}^{L} = {3T}^{H} \\ {T}^{L} - {T}^{H} = 6mg \\ {T}^{H} = 3mg \\ 3mg = \frac{mv²}{r} - mg \\ 4gr = v² \\ v = \sqrt{4gr}$

so the answer is ⤵️

$v = \sqrt{4gr}$

Hope it helps you