Question

A particle is projected with speed 2m/s at an angle 60° with the horizontal in vertical plane . It's radius of curvature after time
$\sf \triangle \: t = \frac{1}{5 \sqrt{3} } \: s$
will be ? (g = 10m/s²)

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Answer 1

Answer:

A particle is projected with speed 2m/s at an angle 60° with the horizontal in vertical plane . It's radius of curvature after time

$\sf \triangle \: t = \frac{1}{5 \sqrt{3} } \: s$

will be ? (g = 10m/s²)

Answer 2

Answer:

The radius of curvature is 0.266m

Explanation:

Given,

A particle is projected with a velocity of 2m/s at an angle of 60° with horizontal

u=2m/s and θ=60°

To find radius of curvature at any instant,we use the following relation

$a = gcosθ = \frac{v {}^{2} }{r} \\ = > r = \frac{v {}^{2} }{gcosθ} \\ here \: v {}^{2} = v_{x} {}^{2} + v_{y} {}^{2}$

$v_{x} = ucosθ = 2cos60 {}^{0} = 1ms {}^{ - 1} \\ v_{y} = usinθ - gt \\ = 2sin60 {}^{0} - (10)( \frac{1}{5 \sqrt{3} } ) = \frac{1}{ \sqrt{3} } ms {}^{ - 1} \\ = > v {}^{2} = 1 {}^{2} + ( \frac{1}{ \sqrt{ 3 } } ) {}^{2} = \frac{4}{3} ms {}^{ - 1}$

Therefore,by substituting this value we can find radius of curvature after this given time

$r = \frac{ \frac{4}{3} }{10cos60 {}^{0} } = \frac{4}{15} = 0.266m$

Therefore radius of curvature after given time is obtained as 0.266m

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