4.17 A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?

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    Vishal Baghel | Contributor-Level 10

    4 months ago

    4.17 Given, the length of the string, l = 80 cm, No. of revolution = 14, Time taken = 25 s

    We know Frequency, v = No.ofrevolutiontimetaken = 1425 Hz

    Angular frequency ω = 2?v = 2 × 227 ×1425 rad/s = 3.52 rad/s

    Centripetal acceleration ac = ω2 r = 3.522×80100 m/s2 = 9.91 m/s2

    The direction of acceleration is towards the centre.

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A
alok kumar singh

Please find the solution below:

 

 

 

V
Vishal Baghel

  a = v 0 2 R

after 10 kicks, v = v 0 + 1 0 v 0 1 0 0 = 1 . 1 v 0

R 1 = 1 . 2 1 R

 

V
Vishal Baghel

v? = 3tî v? = 24cos 60°î + 24sin 60°? = 12î + 12√3?
v? = v? – v? = (12 – 3t)î + 12√3?
It is minimum when 12 - 3t = 0 ⇒ t = 4sec

V
Vishal Baghel

ω = θ² + 2θ
α = (ωdω)/dθ = (θ² + 2θ) (2θ + 2)
At θ = 1rad.
ω = 3rad/s and α = 12rad/s²
a? = αR = 12 m/s² a? = ω²R = 9 m/s² A? = √ (a? ² + a? ²) = 15 m/s²

V
Vishal Baghel

a? = v? ²/4r
a_A? = (v? ²/r²) × r = v? ²/r
a_A = 3v? ²/4r

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