Question

give answer with full explainatio ❗



A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius
1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the
string❓What is the maximum speed with which the stone can be whirled around if
the string can withstand a maximum tension of 200 N ❓​

Answer 1

Answer:

Maximum speed of the stone is 34.64 m/s.

Explanation:

Given,

Mass of the stone, m = 0.25 kg

Radius of the circle, r = 1.5 m

Number of revolution per second, n = 40/60 = (2/3) rps

Angular velocity, ω = v/r = 2πn ………………….(i)

The centripetal force for the stone is provided by the tension T, in the string, i.e.

T = mrω²

  = 0.25 * 1.5 * (2πn)²

  = 0.25 * 1.5 * (2 * 3.14 * 2/3)²

  = 6.57 N

Then,

Given Maximum tension in the string = 200 N.

$T_{max} = \frac{mv^2_{max}}{r}$

$v_{max} = \sqrt{T_{max}} * \frac{r}{m}$

⇒ √200 * 1.5/0.25

⇒ √1200

⇒ 34.64 m/s

Therefore, the maximum speed of the stone is 34.64 m/s.

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Answer 2

$\boxed{\bf{\blue{Find \longrightarrow }}}$

Tension in the string

the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N

$\boxed{\bf{\pink{Given \longrightarrow }}}$

Mass of the stone [m] = 0.25 kg

Radius  [r] = 1.5m

Revolution= 40 rev./min

$\boxed{\bf{\red{Formulas \ used \longrightarrow }}}$

$\sf{\green{ \omega =2\pi n}}$

$\sf{\green{T=m \omega^2r}}$

where,

ω = Angular velocity.

n= no. of revolution.

T= Tension in the string.

m= Mass.

$\boxed{\bf{\orange{SoLution \longrightarrow }}}$

Number of revolution per second, $\sf{n=40/60=2/3}$

The centripetal force on stone is provided by the tension T, in the string,

that means,

$\sf{T=m \omega^2r}$

Put values in the formula we get,

$\sf{ T =0.25\times 1.5\times (2\times 3.14\times \dfrac{2}{3} )^2 }$

$\bf{ T=6.57 N}$

_________________________

Maximum tension in the string,  Tension maximum = 200 N

$\sf{ T_{max}= \dfrac{mv^2_{max}}{r}}$

$\sf{ v_{max} = (\dfrac{T_{max} \ r }{m} )^\dfrac{1}{2} }$

$\sf{ v_{max} = ( \dfrac{200 \times 1.5}{0.25}) ^\dfrac{1}{2} }$

$\sf{v_{max} = ( \dfrac {300}{0.25} )^\dfrac{1}{2}}$

$\sf{v_{max}= (1200)^\dfrac{1}{2} }$

$\sf {v_{max}= 34.64}$

$\bf{\underline{\therefore, max.\ speed\ of\ the\ stone\ is\ 34.64 m/s.}}$

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