Question

A ball 3 cm in diameter and 300 g in weight is attached to the
end of a thin string 46 cm long. If it is rotated uniformly in a
horizontal circle at the rate of 15 revolutions/s, what is the
tension in the string?

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I will mark it as brainliest..​

Answer 1

Answer:

Mass, m=0.2 kg

Radius, r=1 m

Angular speed, ω=90 rpm

⟹ω=

60

90

⟹ω=1.5 rps

⟹ω=3π rad/s

(a) Angular velocity, ω=3π rad/s

(b) Centripetal acceleration. a=rw

2

⟹a=1×3π×3π

⟹a=9π

2

m/s

2

(c) Centripetal Force, F=ma

⟹F=0.2×9π

2

⟹F=1.8π

2

N

(d) Tension in string, = Centripetal Force

⟹T=F=1.8π

2

N

Answer 2

Given:

A ball of 3 cm diameter and 300 grams of weight is attached to the end of a string and rotated in a horizontal circle at the rate of 15 revolutions per second.

To find:

Tension in the string.

Calculation:

The tensionin the stream will be equal to the centripetal force experienced by the ball during horizantal circular trajectory.

$\rm{ \therefore \: T = \dfrac{m {v}^{2} }{r} }$

$\rm{ = > \: T = \dfrac{m { (\omega r)}^{2} }{r} }$

$\rm{ = > \: T = \dfrac{m { \omega}^{2} {r}^{2} }{r} }$

$\rm{ = > \: T = m { \omega}^{2} r}$

$\rm{ = > \: T = \dfrac{300}{1000} \times { ( 15 \times 2\pi)}^{2} \times \dfrac{(46 + 1.5)}{100}}$

$\rm{ = > \: T = 0.3\times { (30\pi)}^{2} \times 47.5}$

$\rm{ = > \: T = 0.3\times 900 {(\pi)}^{2} \times 0.475}$

$\rm{ = > \: T = 12.825\times {(\pi)}^{2} }$

$\rm{ = > \: T = 126.57 \:N }$

$\rm{ = > \: T = 0.1265 \:kN }$

So, final answer is:

$\boxed{\bf{ \: T = 0.1265 \:kN }}$