Question

A Spaceman in training is rotated in a seat at the end of a horizontal arm of length 5m. If he can withstand acceleration upto 9 g then what is the maximum number of revolutions per second permissible ? (Take g = 10 m/s2)
(1) 13.5 rev/s (2) 1.35 rev/s (3) 0.675 rev/s (4) 6.75​

Answer 1

Explanation:

It is given that,

Length of the horizontal arm, l = r = 5 m

Centripetal acceleration, $a=9g=90\ m/s^2$

We need to find the maximum number of revolutions per second permissible. The formula for centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

$v=\sqrt{ar}$

$v=\sqrt{90\times 5}$

v = 21.21 m/s

One revolution is equal to, $d=2\pi r$

$d=2\pi \times 5=31.41\ m$

The time taken for one complete revolution is :

$t=\dfrac{d}{v}$

$t=\dfrac{31.41}{21.21}$

t = 1.48 s

So, number of revolutions is, $n=\dfrac{1}{t}$

$n=\dfrac{1}{1.48}=0.675\ rev/s$

So, the maximum number of revolutions per second is 0.675. Hence, this is the required solution.

Answer 2

Answer:

n = 0.675 rev/sec

Explanation:

Given: r = 5m, a = 9g = 90 m/s²

To find: No. of revolutions (n) = ?

Solⁿ: since acceleration will be centripetal acceleration.

Therefore,

a

$a_{c} \: = \: \frac{ {v}^{2} }{r}$

therefore

$v \: = \sqrt{a_{c} \: \times r } \\ = \: \sqrt{90 \times 5} \\ = \sqrt{450} \\ v = 21.21$

since v = rw

therefore,

$w \: = \frac{v}{r} \\ = \frac{21.21}{5} \\ = 4.242 \:$

now as w = 2πn

therefore

$n \: = \frac{w}{2\pi} \\ = \ \frac{4.242}{6 .28} \\ = 0.675 \: \: \: rev /sec$

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