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Q.A man is standing on a rough(`mu` = 0.5) horizontal disc rotating with constant...

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user8235304 | Student, Grade 11 | Valedictorian

Posted July 8, 2013 at 11:51 AM via web

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Q.A man is standing on a rough(`mu` = 0.5) horizontal disc rotating with constant angular velocity of 5 rad/sec. At what distance from centre should he stand so that he does not slip on the disc?

A)  R `<=` 0.2 m

B)  R `>` 0.2 m

C)  R `>` 0.5 m

D)  R `>` 0.3 m

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llltkl | College Teacher | Valedictorian

Posted July 8, 2013 at 1:44 PM (Answer #1)

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The man feels centrifugal force due to rotational motion, while centripetal force owing to friction on the platform he is standing keeps him stable.

The moment centrifugal force due to rotation surpasses frictional resistance force, he slips out.

Centrifugal force due to rotation is given by `F=m*omega^2*r` , where, r is his distance from the center of the disc.

Frictional resistance force is given by `mu*n=mu*mg`

The man slips when `Fgt=mu*n`

In the limiting case of these two forces being equal,

`mu*m*g =m*omega^2*R`

`rArr mu*g=omega^2*R`

Plugging in the values,

`0.5*9.81=5^2*R`

`rArr R=(0.5*9.81)/25=0.1962 m = 0.2 m` (approximately)

Therefore, the man should stand within a distance of 0.2 m from the centre so that he does not slip on the disc. Correct answer is option A).

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