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Q.A man is standing on a rough(`mu` = 0.5) horizontal disc rotating with constant...
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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,
Plugging in the values,
`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).
Posted by llltkl on July 8, 2013 at 1:44 PM (Answer #1)
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