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a 70 kg man leans against a hollow cylindrical wall with radius 3m.coefficient of friction between clothing and wall is 0.15.find minimum speed of rotation of cylinder so that the man doesnt fall when the floor is removed?

pk bose , 13 Years ago
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anser 3 Answers
rupali sehgal

Taking the man to be the system,

Normal force supplies the necessary centripetal force.

therefore, N=mv2/R

Also, to ensure that the person does not fall, frictional force should be equal to the person''s weight.

therefore, F= mg

=> µs.N=mg

from (1)

µs.mv2/R=mg

µs(Rω)2/R=g                 {v=Rω}

(0.15)(3)(ω)2 =10

ω2= 10/0.45= 22.22

ω = 4.71 rad/sec2

 

Hope it helps. Do approve if satisfied by the answer!

All the best!

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Last Activity: 13 Years ago
pk bose

thanx a lot !! helped me a lot ! thank u!

Last Activity: 13 Years ago
rupali sehgal

Your welcome!

Last Activity: 13 Years ago
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