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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?

```
8 years ago

rupali sehgal
19 Points
```
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

Hope it helps. Do approve if satisfied by the answer!
All the best!
```
8 years ago
pk bose
4 Points
```							thanx a lot !! helped me a lot ! thank u!
```
8 years ago
rupali sehgal
19 Points
```							Your welcome!
```
8 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions