A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?

Asked by Abhisek | 1 year ago |  129

##### Solution :-

Mass of the man, m = 70 kg

Radius of the drum, r = 3 m

Coefficient of friction between the wall and his clothing, μ = 0.15

Number of revolution of hollow cylindrical drum = 200 rev/min

$$\dfrac{200}{60}$$$$\dfrac{10}{3}$$ rev/s

The centripetal force required is provided by the normal N of the wall on the man

N =$$\dfrac{mv^2}{R}$$ = mω2R

When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the man (mg) acting downward is balanced by the frictional force acting vertically upwards.

The man will not fall if mg ≤ limiting frictional force fe (μN)

mg ≤ μN

mg ≤ μ (mω2R)

ω2 ≥ $$\dfrac{g}{Rμ}$$

Therefore, for minimum rotational speed of the cylinder

ω2 = $$\dfrac{g}{Rμ}$$$$\dfrac{10}{(0.15 \times 3)}$$ = 22. 2

ω =$$\sqrt{22.2}$$= 4.7 rad/s

Answered by Pragya Singh | 1 year ago

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