Question #27224

A hollow cylinder of radius 10 cm rotates about its axis which is vertical. A small body remains in contact with the inner wall if the frequency of rotation is 200 per minute but falls at lower frequencies. Find the co-efficient of friction between the body and the cylinde

Expert's answer

QUESTION:

A hollow cylinder of radius 10cm10\mathrm{cm} rotates about its axis which is vertical. A small body remains in contact with the inner wall if the frequency of rotation is 200 per minute but falls at lower frequencies. Find the co-efficient of friction between the body and the cylinder

SOLUTION:

Let's draw a sketch:



According to the Newton's second law of motion:


{FN=ma(projection onto x axis)Ffrmg=0(projection onto y axis)Ffr=μFNa=v2R=2v2d=2(2πvd/2)2d=2π2v2d{FN=2mπ2v2dFfr=mgFfr=μFN\left\{ \begin{array}{l} F _ {N} = m a \text {(projection onto } x \text { axis)} \\ F _ {f r} - m g = 0 \text {(projection onto } y \text { axis)} \\ F _ {f r} = \mu F _ {N} \\ a = \frac {v ^ {2}}{R} = \frac {2 v ^ {2}}{d} = \frac {2 (2 \pi v d / 2) ^ {2}}{d} = 2 \pi^ {2} v ^ {2} d \end{array} \right. \Rightarrow \left\{ \begin{array}{l} F _ {N} = 2 m \pi^ {2} v ^ {2} d \\ F _ {f r} = m g \\ F _ {f r} = \mu F _ {N} \end{array} \right. \Rightarrowmg=μFNm g = \mu F _ {N}mg=μ2mπ2v2dm g = \mu \cdot 2 m \pi^ {2} v ^ {2} dμ=g2π2v2d\mu = \frac {g}{2 \pi^ {2} v ^ {2} d}μ=0.45\mu = 0. 4 5


ANSWER:

0.45


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