Question

A mosquito is sitting on an LP record disc rotating on a turntable at 33⅓ revolutions per minutes. The distance of the mosquito from the center of the turntable is 10 cm. Show that the friction coefficient between the record and the mosquito is greater than π²/81. Take g= 10 m/s².

Answer 1

No. of Revolutions made in 1 minutes = 100/3

∴ Frequency = 100/180 s⁻¹

⇒ Frequency = 5/9 s⁻¹

∴ ω = 2πf = 3.49

Distance or radius = 10 cm = 0.1 m.

In such cases,

f ≥ mv²/r

⇒ μmg ≥ mv²/r

⇒ μg ≥  v²/r

μ ≥  v²/rg

⇒ μ ≥ ω²rg

⇒ μ ≥ (3.49)² × (0.1) × 10

⇒ μ ≥ 12.19

On the other hand,

π²/81 = 0.1219

∵ μ  is greater than this value.

Hence we can say that we have proved the given Condition .

Hope it helps.

Answer 2

HEYA_THERE♥

UA ANSWER

Angular speed of the mosquito,
ω=33⅓ rpm =33⅓x2π/60 =200π/180 =10π/9 rad/s.  
Radius, r=10 cm =0.10 m.   
Outward force on the mosquito =mω²r   (where 'm' is mass of the mosquito).
=m.(100π²/81)x0.10=10π²m/81
Weight of mosquito = mg = 10m N =Normal force
So Limiting frictional force
=µ.10mN 

IT WILL HELP♥

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