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Physics

Mechanics

Easy

Question

$A space c y l i n d e r space r e s t s space o n space a space h o r i z o n t a l space r o t a t i n g space d i s c space a s space s h o w n. space D i s tan c e space b e t w e e n space t h e space a x e s space o f space t h e space d i s c space text and cylinder is end text straight R text . Coefficient of friction between disc and cylinder is end text mu greater than D over h. text Where end text D equals 2 r text is end text text diameter of cylinder. Value of end text omega text such that cylinder falls off the disc will be end text$

$\sqrt{\frac{2 r g}{h R}}$
$\sqrt{\frac{4 r g}{h R}}$
$\sqrt{\frac{2 R g}{h r}}$
$\sqrt{\frac{r g}{h R}}$

The correct answer is: $square root of fraction numerator 2 r g over denominator h R end fraction end root$

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