Question

plz solve this question

Answer 1

hey mate
here's the solution

Answer 2

Hi,

friend this prob is easy....

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just read through my explaination...... and do c the attachment.... :)

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for this body falling from a rotating wheel( equ. to pully) ....

there is a torque acting on the wheel ... i.e; the weight of the body....

due to which the wheel rotates and gives extra acceleration to the body in addition to acc. due to gravity

and here as u can see... axis of rotation is about an axis tangent to the wheel perpendicular to its plane.....

therefore ; new m.o.i = I + mr^{2}[/tex]

if I was the m.o.i of wheel initially;...

Sol.

τ = I α and a = r $\alpha$

where

τ = torque (force x ⊥ dist. frm axis)

I = M.O.I

α = angular acceleration

a = linear acc.

r= perp. dist. frm axis. of rot.

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mg x r = (I + mr^{2}[/tex]) x a/r

solving for 'a'....

we get

a = $\frac{mgr^2}{I+mr^2}$

and using ;

v^{2} = u^{2} + 2 a x h ( as body moves thorugh height h)

we get v= $(\frac{2mgr^2h}{I+mr^2} )^{1/2}$

and as we already knw ...

v= r ω ( where ω is the angular velocity)

we get

ω = $( \frac{2mgh}{I+mr^2}) ^{1/2}$

which is the answer

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Ps. enjoy ;)