IIT JEE , AIEEE Forum - Physics , Mechanics

chandu.coool

A ring of mass m can slide over a rough vertical rod with coe of friction

as shown.The ring is connected to a spring of force constant k=4mg/R . 2R is the natural length of the spring. The other end of the spring is fixed to the groung at a horizontal distance 2R. from the base of the rod. If the mass is released at a height of 1.5 R , and the velocity on reaching the ground is

(3gr) . Find

.

ans = 0.549

hhitesh_1

hey is thi9s 3R/2 the length of the rod

hhitesh_1

just explain the question a bit more

vivsarda

jst energy balance will give u the answer yaar!!!
intial energy = final enegy
1/2kx sqr + mgh = 1/2mvsqr
x is the extension in the string!!!
solve the eqn to get the ans!!
hope it help!!!
RATE IF USEFUL!!!

vivsarda

sry yaar made mistake dint read the question actaully the loss in energy is WD by friction
With that u can get the value of myu
SRY YAAR!!!
NEWAYS NE DOUBTS NUDGE ME!!!

feynmann

See , it is a long prob . So hang on

Let the ring is at a distance x above the gnd

So extension in the spring = sqrt( x ^2 + ( 2 R ) ^2 ) - 2R

Tension in the spring = 4mg/ R * extension

Let the angle made by the spring to the horizontal be @

So , The comp Fsin @ adds the accelaration of the ring ( along with mg ) & comp. Fcos @ gives the normal force on the spring . So the frictional force is

F cos @

Now , from geometry

cos @ = 2R / sqrt( x^2 + ( 2R)^2 )

sin @ = x / sqrt( x^2 + ( 2R ) ^2 )

Now the eqn of motion for the ring is

mv dv/dx = mg - Fcos @ + F sin @

after putting the values and on simplification this becomes v dv = g dx - 8 g { dx - 2r dx / sqrt ( x^2 + (2R)^2 } + 4g/R { x dx - 2R x dx/ sqrt ( x^2 + ( 2R)^2 }

integrating the above from 0 to 3 R / 2 and putting v^2 = 3 g R we get

= 1 / { 16 ( 1.5 - ln 4 ) } = 0.549

feynmann

feynmann

Got it ?

bhupesh

Dear

your friend have replied it correctly , I will explain it a little . You can sole the question usng Work enrgy theorem ( Try solving , i.e. work done due to all the forces must be equal to the change in K.E ) Forces acting

( 1 ) Friction ( Mu time normal force , normal force on the ring is due to the horizontal component of spring fforce .

( 2 ) Gravitational force acting vertically down ward .

( 3 ) Vertical component of spring force which will add on to gravitational force ( i.. mg of the ring )

Point yo be noted is that friction force will be variable so you have to do an integration to find out the work done due to friction .

** Friction force is variable since it depends on the normal force which in this case is provided by the spring force which in turn is changing due to the changing length of the spring .

Try solving you will get the answer .

karthik2007

Excellent work feymann

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