IIT JEE , AIEEE Forum - Physics , Mechanics

rakesh61

A uniform pole 50 kg is placed vertically on a rough surface of

=0.3.Lenght of pole is 8 m.A 500 N horizontal force is given to the pole suddenly at a height of 2m above the ground.Determine the acceleration of the topmost point of the pole..(g=9.8)....

i promise to rate you

catch_arnnie

i'm not 100% sure whether its correct, but this is how i've done the solution

solution::

here, i've taken rotation about the lower most point, so, I = mL²/3

let F = 500 N acting on rod at height of 2m from ground

let f be the frictional force

=> f =

m g = 0.3 50 9.8 = 147 N

writing the torque equation::

(F-f) * 2 = mL²/3

where

is the angular acceleration & L is length of the rod

(353 * 6)/(50 * 64) =

=>

= (353 * 6)/(50 * 64)

let acceleration of the topmost point be a

=> a = L

=> a = 8 *

=> a = (353 * 6)/(50 * 8)

=> a = 5.295 m/s²

plz correct me if i'm wrong anywhere...

nick

well according to me,
let us consider torque about the point in contact with the ground
this is to avvoid taking frictional force,

now, F*2=I@ @ is angular acc.

I=ml^2/3

500*2=50*64/3 @
therefore @= 15/16

at the topmost point, a=L@=8*15/16=7.5m/s^2

i may be wrong but tell me my mistake...

waterdemon

firstly

f=

mg=0.3*50*10=150N

then

(F-f)*2 = I@

(F-f)*2 = ml^2/@

350*2 = 50*8*8*@

700/400*8 = @

7/4*8 = @

@=7/32

Now

accn at the top will be

a = L@

a = 8*@

a = 8*7/32

a = 7/4

a = 1.75 m/s^2

see if I am right or not.

splashprabhu

all u guys r forgettin linear acceleration!!!

a(top) = a - @L/2....in d direction of force...... a = linear accn. of centre of mass & @ = angular accn abt d centre of mass

i'm takin g = 10..for ease of calculation

a = (F-

mg)m = (500 - 0.3*50*10)/50 = 7 m/sec^2

@ = (F*2 -f*4) / I ....I = m(L^2)/12 =

@ = (1000 - 600)*12/(50*64) = 3/2

a(top) = 7 - 4*3/2 = 1 m/sec^2 ...in d direction of force

hope i get rated higher..lol

splashprabhu

a small correction

ma = (F - f)

f is d frictional force...

rohith291991

is the answer 1 m/s² in the direction of the force??

splashprabhu

i edited my post a bit..... now its absolutely rite!! try n find a mistake in dat ..

ya rohith u r rite!!

rohith291991

since the net force is 350N... the bottom point is NOT hinged and the body undergoes both rotation and translation...ok in the frame of center of mass.. the net torque acting=500x2 clockwise - 350x4 anti clockwise.. =400Nm clockwise... but I

=

therefore

=

/I=(400*12/50*8²)=3/2...therefore in center of mass frame the acc of top most point is 4x(3/2)=6 m/s²against the direction of the force ...now with respect to ground frame.. the center of mass has a acc of 350/50=7m/s² in the direction of the force therefore the net acc =7-6=1m/s² in the direction of the force.. if im wrong someone PLZ point out..

spt_001

ans:-0.88

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