IIT JEE , AIEEE Forum - Physics , Mechanics - A POINT MOVES WITH A VELOCITY v = kt WHERE k = 0.5 m/s ^2 .FIND THE TOTAL ACCELERATION OF

shailesh_45

A POINT MOVES WITH A VELOCITY v = kt WHERE k = 0.5 m/s ^2 .FIND THE TOTAL ACCELERATION OF THE POINT AT THE MOMENT WHEN IT HAS COVERED THE n^thFRACTION OF THE CIRCLE AFTER BEGNING OF THE MOTION, WHERE n = 1 /10.

Decoder

acceleration..tangential is constant...
angle covered is 2pi/10... velocity depends linearly on time..
answer is pi\5....

aditi_g

@ decoder....
wat did u do???
we were supposed to calculate the net accelaration......
wat have u calculated and how??
dont u think we were supposed to find the resultant of radial and angular accelaration

rhd92781

suppose radius of circle is r
distance covered in nth fraction = 2(pi)rn = 2(pi) r/10
tangential acc = k
at that point, v^2=2as = 2k * 2(pi)r/10
hence centripetal acc = v^2/r = 4(pi)k/10

net acc = rt( k^2 + (4pik/10)^2)

shailesh_45

ans is 0.8

rhd92781

shailesh_45 try putting the values in ma ans. it comes 0.803

aditya_arora04

Hi,

Look,

Here we hav to calculate the total acc., i.e this tangential and centripetal accleration.

Tangential acceleration : dv/dt = k

Centripetal acceleration : v²/r = k²t²/ r ------ 2

Now,

Angle covered = pie/5

So,

pie/5 = 0 + 1/2 at²/r [ this equation is newton's second equation for circular motion ]

k²t²/r = 2pie/5

So, use the formula for resultant and get the answer.

Put it in 2 to et the answer

shailesh_45

i did not understand how you got centripetal accln

ramyani

centripetal force = m v²/r

centripetal accln = centripetal force / mass = v²/r

given v = kt

so, centripetal accln = k²t²/ r

aditya_arora04

Oops , i did a mistake !!!!!

Centripetal acc.is as follows :

Centripetal force = mv²/r

So, centripetal acc. = v²/r

Now, v is tangential velocity, which is kt.

So, centripetal acc. [ CA ] = k²t²/r

Now, acc. to newton's second law for circular motion.

theta = ut + 1/2 at² [ u is angular velocity here and a angular acc. ]

angular distance covered = 2pie / 10 = pie/5

u is 0, a is k/r. So, kt²/r = 2pie/5

t² = 2pie*r/5k

Put in centripetal acc.

We get : CA = 2(pie)k / 5

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