68 - Rotational mechanics Questions Answers

IN THE PREVIOUS QUES ( THAT WAS FROM ROTATn TOPIC , SUBMITTED 5 DAYS AGO) , I AM STILL NOT GETTING THE SOLn  SUBMITTED BY YOU.

Asked By: SARIKA SHARMA
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Joshi sir comment

if this is the question based on number of rotations then read it carefully

a coin placed on a gramophone record rotating at 33rpm flies off d record , if it is placed at a dist. of more than 16cmfrom d axis of rotation . If d record is revolving at 66rpm , the coin will fly off if it is placed a distance not less than ?

ans=4cm

Asked By: SARIKA SHARMA
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Joshi sir comment

coin will fly off, if it will gain the required centrifugal

so apply mrω2 = mRW2

 

FROM REST A BODY ROTATE  120 rotation/min  FOR 20 sec THEN IT ROTATE FOR 5sec  WITH SAME  ROTATION . FIND  HOW MANY ROTATn IT WILL MAKE?

Asked By: SARIKA SHARMA
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Joshi sir comment

ω = ω0+ αt

so 4π = 0 + α20       so α = π/5

so total angular displacement = 1/2 π/5 202 = 40 π  = 20 rotation

and with angular velocity 2 rotation/sec total rotations in 5 sec. = 10 

so answer will be 20+10 = 30

IN THAT QUES OF CENTRE OF MASS BOTH THE OPTION ARE GIVEN THEN  IN EXAM HOW WE KNOW THAT WHICH ONE IS CORRECT. AS WE USE Vm  /M+m =V OF PLANK & m= mass of man

Asked By: SARIKA
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Joshi sir comment

I could not understand, what the question is,

A SYSTEM OF m1 &m2 THE PARTICLE m1 IS PUSHED TOWARDS THE CM OF PARTICLES THRU DIST. d BY WHAT DIST. WOULD m2 MOVE SO AS TO KEEP MASS CENTRE OF PARTICLES AT THE ORIGINAL POSITION

SIR, I WANT TO KNOW Y ans is m1d/m2  WHY NOT m1d/ m1+m2?

Asked By: SARIKA
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Joshi sir comment

with centre of mass as reference point m1d = m2x so x can be calculated.

sir,  can u please post a topic on rigid body rotation ? mainly heip to solve question please.  with some examples.

Asked By: SHUBHAM VED
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Joshi sir comment

Yes, I will do very soon.

Why earth moves in orbital motion around the sun ...? why not in circular motion...? 

Asked By: HARISH
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Joshi sir comment

Since every heavenly body like sun, earth, moon are rotating freely in free space, there is no body which is placed on a base, so rotation of every body depends on all the surrounding bodies. It means its track will also depend on the attraction forces given by all the surrounding bodies. 

Two cylinders are twisted by the same external torque.If the ratios of their modulous of rigidity , radii & lengths are 1:2, 2:3 & 3:4 respectively. Then find the ratio of  angle of twist ??

Asked By: AMIT DAS
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Joshi sir comment

Let the angle rotated due to torque = θ and angle of twist (shear) = δ

then arc covered = rθ   (here r is radius)

now for twist of wire δ = rθ/l  (here l is length)

according to formula η = F/Aδ  and torque ζ = kFr    (here k is constant of proportion)

we get ζ = kηAδr 

Use the formula and get the answer, you have to calculate ratio of δ's so k is not required

 

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