IIT JEE , AIEEE Forum - Physics , Mechanics - H.C.Verma

varun.tinkle

could anyone solve sum no 57-64

ramyani

Here is 59 solved by Biki.

let angle made =

(as shown in figure)

let v = velocity just before falling

So at an instant just before falling---

mv²/r = mg.cos

_____(1).........where m = mass, r = radius of the sphere

or v²/r = g.cos

_____(1)

again change in K.E. = (1/2)mv² - 0 ......as initial velocity = 0

and loss in P.E. mg(r - r.cos

)

So (1/2)mv²= mg(r - r.cos

)

or v²/2 = gr (1 - cos

)

or v²/r = 2g(1 - cos

)

or g.cos

= 2g - 2g.cos

or 3g.cos

= 2g

or cos

= 2/3

or

= cos^-1(2/3)

varun.tinkle

could u solve the other sums

ramyani

Here u find another of Biki. Q.56.

Consider the first figure ...

In this fig. we will find out what should be the velocity of the bob at the lowest position so that the string becomes slack ( i,e, T = 0 ) at the horizontal diameter..

Now the weight acts perpendicular to the tension at this position and so has no component along the string....

So taking v =velocity at that point and T = tension and l = length of string....

we have T = mv²/l..

Now let u = velocity of the bob at lowest position so that the bob acquires velocity = v at the horizontal diameter....

So ... 1/2.mu² = 1/2mv² + mgl

or v² = u²- 2gl

Using this in equ^n of T.... we have

T = m(u² - 2gl)/l

If string becomes slack.... T = 0

i,e, m(u²- 2gl)/l = 0

or u² - 2gl = 0

or u =

2gl

So we see that if velocity at lowest position should be equal to

2gl such that the string becomes slack at the horizontal diameter...

And see HCV-1 page 128 , example 8 where you will find that the velocity at lowesrt position such that the bob complets a full circle is 5gl .

So we conclude that.....

if velocity is less than 2gl, then string becomes slack below the horizontal diameter.

if velocity = 2gl, then string becomes slack at horizontal diameter..

if velocity is greater than 2gl but less than 5gl, then string becomes slack in the region between the horizontal diameter and the topmost position....

Taking this into consideration.... the question says that in the question the string becomes slack between the horizontal diameter and the topmost position...as velocity =

3gl...

Now consider fig. 2

Let velocity at lowest position = u =

3gl

velocity at the point of slackening = v

So ...

taking the angles as shown in fig.

we write.....

T + mg.cos

= mv²/l

for string to become slack ... T = 0

i,e, mg.cos

= mv²/l

or gl.cos

= v² _________ (1)

And using energy conservation .....

1/2mu²= 1/2mv² + mg( l + l.cos

)

i,e, v² = u² - 2gl ( 1 + cos

)

or v² = 3gl - 2gl - 2gl.cos

or v² = gl - 2gl.cos

________(2)

Using (2) in (1)...

gl.cos

= gl - 2gl.cos

or 3gl.cos

= gl

or cos

= (1/3)

or cos (180 -

) = (1/3) .........[ as

= 180 -

]

or - cos

= (1/3)

or cos

= (-1/3)

i,e,

= cos^-1(-1/3)

varun.tinkle

pls anyone solvew these sums

uday_zingtudor

51)Since the potential energy remains constant, Equate the energy in the spring at instant to the kinetic energy at the vertical position.

length of the spring at the given instant is h/cos37 = 5h/4

That means the elongation is h/4.

(1/2)kh²/16 = (1/2)mv²

You get the answer to be v=(h/4)(sqrt(k/m))

~Cheerio!!!

anish292

pls solve the others. neone??

shailesh_45

anyone give hint for 52

anish292

no one there??

sid.shah.90

Ok, here's for 52

The velocity after falling through height h, i.e

2gh should be equal to minimum velocity for a mass to complete a vertical loop when in horizontal position =

2gR

Here R = l

Thus, h = l

varun.tinkle

62)

a)

the work done in projecting the block up the incline=mglsin@

the work done in projecting up the sphere=change in potentiol energy of the system=mgr(1-cos@) from the geometry of the figure

1/2mv^2=mgr(1-cos@)+mglsin@

solving it

v= root 2g(r(1-cos@) +lsin@)

b)

1/2m(2v)^2=mgr(1-cos@)+mglsin@+1/2m(v1)^2

mgr(1-cos@)+mglsin@=1/2mv^2

at the top n+mg=mv^2/r

solving the above equations we get the force acting as

6mg(1-cos@+l/rsin@)

c)

let the particle make an angle@ when it leaves the sphere

mgcos@=mv^2/r

mgr(1-cos@)=1/2mv^2

solving it we get cos inverse 2/3

plz rate me if u find me useful

cheers!!!!!!

smartboy198062

51)

let the block A start loosing contact with the surface below it at A' after traveling a distance x( fig 1)
in this process the block b will shift from B to B' such that BB' =AA'=x ( as string is inextensible). and hence there is a loss of gravitational potential energy=mgx

This energy is stored in the spring which is stretched by

L and partly appears as kinetic energy of block A and block B. so by conservation of energy we have

mgx=1/2mv^{2} + 1/2mv^2+ 1/2 k(

L)^2

this gives v^2=gx- k/2m(

L)^2- --------------------1

now for vertical equilibrium of A at A'

N+Fcos

=mg
but F=k(

L)
and N=0

so k(

L)cos

=mg---------------------------------------------------------2

further by geometry we get

(

L)=L/cos

-L=L[1/cos

-1]--------------------------3

so substituting the value of (

L) from eqn 3 in 2 and solving for cos

, we get

cos