IIT JEE , AIEEE Forum - Physics , Mechanics

waterdemon

A Solid sphere of Radius 'R' rolls without slipping in a cylindrical vessel

of Radius '5R' . Find the time period of small oscillations ?

Explain each and every step and especially the answer part.

maverick83

when sphere is at anglr thita position then resolve mg into two components such that mgcos

balance the normal reaction and mgsin

acting towards the mean position
let f= mgsin

when

is small sin

=

f=mg

=mg(-x)/4R (because mg acts at center of mass)........(1)
as f

-x (x is arc )
therefore f is restoring force
f=-kx......... (2)
compare 1 and 2
and solve it....

waterdemon

Give the answer complete.

Solve it till the end only then will I rate you.

Cheers !!!!!!!!!!!!!!!

maverick83

solve it or leave it ,i don't want any rating from you ........
bbye........

waterdemon

You are just so offensive .

Dude You don't want the rating but you can atleast solve it complete

Never leave a qyestion half solved.

Cheers !!!!!!!!!!!!!!!!

karthik2007

If you want the complete solution, here it is. maverick has solved most of it anyway.

Let the ball be at some small distance x from the mean position. Now, we have to find the restoring force. The only force acting on the ball is force due to gravity (component). Resolving, we get the force as mgsin

which restores the ball to the mean position, and mgcos

which balances N.

Since you have given in the problem that the oscillations are small, sin

=

.

In the figure, we get sin

= x/(5R-R) = x/4R. (Reason for this subtraction is because mg acts only at COM)

so the force required is f = -mgx/4R. => f

x.

=> a = f/m = -gx/4R

Now, you know that any simple harmonic motion can be represented by a = -w²y
(where w = 2

/T) and y is the displacement.

Comparing, we get w² = g/4R = > w = ^{[ ]}

g/4R => 2

/T =

g/4R

Thus, 1/T =

g/4R / 2

or T = 2

4R/g.

waterdemon

No problems Maverick just wanted to tell you that you were going wrong .

Hey Karhtik just saw that you had replied , so am editing watch the

answer I have solved it.

So here is the solution

For pire rolling to take place

V = R

₀ = Angular Velocity of centre of mass of sphere about point O.

₀ = V/4R = R / 4R = /R

₀ = /R

d₀/dt = 1/4 [ d/dt ]

₀ = /4

For pure rolling

= a / 4

a = (gSin) / [1+ I/MR²]

Since I = 2MR²/5

a = 5gSin/7

₀= 5gSin/28R

For Small oscillations

Sin =

₀ = (-5g/28R)

T = 2 28R/5g

Hence the answer .

Hope you got it Maverick.

Rate if you liked it.

Cheers !!!!!!!!!!!!!!

maverick83

Here the complete solution but check if it is right or wrong.....

when sphere is at anglr thita position then resolve mg into two components such that mgcos

balance the normal reaction and mgsin

acting towards the mean position
let f= mgsin

when

is small sin

=

f=mg

=mg(-x)/4R (because mg acts at center of mass)........(1)
as f

-x (x is arc )
therefore f is restoring force
f=-kx......... (2)
comparing 1 & 2
we find k=mg/4R.................(3)
as f = ma (a= acceleration)
as in SHM a= -

²x =-kx/m
therefore

²= k/m
as

=2pi/T
T=

/2pi
T=(1/2pi)^{[ 2]}

m/k
by putting value of k from equation .....3
we get T = (1/2pi)

4R/g

cutepooja

if u knew da soltn then were u askin to test others thts wrong

maverick83

OK NICE SOLUTION WATER DEMON I HAVE GIVEN YOU THE SALUTE......
..

waterdemon

Puja you are so wrong if you think that I new the solution and put

up the question for testing others.

No , I do it everyone does it.

When I tried solving this question I made many tries and got the

Question right after attempting it about 5-6 tries.

Just wanted to test where I stand.

So don't think so.

Cheers !!!!!!!!

nick

fine work here is another

let w1 be the angular velocity of small sphere ,

w2 be the angular velocity of big sphere.

now using the centre of mass eq.

w1R=w2(5R-R) (think over this people....).......................1

now writing the energy eq.

1/2 Iw1^2 +1/2mv1^2 +mg(5R-R)(1-cos@)= k

here w1 is the angular velocity of small body
v1 is the linear velocity
@ is theta
k is a constant as energy cannot be destroyed

now substituting values
I=2/5mR^2
w1=w2(5R-R)/R^2....................from 1
v1=w2(5R-R) ( as it is W.R.T. the big sphere)

putting the values in the above expression and simplifying it,
we have

1/5 (w2)^2(4R)^2 + 1/2(w2)^2(4R)^2 +g(4R)-g4Rcos@=k

now as this is conservative force ,d/dt of the expression will be 0........

hence differentiating(very easily)
(g)(4R) will become 0
w2^2 will be 2w2 a' where a' is angular acc.
sin@==@(very small)
.....

the final eq. is
28/5a'=g@

a'=5g/28R @

there fore

T=2pi root 28R/5g

which is the ans... try using the diiff. method whenever possible....
rate ma effortsssssssssssssss

Ask & Discuss Questions with Community & Experts