There are two balls touching each other circumferencically. The radius of the big ball is 4 times the diameter of the small all. The outer small ball rotates in anticlockwise direction circumferencically over the bigger one at the rate of 16 rev/sec. The bigger wheel also rotates anticlockwise at N rev/sec. What is 'N' for the horizontal line from the centre of small wheel always is horizontal

Ans N=64

Hi,

Can you specify how did you get this N=64.

Hi

plz can u tell the procedure how u got N=64.

 i think N should be 2.

For the line joining center to be horizontal ,

both balls surface should rotate at same speed.

Circumference of small ball= pie*d

-----=--------   bigger ball= 8 *pie*d

so if you calculate if small ball is moving at 16 rpm

to keep surface horizotal big ball should be moving at 16/8= 2 rpm.

Revoultion o f small ball=16

2*22/7*r=16

r=8/(22/7)

radius of big ball=4d

=32/(22/7)

2*(22/7)R=N

R=32/(22/7)

N=2*(22/7)*32/(22/7)

N=64rev/sec

@Jone : Can you please clearify

small ball takes 16 revolutions per second
i.e in one second it moves distance = 16 * (circumference)
i.e 16 * (22/7 d)


then how are you equating circumference and no of revolutions.
2*(22/7*r) = 16 ???
I don't get it !

I calculated somehow 2 get the answer 2 b 64rps....
for small ball: Speed=dist/time big ball: Speed=Dist/time
16=2*pi*r N= 2*pi*R
but R=4r
So, N=8*pi*r
pi*r=8
Substitute pi*r for N;
N=8*8
=64rps

2*pi*r*16=2*pi*4*r*N
//as both boll will cover same distence..
N=rev of big ball
=>N=16/4=4...ans

16*8=128 is ans..

n2/n1=d2/d1
16/N=d2/(2*4*d2)
N=16*8
N=128

i thing its 2rev/sec.
smaller ball has the radius R.
bigger ball has the radius 8R.
if smaller ball travel distance=16R(as it makes 16 rev per sec),the bigger one has to make(8R+8R=16R).

therefore,the bigger one has to travel twice its own radius,so the answer is 2 rev/sec.