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

Showing Answers 1 - 7 of 7 Answers

seetha

  • Apr 1st, 2006
 

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

  Was this answer useful?  Yes

Rahul Chandrashekar

  • Apr 11th, 2007
 

According to my calculation's the answer should be 2

Reason :

r1 = 8 r2

distance travelled in 16 revoulutions for 1 sec =
16 (2pi r2)

r2 = (1/8) * r1

sub in the equation

hence we get 2(2 pi r1) is the distance travelled by the big ball in 1; hence it rotates 2 times in one second; so N = 2

Here w (angular velocity is given) which is inversely proportional to radius
r.


i.e w α 1/r


Let w1,r1,w2,r2 be measures of Small circle and big circle respectively
hence, w1 α 1/r1......................(1)
& w2 α 1/r2....................(2)
taking ratio of eqn 1 & 2we get
w1/w2=r2/r1
hence 16/N=8*r1/r1.............(r2=8r1)
on solving,
N=2


  Was this answer useful?  Yes

Give your answer:

If you think the above answer is not correct, Please select a reason and add your answer below.

 

Related Answered Questions

 

Related Open Questions