Problems 60+

Problem No. 76

Problem No. 76

by 123iitjee.com Support -
Number of replies: 0
Chain of bicycle moves through two sprockets, a big and a small, the bigger one is towards the paddle side and the smaller one is towards the rear wheel side (refer figure).

Consider a bicycle in which the bigger sprocket has:
Teeth: N1, Radius: r1, Angular speed: \(\omega_1\)

And the smaller sprocket has:
Teeth: N2, Radius: r2, Angular speed: \(\omega_2\)

Let the linear speed of chain be v.

Find a relationship between \(\omega_1\) and \(\omega_2\).

Chain Spocket

Cracked By:
  • Subhrashis Guha Niyogi, Class 12, B. E. College Model School, Howrah (provided full answer with solution). Subhrashis has qualified in IIT JEE 2008.
  • Lokesh Sardana, 12th Pass, D.A.V. School, Fatehabad, Haryana (provided relation of \(\omega\) with r)
  • Kartik M Mankad, 11th, Delhi Public School, Vadodara (provided relation of \(\omega\) with r).
Solution

Solution provided by Subhrashis Guha Niyogi (edited version):

Since chain is moving with constant linear speed,

\(\omega_1 \times r_1= \omega_2 \times r_2\)

So, \(2\pi r_1 \times w_1= 2\pi r_2 \times w_2\)

So, \(circumference \times \omega = constant\)

For chain to fit properly on both sprockets, teeth must be equally spaced.

So, \(circumference\alpha N\)

So, \(N\times \omega = constant\)

Hence, \(N_1\omega_1=N_2\omega_2\)