Numerical - Radius of gyration, given periodic time - Theory of Machines - 1

Q.  A connecting rod has periodic time of 1.5 sec/cycle for one oscillation. Calculate radius of gyration, when centre of gravity is at a distance of 500 mm from the small end.
- Published on 06 Aug 15

a. 0.1325 m
b. 0.1719 m
c. 0.0295 m
d. 0.555 m

ANSWER: 0.1719 m

Related Content

 

    Discussion

  • Sravanthi   -Posted on 19 Oct 15
    Given: 500 mm is the distance of centre of gravity from small end, period of oscillation = 1.5 sec/cycle

    Periodic time of oscillation is given as tp = 2π √ [(k2 + l2) / gl]

    Substituting the given values, we get

    1.5 = 2 π [√ [(k2 + l2) / gl]
    1.5 = 2 π [√ [(k2 + 0.52) / 9.81 x 0.5]

    Radius of gyration (k) = 0.1719 mm

➨ Post your comment / Share knowledge


Enter the code shown above:

(Note: If you cannot read the numbers in the above image, reload the page to generate a new one.)
▲