Mechanical Engineering - Hydraulics and Fluid Mechanics - Discussion
Discussion Forum : Hydraulics and Fluid Mechanics - Section 5 (Q.No. 42)
42.
The metacentric height of a ship is 0.6 m and the radius of gyration is 4 m. The time of rolling of a ship is
Discussion:
8 comments Page 1 of 1.
Amit said:
1 decade ago
Let radius of gyration: K.
Meta centric height: h.
Time period of oscillation: T.
And acceleration due to gravity: g.
Then time period of oscillation T=2πK/sqrt(gh).
Meta centric height: h.
Time period of oscillation: T.
And acceleration due to gravity: g.
Then time period of oscillation T=2πK/sqrt(gh).
(1)
VISHAL said:
1 decade ago
Time Period = 2*3.14*K / ROOT OF (g * h ).
= 2* 3.14 *4 / ROOT OF (9.81 *0.6).
= 10.38 S.
= 10.4 Sec.
= 2* 3.14 *4 / ROOT OF (9.81 *0.6).
= 10.38 S.
= 10.4 Sec.
Satya said:
1 decade ago
Time period (T) = 2*Pi*K/sqrt(gravitaional constant * metacentric height).
Masavvar said:
7 years ago
Formula :- 2π√(k^2÷hg).
where g=9.81, h=.6, k=4, π=3.14.
where g=9.81, h=.6, k=4, π=3.14.
Jayesh mali said:
7 years ago
T = 2πk/√hg.
= 2 * 3.14 * 4/√9.8 * 0.6.
= 10.4 s.
= 2 * 3.14 * 4/√9.8 * 0.6.
= 10.4 s.
Zack said:
6 years ago
Where do you all find the solutions to the problems?
K k mittal said:
5 years ago
t =2π √ k2/gh.
Nandkishor said:
7 years ago
Thank you @Amit.
(1)
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