Civil Engineering - Hydraulics - Discussion

Discussion Forum : Hydraulics - Section 3 (Q.No. 2)
2.
A cylindrical vessel 40 cm high is revolved about its vertical axis so that the water touches the bottom when it just spills out. If the radius of the cylinder is 5 cm, the angular velocity of rotation, is
2 rad/sec
3 rad/sec
4 rad/sec
5 rad/sec
none of these.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
23 comments Page 2 of 3.

Shubham said:   7 years ago
Formula is.

Vessel height =( angula velocity"2 * radius"2)/2g.

Deepak said:   7 years ago
According to me, it is 125.28 radian/sec.

Anurag said:   7 years ago
12.52 rad/sec is the right answer.

Vasudev said:   6 years ago
I also got 12.52. Here h equals 40 cm equals omega square x r square which is 5 cm divided by 2x 10 m per sec square.

Muhammad Saqib said:   6 years ago
Explain how 2rad/sec correct? Please explain me.

Lalit pathade said:   6 years ago
Vessel height =( angula velocity^2 * radius^2)/2g.
(40) =(angular velocity ^2 * (sqrt(5)) ^2) /(2*9.81).
Angular velocity = 2rad/sec.

Kiran said:   6 years ago
@ Deepak. You are close to the answer. Just covert the units. 125.28 is in rad/min. Divide by 60 to get 2 rad/sec.

12.28/60 ~ 2 rad/sec.

KANYA said:   6 years ago
@Kiran, 12.28÷60 = 0.2 not 2.

Sunil Padhy said:   5 years ago
We know,

If a cylinder is rotated with constant angular velocity (v)by the application of external torque. It is called forced vortex motion (V).

Or H = (v^2 x R^2 )/(2g).
Then v= π{2gH/( R^2)}.
To get rad/sec multiple (π/180).

SLR said:   5 years ago
Correct ans is 125.28 rad/sec.

Equate all in meters for ease.
The equation is:
H - y(min) = (w^2 x R^2)/(2g)
So, here y(min) =0.
Therefore 0.4 = (w^2 x .0005)/(2 x 9.81)
Solve it, you directly get an answer in rad/sec. There's nothing to convert.
After solving. w= 125.28 rad/sec.

You need to convert to rad/sec if you get an answer in degree.
180°= π rad/sec
1°= π/180 rad/sec.


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