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
Discussion:
23 comments Page 1 of 3.
Inayat Ullah Kakar said:
9 months ago
Formula:
H = w^2× r^2/2g
H = 40cm=0.4m
r = √5,
r^2 = 0.0005m.
Rearrange the above equation:
w^2 = H×2×g/r^2
= 0.4×2×10/0.0005
= 16000
w = √16000.
w = 126 rad/ minutes.
Now rad/sec;
w = 126/60
w= 2 rad/sec
So, the given answer is correct.
H = w^2× r^2/2g
H = 40cm=0.4m
r = √5,
r^2 = 0.0005m.
Rearrange the above equation:
w^2 = H×2×g/r^2
= 0.4×2×10/0.0005
= 16000
w = √16000.
w = 126 rad/ minutes.
Now rad/sec;
w = 126/60
w= 2 rad/sec
So, the given answer is correct.
BILAL KHAN Zimri said:
3 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.
After solving. w= 125.28 rad/sec.
And Just covert the units. 125.28 is in rad/min. Divide by 60 to get 2 rad/sec, we will get the answer.
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.
After solving. w= 125.28 rad/sec.
And Just covert the units. 125.28 is in rad/min. Divide by 60 to get 2 rad/sec, we will get the answer.
(3)
Deepankar said:
4 years ago
According to my knowledge, the answer is 125.28 rad/sec.
(1)
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.
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.
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).
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).
KANYA said:
6 years ago
@Kiran, 12.28÷60 = 0.2 not 2.
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.
12.28/60 ~ 2 rad/sec.
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.
(40) =(angular velocity ^2 * (sqrt(5)) ^2) /(2*9.81).
Angular velocity = 2rad/sec.
Muhammad Saqib said:
6 years ago
Explain how 2rad/sec correct? Please explain me.
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.
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