OK, let's recap what you've done so far:
niket548 said:
the power needed to accelerate a cylinder of weight 17000 lb from 0 to 3000 rpm in 90 seconds, exerting a force on the edge of the cylinder, 1.5 ft from the axle?
You found the angular acceleration required,
niket548 said:
Aceee = Change in angular velocity / time
Angular velocity = 2xPi (3.14) 3000/60 = 314rad / sec
acee = ( 314-0) / 90
= 3.48 Rad / sec2.
3000\text{ rpm}\times\frac{2\pi\text{ rad}}{\text{rev}}\times\frac{\text{min}}{60\text{ s}} = 314\frac{\text{rad}}{\text{s}}
\alpha = 314\frac{\text{rad}}{\text{s}} / 90\text{ s} = 3.5\frac{\text{rad}}{\text{s}^2}
(here we're using lb as a unit of mass)
Then you found the torque required,
niket548 said:
If I use in Torque = Moment of inertia x Alpha (angular accee)
= 1/2(Radiusx radius 0x Weight X 3.48 Rad/sec^2
= 1/2x1.5x1.5x17000x3.48
= 66555 lb-ft^2 / sec^2
\tau = \frac{1}{2}MR^2\alpha = \frac{1}{2}(17000\text{ lb})(1.5\text{ ft})^2\left(3.49\frac{\text{rad}}{\text{s}^2}\right) = 67000\frac{\text{lb ft}^2}{\text{s}^2}
(note that the radian is a "dummy unit" - it's really just equal to 1, so you can remove it)
Then you attempted to find the power by multiplying torque by angular speed, but you forgot about what OmCheeto pointed out:
OmCheeto said:
I would imagine it is your "Power = torque x omega" that has thrown things off.
Omega changes from zero to 314 rad/sec over the 90 seconds.
"Power = torque x omega" may only be good for steady state conditions.
What you need to do is multiply the torque by the angular speed at each individual time and add up all those contributions to the torque - in other words, do an integral. In this
particular case, though, since the torque is constant, you can just find the
average angular speed and multiply that by the torque. But don't use the final angular speed!
niket548 said:
becoz i find torqe = i x alpha
= lb-ft^2 X Rad /Sec^2
Torque unit is lb.ft
Now i am cofusing about units ?
Remember that we are using lb as a unit of mass! Torque has to have a unit of
\frac{(\text{mass})(\text{distance})^2}{(\text{time})^2}
Notice that lb*ft doesn't work, since that's just mass*distance. You need lb*ft^2/s^2.
In some other places you may see people write torque in units of lb*ft (foot-pounds), but what they really mean is lbf*ft, foot-pounds-of-force. The lbf is a unit of force,
\text{lbf} = 32.2\frac{\text{lb}\cdot\text{ft}}{\text{s}^2}
In order to calculate the power required in units of horsepower, you will need to multiply the torque,
67000\frac{\text{lb ft}^2}{\text{s}^2}
by the correct angular speed. There are two ways to do this. Either you could leave the angular speed in revolutions per minute, then you will get an answer of
\text{(number)}\frac{\text{lb ft}^2}{\text{s}^2}\times\frac{\text{rev}}{\text{min}}
You can then use the conversion factor
\text{hp} = 5252\text{lbf}\cdot\text{ft}\times\frac{\text{rev}}{\text{min}}
You will
also need to use the definition of the foot-pound-of-force (four equations up) as a conversion factor.
The alternative way is to convert the angular speed into radians per second. If you do that, you will get an answer of
\text{(number)}\frac{\text{lb ft}^2}{\text{s}^3}
(remember that the radian is a "dummy unit"). From here, you can use the conversion factor
\text{hp} = 17696\frac{\text{lb ft}^2}{\text{s}^3}
which you could find from
Google, among other sources.
