Thank yo very much diazona.
"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...
cylinder weight: 17000 lb weight , Diameter = 36"= 3 ft, length = 5 ft. , 3000 rpm , time=90 sec.
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?
See I done this calculation...
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?
can u help me from starting?
If torque is in lbf·ft and rotational speed in revolutions per minute, the equation gives power in ft·lbf/min. The horsepower form of the equation is then derived by applying the conversion factor 33000 ft·lbf/min per horsepower:
hp = torque x rpm / 5252
Hi Tiny tim ,
Hp = Torque x RPM / 5252
= 66555x 2x 3.14x 3000 / 5252 = 3979 .10 hp
is that correce ?
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 ?
Thank you...
Hi Tiny tim and omcheeto ,
Thanks for reply. so i am confusing where can i use that angular acce .
If I use in Torque = Moment of inertia x Alpha (angular accee)
= 1/2(Radiusx radius 0x Weight X 3.48 Rad/sec^2
=...