- #1
warmfire540
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A 30-cm diameter circular saw blade has a mass of 0.9 kg distributed uniformly in a disc.
(a) What is its rotational kinetic energy when it is operating at 4000 rpm?
(b) What average power must be applied to bring the blade from rest to its operating 4000 rpm?
Just making sure I'm doing this right...
a. Inertia=(1/2)mr^2
I=(1/2)(.9)(.15)^2
I=.0101
w=418.88
K=(1/2)Iw^2
K=(1/2)(.0101)(418.88)^2
K=886.08
b. P=average force
torque=r*F
torque=I*ang (ang being angular velocity)
so:
I*ang=r*f
F=I*ang/r
ergh...now i come to a block!
I don't know how to find acceleration without time..or how to find Force..am i on the right track here?
any help please! thanks!
(a) What is its rotational kinetic energy when it is operating at 4000 rpm?
(b) What average power must be applied to bring the blade from rest to its operating 4000 rpm?
Just making sure I'm doing this right...
a. Inertia=(1/2)mr^2
I=(1/2)(.9)(.15)^2
I=.0101
w=418.88
K=(1/2)Iw^2
K=(1/2)(.0101)(418.88)^2
K=886.08
b. P=average force
torque=r*F
torque=I*ang (ang being angular velocity)
so:
I*ang=r*f
F=I*ang/r
ergh...now i come to a block!
I don't know how to find acceleration without time..or how to find Force..am i on the right track here?
any help please! thanks!