(a) A 0.35 m diameter grinding wheel rotates at 2300 rpm. Calculate its angular velocity.
rad/s
(b) What is the linear speed of a point on the edge of the grinding wheel?
m/s
What is the acceleration of a point on the edge of the grinding wheel?
Magnitude
m/s2
2006-08-01 09:30:12 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
a)w=2Pif = 2 (3.14…) 2300/60= 241 rad/sec
b) V=2Pi R f = w R=
=241 (.35) = 84 meters/sec
c)Cenripital acceleraration a=
a=v^2/R=((84)^2)/(0.35)= 20160 meter/sec squared
I hope that helps.
2006-08-01 10:02:53 · answer #1 · answered by Edward 7 · 1⤊ 0⤋
seriously, any basic physics book would have the equations for that written out exactly how you would use them.
Open your book and look at the first equation in the rotational motion section then plug and chug the equation... easy.
2006-08-01 11:17:29 · answer #2 · answered by AresIV 4 · 0⤊ 1⤋
I think it is br549
2006-08-01 09:34:37 · answer #3 · answered by jumpinjackflash 1 · 0⤊ 1⤋