A machine shop grinding wheel accelerates from rest with a constant angular acceleration of 2.1rad/s^2 for 7.2s and is then brought to rest with a constant angular acceleration of -4.8 rad/s^2 .Find the total time elapsed.Find the total number of revolutions turned.
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i found the total time which it t=10.4s. but, i can not figure out how to find \[\Delta \phi\].
a(t) = 2.1
v(t) = 2.1t
s(t) = 1.05t^2
1.05 * (7.2)^2 = 54.432 rads for the first 7.2 seconds i believe
a(t) = -4.8
v(t) = -4.8t + c ; v(0) = 2.1(7.2)
v(t) = -4.8t + 15.12
s(t) = -2.4 t^2 +15.12t + c ; s(0) = 54.432
s(t) = -2.4 t^2 +15.12t +54.432 ; s(3.2) since 10.4 - 7.2 = 3.2
s(3.2) = 78.24
which makes me wonder if my idea is right :)
if so then 54.432 + 78.24 radians
which i have my doubts
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hmmm, hey, I am new in physics and I could not help but to recognize the triangle in you equation, um, could you tell me the name of it and the language it is written on so I can learn the symbols of this language and what they mean? thanks
the triangle means delta or the change in (final-initial) a certain symbol. For example if i say find (delta)x. you will just basically do x(final)-x(initial). same thing goes with (delta)v or (delta)a or any other symbols.
the velocity graphs for these are linear and create a triangle with a base from t=0 to t=10.35 with a high speed of 54.432 at t=7.2
the displacement is equal to the area of the triangle:
if we divide this by 2pi then we can determine how many radians it equals between 0 and 2pi; which is 44.8317
forget the 44, its just 44 rotations back to the beginning. this leaves us with 0.8317 radians as the change of theta from the start to the finish.
and with any luck i figured that correctly :)
always remember these things:
in rotational motion
s is displacement (angle swept in radians)
w is omega in radians
alpha ang acc in rad/s^2
so here by using eqns of uniform acc find out s (angle swept)
for 1 REVOLUTION 2*pi radians swept so for 's' radians?
division by variation method