At t = 0 a grinding wheel has an angular velocity of 26.5 rad/s. It has a constant angular acceleration of 25.0 rad/s2 until a circuit breaker trips at t = 2.17 s. From then on, it turns through 432 rad as it coasts to a stop at constant angular acceleration.
a) when speeding up
x = w0 t + 1/2 alpha t^2 = 26.5*2.17 + 0.5*25*2.17^2=116.4 rad
so total = 116.4 + 432=548.4 rad
b) speed after speeding up = 26.5 + 25*2.17= 80.75
now find alpha
w^2 = w0^2 + 2 alpha dtheta
0 = 80.75^2 + 2*alpha*432
alpha=-7.55
so w = w0 + alpha t
t = 80.75/7.55=10.70 s
so total time is 2.17 + 10.70 = 12.87 s
c) we already found alpha = -7.55 rad/s^2
a)
Between t=0 to t=2.17 sec
Total angle turned,
x=wit+0.5*a*t^2
=26.5*2.17+0.5*25*2.17^2
=116.37 radians
Between t=2.17 and til end
Angle turned=432 radians
Total angle turned=116.37+432=548.37 radians.
b)
angular Velocity at t=2.17 sec
wf=wi+at
=26.5+25*2.17
=80.75 rad /sec
Between t= 2.17 and end
wff^2=wf^2+2*a*angle turned
0=80.75^2+2*a*432
a=7.55 rad /s^2
Time taken to stop=2.17+ 80.75/7.55
=12.86 sec
c) As alredy found in b
Acceleration=7.55 rad/s^2
a. accel is +ve from t=0 to t=2.17s
so angle turned from t=0 to 2.17s is s = u*t +
0.5*a*t^2
s = (26.5*2.17) + 0.5*25.0*(2.17)^2 = 55.5 rad + 58.86 rad = 114.36
rad
Total angle from t=0 to rest = 114.36 rad + 432 rad = 546.36 rad (3
s.f.)
c. velocity at t=2.17s is v = u + a*t = 26.5 + 25.0*2.17
m/s
v = 26.5 + 54.25 m/s = 80.75 m/s
accel from t=2.17s is negative (slowing down)
so a = -v^2 / 2s = 80.75^2 / (2*432) rad/s^2 = 7.546 rad/s^2
(negative)
b. time slowing down is v /a = 80.75 / 7.546 = 10.699
s
so time of stopping is t = 10.699s + 2.17s = 12.87s
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