Question

At t=0 a grinding wheel has an angular velocity of 21.0 rad/s . It has a constant angular acceleration of 33.0 rad/s2 until a circuit breaker trips at time t = 2.40 s . From then on, it turns through an angle 433 rad as it coasts to a stop at constant angular acceleration.

A-Through what total angle did the wheel turn between t=0 and the time it stopped?

B-At what time did it stop?

C-What was its acceleration as it slowed down?

Answer 1

A) By using kinetic equation,

Where, a is angular acceleration

w = w_o + at

w = 21 + 33*2.4

= 100.2 rad/s

theta = theta_o + w_ot + 0.5at^2

= 0 + 21*2.4 + 0.5*33*(2.4)^2

= 145.44 rad

Total angle = 433 + 145.44 = 578.44 rad

C) w^2 = w_o^2 + 2a*theta

0 = (100.2)^2 + 2*a*433

a = - (100.2)^2 /(2*433)

= - 11.6 rad/s^2

Angular acceleration = - 11.6 rad/s^2

B) theta = 0.5at^2

433 = 0.5*(11.6)*t^2

t = sqrt(433*2/11.6)

= 8.64 s

Time at which it stopped = 2.4 + 8.64 = 11.04 s