Question

At time t=0 a grinding wheel has an angular velocity of 28.0 rad/s . It has a constant angular acceleration of 35.0 rad/s2 until a circuit breaker trips at time t = 2.20 s . From then on, the wheel turns through an angle of 437 rad as it coasts to a stop at constant angular deceleration.(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

Answer:

a).$x_{t}=583.3rad$

b).$t_{total}=10.52s$

c).$a=12.62 \frac{rad}{s^2}$

Explanation:

The angular acceleration is constant so we can use the formulas of uniform motion with the model of angular acceleration

a).

$x_{r}=x_{i}+v_{i}+\frac{1}{2}a_{a}*t^2$

$x_{r}=0+28.0\frac{rad}{s}*2.20s+\frac{1}{2}*35.0\frac{rad}{s^2}*2.20s$

$x_{r}=146.3rad$

so the total angle between t=0 and the time it stopped is

$x_{t}=146.3rad+437rad=583.3rad$

b).

$w_{f}=w_{o}+a*t$

$w_{f}=28.0rad/s+35rad/s^2*2.2s=105rad/s$=$w_{o}$

$x_{t}-x_{r}=\frac{1}{2}*(w_{o}-w_{f})*t$

$583.3-146.3=\frac{1}{2}*(105-0)*t$

$t=\frac{437rad}{105\frac{rad}{s}}=8.32s$

$t_{total}=8.32+2.2=10.52s$

c).

$w_{f}=w_{o}+a*t$

$0=105 rad/s+a*8.32s$

$a=\frac{105 rad/s}{8.32s}$

$a=12.62 \frac{rad}{s^2}$