Rotation with Constant Angular Acceleration — Question 7

Question 7

A braking delay consumes part of the available angle

A rotor turning at $10\,\mathrm{rad/s}$ must stop within $15\,\mathrm{rad}$. The brake begins only after a $0.30\,\mathrm s$ delay during which speed remains constant.

1. Find the minimum magnitude of constant braking acceleration.

2. Find total time to stop when the limiting acceleration is used.

Question 7 - Solution

A braking delay consumes part of the available angle

(a) Derivation and calculation

The delay consumes part of the available angular displacement: $$\theta_d=\omega_0t_d=10(0.30)=3\ \mathrm{rad},\qquad\theta_b=15-3=12\ \mathrm{rad}.$$ For braking magnitude $a>0$, $$\frac{\omega_0^2}{2a}\le12\Rightarrow \boxed{a_{\min}=\frac{100}{24}=\frac{25}{6}=4.167\ \mathrm{rad/s^2}}.$$

(b) Result and interpretation

At the limiting rate, $$t_b=\frac{\omega_0}{a_{\min}}=\frac{10}{25/6}=2.40\ \mathrm s,\qquad \boxed{T=t_d+t_b=2.70\ \mathrm s}.$$ The signed braking acceleration is $-a_{\min}$.

Physical check

The total area is $10(0.30)+\tfrac12(10)(2.40)=15$ rad.