## Rotational Speed

Consider a braking system with a 5 m long brake arm has a 200 N force applied to the arm’s end. If the total brake power is 100 kW, what is the rotational speed in rpm?

Hint
Brake Power:
$$\dot{W}_b=2\pi \tau \upsilon =2\pi FR\upsilon$$$where $$\tau$$ is torque, $$\upsilon$$ is rotational speed, $$F$$ is the force at the end of the brake arm, and $$R$$ is the brake arm’s length. Hint 2 $$\upsilon$$ is in $$rev/sec$$ , so a final answer conversion is needed to achieve the rotational speed in rpm. Brake Power: $$\dot{W}_b=2\pi \tau \upsilon =2\pi FR\upsilon$$$
where $$\tau$$ is torque in $$N \cdot m$$ , $$\upsilon$$ is rotational speed in $$rev/sec$$ , $$F$$ is the force at the end of the brake arm, and $$R$$ is the brake arm’s length. Thus,
$$\upsilon=\frac{\dot{W}_b}{2\pi FR}=\frac{100,000W}{2\pi(200N)(5m)}$$$$$=\frac{100,000kg\cdot m^2}{6,280N\cdot m\cdot s^3}=\frac{100,000kg\cdot m^2}{6,280kg\cdot \frac{m}{s^2}\cdot m\cdot s^3}=15.92\:\frac{rev}{sec}$$$
Finally, to convert the rotational speed to rpm:
$$\upsilon =15.92\:\frac{rev}{sec}\cdot \frac{60\:sec}{1\:min}=955\:rpm$$\$
955 rpm