# Angular Acceleration

Need confirmation for this question please: A pulley with Radius of 90mm and a mass moment of inertia 0.28$$kgm^2$$ about its central axis is subjected to a torque of 50Nm about this axis. What is angular acceleration of the pulley?

Soo, I did this: $$\tau=I_o \cdot alpha$$, therefore angular acceleration alpha= 50/0.28= 178.6rad/s^2??

Isn’t that number too high for a pulley please? What am I missing?

Again your basic formula is just fine. $$\tau=I_o \cdot \alpha$$,. Therefore

$$\alpha = \frac{\tau}{I_o}$$

The reason why you are getting such a high acceleration is due to the very high torque $$\tau$$ with respect to the reported mass moment of inertia and radius.

I would crosscheck the values regarding the mass moment of inertia because given the radius and the moment of inertia (by assuming that the pulley is a uniform disc), you can calculate the mass as:

$$m = \frac{2\cdot I_o}{R^2}\approx 70[kg]$$

A pulley with a diameter of 180[mm], weighing 80[kg] is not very likely. So its either a matter of some mistake in the mass moment of inertia or the mass.

• Or the numbers where thrown into the question without regard for how "reasonable" they were. – hazzey Mar 25 at 20:35