Question

A grinding wheel is in the form of a uniform solid disk of radius 7.00 cm and mass 2.00 kg. It starts from rest and accelerates uniformly under the action of the constant torque of 0.600 N m that the motor exerts on the wheel. (a) How long does the wheel take to reach its final operating speed of 1200 rev/min? (b) Through how many revolutions does it turn while accelerating?

Short answer

The grinding wheel reaches to final speed in 1.02 s.

Step-by-step solution

Identification of given data

The radius of uniform solid disk is r = 7 cm

The mass of uniform solid disk is m = 2 Kg

The torque for the grinding wheel is $\tau =0.600N·m$

The final speed of wheel is $\omega =1200rev/\min$.

Conceptual Explanation

The angular acceleration of the disk is calculated by torque of the disk and using this angular acceleration duration to achieve final angular speed of grinding wheel.

Determination of time to achieve final angular speed of grinding wheel

The moment of inertia of uniform solid disk is given as:

$$\begin{gathered} I=\frac{m{r}^2}{2} \\ I=\frac{\left(2kg\right){\left[\left(7cm\right)\left(\frac{1m}{100cm}\right)\right]}^2}{2} \\ I=4.9\times {10}^{-3}kg·m^2 \end{gathered}$$

The angular acceleration of wheel is given as:

$$\begin{gathered} \tau =I\alpha \\ 0.600N·m=\left(4.9\times {10}^{-3}kg·m^2\right)\alpha \\ \alpha =123rad/s^2 \end{gathered}$$

The duration to reach a final speed of grinding wheel is given as:

$$\begin{gathered} t=\frac{\omega }{\alpha } \\ t=\frac{\left(1200rev/\min \right)\left(\frac{2\pi rad}{1rev}\right)\left(\frac{1\min }{60s}\right)}{123rad/s^2} \\ t=1.02s \end{gathered}$$

Therefore, the duration to reach a final speed of grinding wheel is 1.02 s.