Question

Each of the two grinding wheels has a diameter of 6 in., a thickness of 3/4 in., and a specific weight of 425 lb/ft3. When switched on, the machine accelerates from rest to its operating speed of 3450 rev/min in 5 sec. When switched off, it comes to rest in 35 sec. Determine the motor torque and frictional moment, assuming that each is constant. Neglect the effects of the inertia of the rotating motor armature.

Answer 1

Answer:

$0.842\ \text{lb ft}$

$0.1052\ \text{lb ft}$

Explanation:

d = Diameter of wheel = 6 in

r = Radius = 3 in = $\dfrac{3}{12}=0.25\ \text{ft}$

t = Thickness = $\dfrac{3}{4}=0.75\ \text{in}=\dfrac{0.75}{12}\ \text{ft}$

w = Specific weight = $425\ \text{lb/ft}^3$

$t_2$ = Time taken to slow down = 35 s

$t_1$ = Time taken to reach operating speed = 5 s

$\omega$ = Angular velocity = $3450\times \dfrac{2\pi}{60}\ \text{rad/s}$

Weight is given by

$W=2\pi r^2tw\\\Rightarrow W=2\pi\times 0.25^2\times \dfrac{0.75}{12}\times 425\\\Rightarrow W=10.43\ \text{lbs}$

Mass is given by

$m=\dfrac{W}{g}\\\Rightarrow m=\dfrac{10.43}{32}\\\Rightarrow m=0.326\ \text{lb}$

Moment of inertia is given by

$I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{0.326\times 0.25^2}{2}\\\Rightarrow I=0.01019\ \text{lb ft}^2$

Angular acceleration while slowing down is given by

$\alpha_f=\dfrac{\omega}{t_2}\\\Rightarrow \alpha_f=\dfrac{3450\times \dfrac{2\pi}{60}}{35}\\\Rightarrow \alpha_f=10.32\ \text{rad/s}^2$

Frictional moment is given

$\tau_f=I\alpha_f\\\Rightarrow \tau_f=0.01019\times 10.32\\\Rightarrow \tau_f=0.1052\ \text{lb ft}$

Frictional moment is $0.1052\ \text{lb ft}$

Angular acceleration while speeding up is given by

$\alpha=\dfrac{\omega}{t_1}\\\Rightarrow \alpha=\dfrac{3450\times \dfrac{2\pi}{60}}{5}\\\Rightarrow \alpha=72.26\ \text{rad/s}^2$

Motor torque is given by

$\tau_m=\tau_f+I\alpha\\\Rightarrow \tau_m=0.1052+0.01019\times 72.26\\\Rightarrow \tau_m=0.842\ \text{lb ft}$

Motor torque is $0.842\ \text{lb ft}$.