Answer to Question #150068 in Mechanical Engineering for Tugberk Yıldız

Question

An electric motor drives a co-axial rotor through a
single-plate clutch, which has two pairs of driving
surfaces, each of 275mm external and 200mm internal
diameter, the total spring load pressing the plates
together is 500N. The mass of the motor armature and
shaft is 800kg and its radius of gyration is 260mm; the
rotor has a mass of 1350kg and its radius of gyration is
225mm.
The motor is brought up to a speed of 1250rev/min; the
current is then switched off and the clutch suddenly
engaged. Determine the final speed of the motor and
rotor, and find the time taken to reach that speed and
the kinetic energy lost during the period of slipping.
How long would slipping continue if a constant torque
of 50Nm were maintained on the armature shaft? Take
=0.35

Answer 1

Solution

1) Final speed $\omega_3$

$I_1=m_1k_1^2=800*0.26^2=54.08 kgm^2$

$I_2=m_2k_2^2=1350*0.225^2=68.34375 kgm^2$

Conservation of energy

$\omega_1=\frac{2\pi N_1}{60}=\frac{2\pi*300}{60}=130.8997 rad/sec$

$\omega_2=\frac{2\pi N_2}{60}=\frac{2\pi*300}{60}=62.8319 rad/sec$ (Assume the speed of the rotor was originally at 300 rpm)

$I_1\omega_1+I_2\omega_2=(I_1+I_2)\omega_3$

$\omega_3=92.9004 rad/sec= 889.133 rpm$

2) Energy lost

$K.E (before )=\frac{1}{2}*I_1*\omega_1^2+\frac{1}{2}*I_2*\omega_2^2=598215.06 Nm$

$K.E (after )=\frac{1}{2}*(I_1+I_2)\omega_3^2=528267.3961 Nm$

$Lost energy =598215.06-528267.3961=69947.66Nm$

Time taken to reach the speed

$T=\frac{1}{2}W\mu(R_1+R_2)=12906.25 Nmm=12.906 Nm$

Angular acceleration of the rotor

$I_2\alpha_2=T=12.906 \implies\alpha_2=0.18885 rad/sec$

$\omega_3=\omega_2+\alpha_2t \implies t = 159.22 sec$

Now constant resisting torque is 50Nm is applied

Torque on motor = $T_1=-12.906-50=-62.906 Nm$

$\omega_3=\omega_1+\alpha_1t =\omega_1+\frac{T_1}{I_1}t =130.8997-\frac{62.906}{54.08}t$

Considering the rotor

$\omega_3=\omega_2+\alpha_2t= 62.8319+0.18885t$

Equating the two implies that t = 50.35 sec