Question #150068

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

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

Expert's answer

**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

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