bme403_muslec_4_2006

# bme403_muslec_4_2006 - To convert to angle x=r can be...

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Dynamics-Continued Consider the model shown below for the shoulder: The muscle shown is the deltoid muscle. The 3 Kg limb weight is assumed to be at the limb center of mass. From an application of statics Fm * .03=3*9.8*.3 from a summation of moments around the shoulder joint. Fm = 294 N. If Fm decreases from this value say to .75 * 294, the limb will fall and settle at a new angle where the forces balance again. This can be seen from the diagram below. If the angle the arm makes with the horizontal is θ , then 294*.75*.03=3*9.8*.3*cos( θ ) cos( θ )=0.75 θ = 41.4 degrees To consider the dynamics involved in this transition the muscle model described earlier must be added.

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Fm= Fmax-k1 x – k2 dx/dt where k1= 5800 N/m and k2 = 2000 N/m/sec are representative values for skeletal muscle. k2/k1 can be interpreted as a time constant which in this case is short(0.34 sec).
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Unformatted text preview: To convert to angle x=r can be substituted in resulting in: Fm = Fmax-5800*.03 2000*.03 d /dt = Fmax-174 60 d /dt Then assuming small angle where cos( ) can be approximated by 1.0 the following moment balance can be made: moments = I d 2 /dt 2 Fm*.03 mg *.3 =- m L 2 d 2 /dt 2 Fmax*.03 5.22 1.8 d /dt 8.82 = -0.27 d 2 /dt 2 .27 d 2 /dt 2 + 1.8 d /dt + 5.22 = 8.82 Fmax*.03 = (5.22/.27 (1.8/(2*.27)) 2 ) 1/2 = 2.87rad/sec f=2.87/6.28=0.46 Hz This can be compared to the pendulum natural frequency which is =(9.8/.3) 1/2 = 5.72 rad/sec or 0.91 Hz The plot of the time solution (normalized) is shown below, the A coefficient was set to 1.0 and the phase at time zero was set to 0. 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-0.2 0.2 0.4 0.6 0.8 1 1.2 X TIME (SEC)...
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## This note was uploaded on 10/20/2009 for the course BME 403 at USC.

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bme403_muslec_4_2006 - To convert to angle x=r can be...

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