Question

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We wish to determine the moment at the shoulder that is required to perform the arm motion (shoulder abduction) depicted in Figure 3 below. This motion may be modelled with a simple, single- element linkage system as shown in the figure. The shoulder joint is represented by a simple pin support centred at O. The arm and the carried weight are represented as a rigid body consisting of a rod and a cuboid. In the rigid body model, the length of the rod, L is 50 [cm), its mass is 4.2 [kg], and its centre of mass is located 25 [cm] from point O. The relevant dimensions of the cuboid are b = 15 [cm] and c = 10 [cm]. and it has a mass of 2 [kg] which acts 57.5 [cm] from the point O. c = 10 [cm] b = 15 [cm] L = 50 [cm] Мо 135° 0° Figure 3: Shoulder abduction.

Part (a) [2 marks]

The angular displacement of the rigid body ranges from θ = 0° (the vertical as shown in the figure) to θ = 135° and can be modelled using simple harmonic motion. Assuming a rate of 20 [reps/min], write down an expression for angular displacement, θ [rad] as a function of time, t [s]. You may assume that the motion starts with an angular displacement of 135°.

Hint: The angular displacement, θ can be expressed as θ= θ_offset + θ₀ cos(φt), where θ_offset is an offset angle. When θ_offset is added to the simple harmonic motion expression, it forces θ to range from 0° to 135°.

Part (b) [2 marks]

Determine the angular velocity, ω [rad/s] and the angular acceleration, α [rad/s²] as a function of time, t [s].

Part (c) [4 marks]

Determine the mass moment of inertia for the rod-cuboid system.

Part (d) [2 marks]

Determine the radius of gyration for the rod-cuboid system.

Part (e) [4 marks]

Compute the moment, M_O that must be generated at the shoulder joint at t = 3.375 [s].

Part (f) [8 marks]

Compute the magnitude of the reaction forces at the shoulder joint (parallel and perpendicular to the moving arm segment) at t = 3.375 [s].

Part (g) [4 marks]

The moment M_O is actually generated by a muscle. The upper attachment of this muscle is located μ_u= 1 [cm] above point O, while the lower attachment is located μ_d= 5 [cm] from point O along the arm (see Figure 4 below). Convert the moment M_O you have calculated in part (f) into a corresponding muscle force, F_muscle at t = 3.375 [s].

Answer 1

(a) Assuming he is doing 20 reps/min, time taken to do one rep is (60/20)=3 second which is of course the time period.

As he is not doing a full revolution. He starts from θ = 0° and ends at  θ = 135°,again he starts going back to  θ = 0° ,and so the angular frequency of the rigid body is .

2л 3

Now,as the initial value of θ is 135°. So   $\boldsymbol{\theta (t)=\frac{3\pi }{8}(1+\cos (\frac{2\pi t}{3}))}$, where  $\theta (t)$ is measured in radian and t is in seconds.

(b) The angular velocity is just the derivative of $\theta (t)$ . So  
7T 2 27t wt) sin 4 3
, where  $\omega (t)$ is measured in rad/sec. Similarly the the angular accelarartion is derivative of the angular velocity. So -

  $\boldsymbol{\alpha (t)=-\frac{\pi ^{3}}{6} \cos (\frac{2\pi t}{3})}$

(c) Moment of inertia of the rod taking O as origin is $I=\frac{1}{3}ML^{2}=0.35\; kg.m^{2}$ And moment of inertia of the cuboid taking O as origin is $I=\frac{1}{6}Ma^{2}+Md^{2}=0.67\; kg.m^{2}$ So moment of inertia of the system taking O as origin is $\boldsymbol{1.02\; kg.m^{2}}$ .

(d) Radius of gyration of the rod-cuboid system is $\boldsymbol{k=\sqrt{\frac{1.02}{4.2+2}}=0.405\; m}$ .

(e) At t = 3.375 s $\alpha (3.375)=-3.65$ , and so torque genrated in system is $\tau =I\alpha (3.375)=-3.727$ .  From simple calculations we can show that moments generated by the rod is 19 N.m and the cube is 10.4 N.m. So moment generated at shoulder joint is $\boldsymbol{19+10.4-(-3.727)=33.13\; N.m}$ .

(f) As total force acting downward on the shoulder is $(4.2+2)\times 10=62\; N$ . Reaction force at the shoulder joint acting upward is 62 N. And the centrifugal force will act outward along the rod.