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Kinematics and dynamics of rigid bodies 73<br />

Z 3<br />

Z 1<br />

Body 1<br />

Body 2<br />

Y 2 ,Y 3<br />

G 3<br />

θ<br />

Y 1 X<br />

Body 3<br />

O 3<br />

1<br />

X 1<br />

{ω 3 } 2/2<br />

{ω 2 } 1/1<br />

Fig. 2.37<br />

Swing arm suspension dynamics<br />

with a constant angular velocity of 100 rad/s about the axis of the revolute<br />

joint representing the wheel bearing.<br />

The following data may be used to represent the mass properties of the road<br />

wheel:<br />

m 3 16 kg<br />

I 31 I 3xx 0.5 kg m 2<br />

I 32 I 3yy 1.0 kg m 2<br />

I 33 I 3zz 0.5 kg m 2<br />

In order to determine the reaction torque on the wheel bearing when the<br />

axle is still in a horizontal position, 0, we can use the following to give<br />

us the required angular velocity vector { 3 } 1/3 :<br />

⎡10⎤<br />

⎡ 0 ⎤<br />

⎡ 10 ⎤<br />

{ } / <br />

⎢<br />

2 12 0<br />

⎥<br />

rad/s { } / <br />

⎢<br />

23 100<br />

⎥<br />

rad/s { } / <br />

⎢<br />

13 100<br />

⎥<br />

rad/s<br />

⎢ ⎥<br />

3 ⎢ ⎥<br />

3 ⎢ ⎥<br />

⎣⎢<br />

0 ⎦⎥<br />

⎣⎢<br />

0 ⎦⎥<br />

⎣⎢<br />

0 ⎦⎥<br />

In the absence of angular acceleration equation (2.233) can be adapted to<br />

give for this problem<br />

Σ{ M3} 13 / [ 3] 13 / [ I3] 13 / { 3}<br />

13 /<br />

(2.238)<br />

Expanding this gives<br />

∑<br />

⎡M<br />

⎢<br />

⎢<br />

M<br />

⎣⎢<br />

M<br />

3x<br />

3y<br />

⎤<br />

⎥<br />

⎥<br />

⎦⎥<br />

3z<br />

1/3<br />

⎡ 0<br />

⎢<br />

⎢ <br />

⎢<br />

⎣<br />

<br />

<br />

0<br />

<br />

3z<br />

3y<br />

3z<br />

3x<br />

3y<br />

3x<br />

<br />

<br />

0<br />

⎤<br />

⎥<br />

⎥<br />

⎥<br />

⎦<br />

1/3<br />

⎡I3xx<br />

0 0 ⎤<br />

⎢<br />

⎢<br />

0 I<br />

⎥<br />

3yy<br />

0<br />

⎥<br />

⎣⎢<br />

0 0 I ⎦⎥<br />

3zz<br />

3/3<br />

⎡<br />

⎢<br />

⎢<br />

<br />

⎣⎢<br />

<br />

3x<br />

3y<br />

⎤<br />

⎥<br />

⎥<br />

⎦⎥<br />

3z<br />

1/3<br />

(2.239)

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