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Chapter 2

Q. 2.11

The system of Figure 2.24 moves in a horizontal plane.
(a) Determine the equivalent mass when x (the displacement of the 2 kg block) is used as the generalized coordinate.
(b) Determine the equivalent moment of inertia when \theta (the clockwise angular rotation of the disk) is used as the generalized coordinate.

Step-by-Step

Verified Solution

During the solution of Example 2.10, it is determined that if y is the upward displacement of the 1 kg block, then y=\frac{3}{2} x and \theta =\frac{x}{r}=\frac{x}{0.1m}=10x .The total kinetic energy is the kinetic energy of the blocks plus the kinetic energy of the disk:

T= \frac{1}{2}\left(2kg\right)\dot{x}^{2} + \frac{1}{2}\left(1kg\right)\dot{y}^{2}+ \frac{1}{2}\left(0.04kg\cdot m^{2}\right)\theta ^{2}

=\frac{1}{2}\left(2kg\right)\dot{x}^{2}+ \frac{1}{2}\left(1kg\right) \left(\frac{3}{2} \dot{x}\right)^{2}+ \frac{1}{2}\left(0.04kg\cdot m^{2}\right)\left(10 \dot{x}m^{-1}\right)^{2}

=\frac{1}{2}\left(8.25kg\right)\dot{x}^{2}                (a)

Thus, the equivalent mass is 8.25 kg.

(b) During the solution of Example 2.10, it is shown that y=\frac{3}{2} r\theta =\frac{3}{2}\left(0.1m\right) \theta

T=\frac{1}{2}\left(2kg\right)\dot{x}^{2}+\frac{1}{2}\left(1 kg\right)\dot{y}^{2} +\frac{1}{2}\left(0.04kg\cdot m^{2}\right)\theta ^{2}

=\frac{1}{2}\left(8.25kg\right)\left[\left(0.1m\right)\dot{\theta } \right]^{2} +\frac{1}{2}\left(1 kg\right)\left[\frac{3}{2}\left(0.1m\right)\dot{\theta } \right] ^{2}+\frac{1}{2}\left(0.04kg\cdot m^{2}\right)\dot{\theta}^{2}

=\frac{1}{2}\left(0.0825kg\cdot m^{2}\right)\dot{\theta}^{2}            (b)

Thus, if all of the inertia were concentrated on the disk, the disk would have a moment of inertia of 0.0825 kg \cdot m^{2}.