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Chapter 2

Q. 2.5

The system of Figure 2.13 moves in a horizontal plane. Replace the system of springs by (a) a single spring of equivalent stiffness when x is the displacement of the block of mass 2 kg and is used as the generalized coordinate and (b) a spring of an equivalent torsional stiffness when the clockwise angular rotation of the disk \theta is used as the generalized coordinate.

Step-by-Step

Verified Solution

(a) When the block of mass 2 kg moves through a displacement x, as shown in Figure 2.13, and assuming the cable connecting the block to the disk is inextensible, the point of contact between the disk and the cable have the same velocity. The velocity of the cable \dot{x } is and the velocity of a point on the outer edge of the inner disk is r\dot{\theta }. Thus,

\dot{x}=r\dot{\theta }          (a)

Let y be the displacement of the cable attached to the 1 kg block. Its direction is opposite that of the other block. Assuming the cable is inextensible, the velocity of the cable \dot{y } is the same as the velocity of the point on the disk in contact with the cable which is \frac{2}{3} r\dot{\theta } leading to

\dot{y} \frac{2}{3} r\dot{\theta }        (b)

Equations (a) and (b) are combined, leading to

\dot{y}= \frac{2}{3}\dot{x}         (c)

which is true for all time. Integrating and setting y(0) = x(0) = 0 leads to

{y}= \frac{2}{3}{x}         (d)

The total potential energy developed in the system at an arbitrary time in terms of x is the sum of the potential energies in the springs

V =\frac{1}{2}\left( 3000 N/m\right)x^{2}+ \frac{1}{2}\left(1000 N/m\right) \left(\frac{3}{2}x \right)^{2}

     =\frac{1}{2}\left( 3000 N/m\right)x^{2}           (e)

The equivalent stiffness of a spring placed on the 2 kg block to model the potential energy of the system is 5250 N/m. (b) Using Equations (a) and (b) to give relations between x and \theta and y and \theta leads to the total potential energy in the system, which is written using \thetaas the generalized coordinate as

V =\frac{1}{2}\left( 3000 N/m\right)\left(r\theta \right)^{2}+ \frac{1}{2}\left(1000 N/m\right) \left(\frac{3}{2} r\theta \right)^{2}          (f)

Substituting r = 0.1 m gives

V =\frac{1}{2}\left(52.5\frac{N.m}{r} \right)\theta ^{2}         (g)

Thus, the equivalent torsional stiffness of the system when using \theta as the generalized coordinate is 52.5 N m/rad, which implies that the springs can be replaced by a single torsional spring of stiffness 52.5 N m/rad attached to the pulley.