Initial Motion

If one of the supports of a body at rest suddenly gives way or is removed, the reactions at the other supports are instantaneously altered and the accelerations of the body are changed. To determine the altered reactions immediately after the removal of the support, we note that all the displacements and velocities resulting from the new accelerations are infinitely small and therefore can be neglected. The sudden removal of a support produces just the same effect as the sudden application of a force that is exactly the reverse of that exerted by the support while all other forces remain the same. A suddenly applied force is in no way comparable to an impulsive force. A finite force may alter the acceleration immediately but will require a finite time to generate a finite velocity. An impulsive force, on the other hand, produces finite changes of velocity instantaneously.

Example 7.8. A Cord Is Cut. A uniform horizontal bar of length L and mass m is supported by two vertical cords (Fig. 7.13). The distance between the two cords is d. If one of the cords is suddenly cut, determine the tension in the other cord immediately after the cutting.

Solution: Let T be the tension immediately after the cutting and To before. From equations of statics we know that To must be equal to half the weight of the bar (To = mg/2). Let a be the angular acceleration of the bar immediately after the breaking of the cord. The conservation of moment of momentum about the center of mass dictates that

The equation of motion of the center of mass in the vertical direction is T - mg = m (d/2) a (7.16b)

Solving Eqns. 7.16a and 7.16b for a and T, we find a = -(6 d g)/(3d2 + L2) (7.17a)

If the cords were attached at the ends of the bar before one was cut, then L = d, and T = mg/4. As the bar aligns itself in the vertical direction after the cord is cut, the tension T in the other cord eventually increases to become equal to mg.

Figure 7.13. A horizontal bar is suspended from a ceiling using two cords. The sketch shows the bar at the instant immediately after one of the cords is cut.

Example 7.9. Resisting Gravity. A man of mass m makes a bet that he can hang by his hands from a parallel bar at least for a minute (Fig. 7.14). After 37 s, however, he can no longer stand the pain in his shoulders and lets one hand go. Determine the force exerted by the bar on the other hand of the man. Assume that the distance between the two hands of the man on the parallel bar is d and that the radius of gyration is k.

Figure 7.14. A man grabs a bar with both hands and remains suspended in air for about 35 s. To relieve the tension in his shoulders, he lets go of one hand. With one hand released, he acquires a clockwise angular acceleration a.

Solution: Let T be the vertical force exerted by the bar on the man immediately after he lets go his one hand, and 2To before. Because the resultant force acting on a body at rest is equal to zero, To = mg/2.

The conservation of angular momentum of the body about the center of mass requires that

where a denotes the angular acceleration of the man. The equation of motion of the center of mass of the whole body in the vertical direction is

Note that in this equation the term (d/2) a represents the acceleration of the center of mass. Solving these two equations for a and T, we find:

Thus, immediately after the release of one hand, the man gains angular acceleration in the clockwise direction. The smaller the distance between the hands, the greater is the force exerted on the holding hand.

7.6 Summary

Impulse £ of a force F during a time interval tf — ti is defined as the time integral of the force over tf — ti. The conservation of linear momentum before and after an impulse requires that m vcf — m vci =

in which m is the mass of the body, vf and vci are, respectively, the velocity of the center of mass at tf and ti, and is the resultant impulse acting on the body.

Similarly, the conservation of angular momentum of a rigid body for which the plane of motion is a plane of symmetry yields the following equation:

Ic (Of — m) = /2Mc dt = 2 Ac in which Ic denote the moment of inertia with respect to the mass center, and Mf and Mi are the angular velocities of the body before and after impulse. The term 2Mc is the resultant moment about the center of mass. The term Ac, called angular impulse, is the time integral of the moment Mc.

A force that becomes very large during a very small time is called an impulsive force. Such a force can alter the velocity of a solid during the brief period it acts. Thus in the presence of impulsive forces, the contri bution of finite forces to linear and angular impulse are neglected. A parameter called coefficient of restitution is introduced as a measure of the capacity of colliding bodies to rebound from each other. (See Section 7.4 for the mathematical definition of the coefficient of restitution.)

If one of the supports of a resting body suddenly gives way or is removed, the reactions at the other supports are instantaneously altered. The reason for the alteration is the change in acceleration of the body. However, in this case, the body does not immediately gain velocity as a result of a support giving way or being removed.

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Getting Started With Dumbbells

Getting Started With Dumbbells

The use of dumbbells gives you a much more comprehensive strengthening effect because the workout engages your stabilizer muscles, in addition to the muscle you may be pin-pointing. Without all of the belts and artificial stabilizers of a machine, you also engage your core muscles, which are your body's natural stabilizers.

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