Question

The 15-kg slender bar OA is released from rest in the vertical position and compresses the
spring of stiffness k = 20 kN/m as the horizontal position is passed. Determine the proper
setting of the spring, by specifying the distance h, that will result in the bar having an angular
velocity ω = 4 rad/s as it crosses the horizontal position. What is the effect of x on the
dynamics of the problem?

Answer 1

Principle of work and energy:

If the equation of motion in the tangential direction, is combined with the kinematic equation, the principle of work and energy can be obtained. Thus, it is the sum of initial kinetic energy, and work done is equal to the final kinetic energy. This can also be stated as total work done on a body is equal to the change in kinetic energy of the body.

Potential energy: It is the energy possessed by the body due to its position compared to a datum. It is the energy stored inside the body due to gravity.

Elastic potential energy: Energy stored in the spring due to spring constant.

Kinetic energy: Energy possessed by the body due to virtue of its motion.

Conservation of energy: Total energy in a body remains constant irrespective of the changes in internal energy. Energy can neither be created nor destroyed, but it can change from one form to another form.

In this problem apply the concept of work-energy principle to find the value of required force F.

Fundamentals

The equation for the principle of work and energy is,

Here, is the initial kinetic energy, is the final kinetic energy, V is the potential energy, and is the total work done except gravity.

When no external work is done on the system the equation reduces to:

That is, the total energy remains conserved and the change in kinetic energy is equal to the change in the potential energy of the system.

Potential energy:

Here, m is mass, g is the acceleration due to gravity, and h is the height of the body from the datum.

Elastic potential energy:

Here, k is the spring constant and x is the deflection in spring.

Kinetic energy:

Here, v is the velocity of the object.

Calculate the moment of inertia of the rod about point O.

Here, is the mass moment of inertia of the rod about point O, m is the mass the rod and l is the length of the rod.

Substitute 15 kg for m and for l.

Calculate the change in the kinetic energy of the rod when it released from rest in the vertical position and compresses the spring in the horizontal position.

Here, is the angular velocity of the rod.

Substitute for and for .

When the rod is released from rest it rotates about the hinge O, so for calculating the change in the kinetic energy of the rod its moment of inertia about the hinge is required. Calculate the mass moment of inertia of the rod about O and calculate the change in the kinetic energy of the rod.

Calculate the change in the potential energy of the system.

Calculate the change in the potential energy of the rod as it comes in final position.

Here, g is the acceleration due to gravity.

Substitute 15 kg for m, for g and for l.

Calculate the change in the potential energy of the spring.

Here, k is the spring constant and h is the length by which spring is compressed.

Substitute for k.

Calculate the total change in the potential energy of the system.

Substitute for and for .

 

As the rod comes from initial position OA to final position its centroid of gravity goes down by a vertical distance of , so the change in the potential energy of the rod will be: . Similarly, calculate the change in the potential energy of the spring.

Apply work energy principle to calculate the value of distance h.

Apply work energy principle.

As no external work is done on the system except by gravitational and spring forces, net work done will be zero.

Substitute 0 for , for and for .

As the final position of the bar is horizontal, the same amount of spring will be compressed. Therefore, the value of x does not affect the dynamics problem.

The value of distance h is 54.5 mm and the value of x does not affect the dynamics problem.

Apply work-energy theorem between initial and final state; that is, the total energy of the system remains conserved. Also, as no external work is done on the system except by gravity and spring force, the work done will be zero. As the final position of the bar is horizontal, the same amount of spring will be compressed. Hence, the value of x does not affect the problem.