Equations of Motion w/ spring scale

In summary, the conversation discusses a problem involving a scale with a mass attached to it and a spring. The goal is to find the equation of motion for the mass in terms of displacement and acceleration. After applying Newton's 2nd law and setting up the kinematics, the equation (I0)x" + k(L1)x = mgL22 is derived, with some mistakes in the calculation. It is noted that this problem is not a statics problem and the scale will continue to swing back and forth due to the lack of a damping mechanism.
  • #1
Nikstykal
31
1

Homework Statement


http://imgur.com/a/X7mWA
QC78LeR.png

Homework Equations


1. ΣF = m a
2. Στ = Iθ"

The Attempt at a Solution


First , assuming small motion, all movement of the scale can disregard x components (so the spring stretches only in vertical direction without an impact from the x directional movement). I summed torque around the pin, point O, which would be equal to the mass moment of inertia * theta double dot. so...

mgL2 - kx1 = I0θ" (mass force positive since displacement is positive in the negative y direction)

x1 is the spring movement upwards, however we want to relate that to the x displacement downwards with the mass, which we'll just call x. Now, using like triangles, x1=(L1/L2)x. Then I know x = rθ, differentiate with time twice to get x" = r θ" so then θ" = x"/r, where r = L2. Now we have ...

(I0/L2)x" - k(L1/L2)x= mgL2 ===> (I0)x" + k(L1)x = mgL22

This answer just doesn't feel right. I feel like I may not have correctly applied Newtons 2nd law or maybe didn't correctly set up the kinematics. Any input would be helpful, thanks.
 
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  • #2
You need not take into account the mass moment of Inertia of mass m, since it is a statics problem
 
  • #3
Nikstykal said:

The Attempt at a Solution


First , assuming small motion, all movement of the scale can disregard x components (so the spring stretches only in vertical direction without an impact from the x directional movement). I summed torque around the pin, point O, which would be equal to the mass moment of inertia * theta double dot. so...

mgL2 - kx1 = I0θ" (mass force positive since displacement is positive in the negative y direction)

x1 is the spring movement upwards, however we want to relate that to the x displacement downwards with the mass, which we'll just call x. Now, using like triangles, x1=(L1/L2)x. Then I know x = rθ, differentiate with time twice to get x" = r θ" so then θ" = x"/r, where r = L2.
OK up to here.
Now we have ...
(I0/L2)x" - k(L1/L2)x= mgL2 ===> (I0)x" + k(L1)x = mgL22
Check your math, this has at least 2 mistakes in it. One is a sign error.
Also, what is I0? You need to express this in terms yuu were given.

Also BTW not only is this not a static question but, contary to the problem statement iself, the scale swings forever back & forth since no damping mechanism was introduced.
 
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1. What is an equation of motion with a spring scale?

An equation of motion with a spring scale is a mathematical formula that describes the motion of an object attached to a spring scale. It takes into account the force applied by the spring and other factors such as mass and acceleration to determine the position, velocity, and acceleration of the object.

2. How is the equation of motion with a spring scale derived?

The equation of motion with a spring scale is derived using Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration (F=ma). It also takes into account Hooke's law, which describes the relationship between the force applied by a spring and the displacement of the object attached to it.

3. What are the key variables in the equation of motion with a spring scale?

The key variables in the equation of motion with a spring scale include the mass of the object (m), the force applied by the spring (F), the displacement of the object (x), and the acceleration of the object (a). These variables are typically represented as symbols in the equation.

4. How is the equation of motion with a spring scale used in real-world applications?

The equation of motion with a spring scale is used in various real-world applications, such as measuring the force required to compress a spring, predicting the motion of objects attached to springs in engineering and physics experiments, and designing shock absorbers and other mechanical systems that involve springs.

5. Are there any limitations to the equation of motion with a spring scale?

Yes, there are some limitations to the equation of motion with a spring scale. It assumes that the spring is ideal (i.e. obeys Hooke's law perfectly), neglects factors such as air resistance and friction, and does not take into account the non-linear behavior of some springs at high forces. Additionally, the equation may not accurately predict the motion of objects with complex shapes or in non-uniform gravitational fields.

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