# Small oscillations problem need help work shown!

1. Oct 28, 2009

### johnq2k7

A point of mass slides without friction on a horizontal table at one end of a massless spring of natural length a and spring const k as shown in the figure below. The other end of the spring is attached to the table so it can rotate freely without friction. The spring is driven by a motor beneath the table so that the spring and mass are constrained to move around the origin with angular frequency w(ignore any bending of the spring and assume it always remains radially outward from the origin)

a.) Using Cartesian coordinates, write down the expressions for the kinetic energy of the system.

b.) Change to a polar coordinate system using
x= r*cos (wt)
y= r*sin(wt)
where w is the angular velocity of the mass, and express your eq for the KE as a func. of
these new co-ord. Give also expression for the PE and Langrangian of the sys. (in polar
co-ord)

c.) How many D.O.F do you have for this system? Name them. Note angular velocity w is given, and therefore is const.

d.) Calculate the eq. of motion for r using the Langrange eq. for this system. Is the radial motion of a simple harmonic oscillator?

some of my work for a previous problem:
View attachment SolutionsH3.doc

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 28, 2009

### Staff: Mentor

How about showing your work on *this* problem.

3. Oct 29, 2009

### johnq2k7

my work:

for a.) let v(t)= ( x(t), y(t), z(t))
v'(t)= (x'(t),y'(t),z'(t)
therefore T= 1/2*m*[(x'(t)^2 + y'(t)^2 + z'(t)^2]
i'm not sure if this is correct

for b.) if x= r*cost(wt) and y= r sin(wt) where w is angular velocity

need help here

langrangian is equal to T-U

for c.) need help i believe it's 3 degrees of freedom

d. need help

4. Oct 29, 2009

### jdwood983

If the particle is constrained to exist on a horizontal table, shouldn't $z=\dot{z}=0$? Then you should get

$$T=\frac{m}{2}\left(\dot{x}^2+\dot{y}^2\right)$$

You will need to take the time derivatives of $x$ and $y$ in their polar form and then plug your given $\dot{x}$ and $\dot{y}$ into the kinetic energy above and reduce it as much as possible (remember your trigonometry and that $\frac{d}{dt}(\cos\theta)=-\sin\theta\frac{d\theta}{dt}$)

Why would it be 3 degrees of freedom? The point mass exists only on the horizontal surface, so $$z=0[/itex]. This looks like it should be less than 3. Well you have a conservative force, the spring. So then the potential is $V=\frac{1}{2}kr^2$. Then subtract this from the polar kinetic energy from part (b), there is your Lagrangian. After that, you can use your Euler-Lagrange equations of motion: [tex] \frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\right)-\frac{\partial L}{\partial q}=0$$

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