Question

Please explain where the angular velocity squared comes from in 1d) and thoroughly explain 1d) thanks!

1) In the deepest, darkest reaches of space. far, far away from the influence of other matter and matters of civilization, in a dreadfully isolated and desolate spot known as 'Fresno, two small objects of mass my ind m2 joined by a spring of constant k and natural (unstretched) length Lo, orbit about their mutual center-of-mass with a fixed separation (point-mass-to-point-mass) R in some inertial frame of reference. • la) (5 points) Describe the physics of the situation - in particular, how do we know that the objects orbit around their mutual center-of-mass? Describe the orbital motion of the objects - how does the position of each object relate to that of the other object and the center-of-mass? A quick sketch might help. & Fext = 0, So Acmo, which means Vem Constant. A reference frame attached to the center-of-mass would (in this case) be an inertial frame in the center-of-mass frome, the Center of mass (located directly between the masses) is a fixed cont... the masses orbit in Circles as and the center- of mass, one on either side... M2 . lb) (5 points) Find the orbital radius for each object. Yene EMi Ki Emi Cen(o) + M2 (A) M2 r = RM + M2 - Mit Mz CML r=R-r .. 6.2 R M M 2

Answer 1

See, from the equation given above in b) we have calculated r_{​​​​​​1}​​​ & r_{​​​​​​2} , so now

In order to find the velocity of each block in inertial frame, (taken as centre of mass of the system, since it is un-accelerated)

We use V_{​​​​​​1} = $\omega$ *r_{​​​​​​1} & V_{​​​​​​2} = $\omega$ *r_{​​​​​​2}

And finally when we want to calculate total energy, we are required to add kinetic (K) as well as potential energy (U) of the system.

K = (0.5)*m_{​​​​​​1}*(V_{​​​​​​1})² + (0.5)*m_{​​​​​​2}*(V_{​​​​​​2})²

K = (0.5)*m_{​​​​​​1}*$\omega$²*(r_{​​​​​​1})² + (0.5)*m_{​​​​​​2}*$\omega$²*(r_{​​​​​​2})²

Now substituting, values of r_{​​​​​​1} & r_{​​​​​​2}, we get

K = (0.5)*{m_{​​​​​​1}*m_{​​​​​​2}/(m_{​​​​​​1}+m_{​​​​​​2})}*($\omega$*R)²

And the potential energy of the system is given by the spring potential energy which is given by { E = (0.5)*k*x^{​​​​​​2}​​​}

Where k is spring constant, and x is the extension in the spring

We can see from the diagram, that extension in spring is (R-L_{​​​​o})

Hence U = (0.5)*k*(R-L_{​​​​o})²

Now, finally substituting $\omega$ ,

We know both the particles are performing circular motion, hence the spring force provides the centripetal force

Hence,

m₁*$\omega$²*r₁ = k*(R-L_{​​​​o})

Using value of r_{​​​​​​1} ,

m_{​​​​​​1}*$\omega$²*(R*m_{​​​​2}/(m_​1 + m_{​​​​​​2})) = k*(R-L_{​​​​o})

Hence we get, $\omega$ ² = {(m_{​​​​​​1}+m_{​​​​​​2})*k*(R-L_{​​​​o})}/(m_{​​​​​​1}*m_{​​​​​​2}*R)

Finally substituting $\omega$ ²,

We get the final results as given in answer in d) part.