1. Eight point masses of mass \( m \) are located on the corners of a cube at the cartesian locations: \( ( \pm a, \pm a, 0),( \pm a, \pm a, 2 a) \) a) [5 pts] Find the inertia tensor (use symmetry when possible to simplify any calculations). b) [5 pts] Find the principle moments of inertia by diagonalizing the inertia tensor (if necessary). c) [5 pts] Find the angular momentum of the cube if it spins about the axis \( (1,1,1) \) with angular velocity \( \omega \). d) [5 pts] Find the kinetic energy for the situation described in part c). 2. Consider a triatomic molecule with all atoms collinear and of equal mass \( m \). We model the coupling between the atoms of the molecule with two identical springs with spring constant \( k \). Each atom can only translate in the \( \mathrm{x} \)-direction (the atom does not rotate or translate as a whole). a) [10 pts] Find the normal mode frequencies of the molecule. Hint: use the coordinates \( \mathrm{x}_{1}, \mathrm{x}_{2} \), and \( \mathrm{x}_{3} \) to describe the position of each atom with respect to its equilibrium. The solution will involve diagonalizing a \( 3 \times 3 \) matrix, so you will probably want to use Mathematica b) [10 pts] Qualitatively describe the motion of the atoms for at least one of the normal modes of oscillation. No calculations required.
Added by Ratanic C.
Close
Step 1
1a) Show more…
Show all steps
Your feedback will help us improve your experience
Paul Gabriel and 60 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Find the moment of inertia of the four masses shown in Eig. $10-7$ relative to an axis perpendicular to the page and extending ( $a$ ) through point- $A$ and $(b)$ through point- $B$. (a) From the definition of moment of inertia, $$I_{A}=m_{1} r_{1}^{2}+m_{2} r_{2}^{2}+\cdots+m_{N} r_{N}^{2}=(2.0 \mathrm{~kg}+3.0 \mathrm{~kg}+4.0 \mathrm{~kg}+5.0 \mathrm{~kg})\left(r^{2}\right)$$ where $r$ is half the length of the diagonal: $$r=\frac{1}{2} \sqrt{(1.20 \mathrm{~m})^{2}+(2.50 \mathrm{~m})^{2}}=1.39 \mathrm{~m}$$ Thus, $I_{A}=27 \mathrm{~kg} \cdot \mathrm{m}^{2}$ (b) We cannot use the parallel-axis theorem here because neither $A$ nor $B$ is at the center of mass. Hence, we proceed as before. Because $r=1.25 \mathrm{~m}$ for the $2.0$ - and $3.0$ -kg masses, while $r=\sqrt{(1.20)^{2}+(1.25)^{2}}=1.733$ for the other two masses, $I_{B}=(2.0 \mathrm{~kg}+3.0 \mathrm{~kg})(1.25 \mathrm{~m})^{2}+(5.0 \mathrm{~kg}+4.0 \mathrm{~kg})(1.733 \mathrm{~m})^{2}=33 \mathrm{~kg} \cdot \mathrm{m}^{2}$.
Four hollow spheres, each with a mass of $1 \mathrm{~kg}$ and a radius $R=10 \mathrm{~cm},$ are connected with massless rods to form a square with sides of length $L=50 \mathrm{~cm} .$ In case $1,$ the masses rotate about an axis that bisects two sides of the square. In case $2,$ the masses rotate about an axis that passes through the diagonal of the square, as shown in the figure. Compute the ratio of the moments of inertia, $I_{1} / I_{2},$ for the two cases. a) $I_{1} / I_{2}=8$ b) $I_{1} / I_{2}=4$ c) $I_{1} / I_{2}=2$ d) $I_{1} / I_{2}=1$ e) $I_{1} / I_{2}=0.5$
Four hollow spheres, each with a mass of $1 \mathrm{~kg}$ and a radius $R=10 \mathrm{~cm},$ are connected with massless rods to form a square with sides of length $L=50 \mathrm{~cm} .$ In case 1 , the masses rotate about an axis that bisects two sides of the square. In case $2,$ the masses rotate about an axis that passes through the diagonal of the square, as shown in the figure. Compute the ratio of the moments of inertia, $I_{1} / I_{2},$ for the two cases a) $I_{1} / I_{2}=8$ b) $I_{1} / I_{2}=4$ c) $I_{1} / I_{2}=2$ d) $I_{1} / I_{2}=1$ e) $I_{1} / I_{2}=0.5$
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Watch the video solution with this free unlock.
EMAIL
PASSWORD