A square of mass M and sides of length L has a moment of inertia I0 when rotated about an axis perpendicular to its surface and passing through its center, as shown. Now a lump of clay, also of mass M is attached to one corner of the square as shown. What is the new moment of inertia of the masses about the same axis of rotation? I0+(ML^2)/2 I0+2(ML^2) I0+(ML^2) I0+(ML^2)/4 I0+(sqrt(2)*ML^2)/2
Added by Kelly W.
Close
Step 1
** Show more…
Show all steps
Your feedback will help us improve your experience
Mahendra Kumar and 81 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
Moment of inertia of a rod of length $\mathrm{L}$, mass $\mathrm{M}$ about an axis perpendicular to its length and passing through its center is $\frac{\mathrm{ML}^{2}}{12}$. The moment of inertia about a parallel axis through one end is (a) $\frac{\mathrm{ML}^{2}}{4}$ (b) $\frac{\mathrm{ML}^{2}}{2}$ (c) $2 \mathrm{ML}^{2}$ (d) $\frac{\mathrm{ML}^{2}}{3}$
Center of mass, moments of inertia Find the center of mass, the moment of inertia about the coordinate axes, and the polar moment of inertia of a thin triangular plate bounded by the lines $y=x, y=-x,$ and $y=1$ if $\delta(x, y)=y+1 \mathrm{kg} / \mathrm{m}^{2}$
Multiple Integrals
Moments and Centers of Mass
A thin, uniform rod is bent into a square of side length $a$. If the total mass is $M$, find the moment of inertia about an axis through the center and perpendicular to the plane of the square. ($Hint$: Use the parallel-axis theorem.)
Rotation of Rigid Bodies
Moment-of-Inertia Calculations
Recommended Textbooks
University Physics with Modern Physics
Physics: Principles with Applications
Fundamentals of Physics
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD