The moment of inertia, otherwise known as the mass moment of inertia, angular mass, second moment of mass, or most accurately, rotational inertia, of a rigid body is a quantity that determines the torque needed for a desired angular acceleration about a rotational axis, akin to how mass determines the force needed for a desired acceleration. It depends on the body's mass distribution and the axis chosen, with larger moments requiring more torque to change the body's rate of rotation.
It is an extensive (additive) property: for a point mass the moment of inertia is simply the mass times the square of the perpendicular distance to the axis of rotation. The moment of inertia of a rigid composite system is the sum of the moments of inertia of its component subsystems (all taken about the same axis). Its simplest definition is the second moment of mass with respect to distance from an axis.
For bodies constrained to rotate in a plane, only their moment of inertia about an axis perpendicular to the plane, a scalar value, matters. For bodies free to rotate in three dimensions, their moments can be described by a symmetric 3 × 3 matrix, with a set of mutually perpendicular principal axes for which this matrix is diagonal and torques around the axes act independently of each other.
For this translating block problem, below is the solution. I was wondering why if I took the moment about the center of gravity G, the answer for F would no longer be the same because ##I_G \alpha = -\mu_k N (h/2) + N (b/2) - F*d = 0## because ##\alpha = 0##
$$F = \frac{-\mu_k mg (h/2) +...
Supposing L = 6m, m = 30kg, M = 70kg, h = 2m. Also set the origin at the leftmost edge of the plank.
Free body diagram description:
The center of mass of the plank is at ##\frac{L}{2}##, so considering it as a point mass, it feels the force of gravity ##F_m = mg##.
The person is at some point...
Hi all – this is a repost of an old question I had that was phrased quite poorly, so I hope some pictures help clarify what I was asking for.
I understand how the mass moment of inertia of a rectangle rotating about its width axis (=(1/12)mL^2) is derived by integrating with respect to mass. I...
I'm sorry for a really basic question, but my physics class was just introduced to moments of inertia and I want to try find them in three dimensions
Attempt:
Step 1) Search up 'moment of inertia of a rectangle' and find this:
Step 2) Multiply by the height??? This is definitely wrong
Step...
I'm trying to find the moment of inertia of a human body with (say) the upper arm at angle alpha to the torso and the lower arm at another angle beta, where beta is the angle between the lower and upper arms.
I model the torso as a cylinder of radius R, mass M and the parts of the arm as...
Hey guys do you know how Solidworks allow us to estimate the moment of inertia of the 3D model given the material properties? Well I'm trying to find the moment of inertia of my rotary drum with 1177 peanuts centers inside the drum and obviously I can't draw 1177 peanut centers inside the...
Using the trivial way, you get I = M(r+L)^2. This is definitely correct.
Out of curiosity, can the parallel axis theorem be applied here? I took the axis of rotation to be into the page, then calculated the moment of inertia for the disk, which is I = 1/2Mr^2. Next, I calculated the distance...
In deriving the MOI of a ring about its center perpendicular to plane, our teacher said think of ring as made up of ##dm## units each at ##R## distance from the axis therefore the MOI becomes: $$I=R^2\int{dm}=MR^2$$
If disc is considered to be made up of rods of ##dx## thickness, in this manner...
Here is the homogenous paper rectangle
And if we roll it we get a cylinder with base radius ##a##.
It is not clear to me what "an axis through a diameter of the circular base means".
Let's imagine such as axis is ##\alpha## in the following figure
Then we have...
Given that we're working with an elastic collision we want to populate the following system:
##k_{i} = k_{f}##
##p_{i} = p_{f}##
Solve for kinetic energy just before and after the collision:
##k_{i} = \frac{1}{2}mv_{i}^{2}##
##k_{f} = \frac{1}{2}mv_{f}^{2} + \frac{1}{2}I_{P}^{sys}...
I often encounter the formula: I = (1/12)Mbh^2 when dealing with moment of inertia of rectangles and got confused when I was unable to get the same result when figuring it out with integration. It seems that the axis of rotation used is a line perpendicular to one of the bases and on the plane...
so calculated, the moment of inertia for a rod about an axis at the end of the rod is I = 1/3 * M * L^2
here for case 1: arms to the side
I is calculated to be ##I = 0.224##
for case 2: arms stretched
## I = 1 / 3 * M * L^2 + M * d^2 ## with L = 0.6 m (length of rod) and d = 0.2 (dinstance from...
I was able to derive an equation for acceleration for the case of a fully solid cylindrical shell and then used law of conservation of energy to determine equations for the velocity of a solid and hollow cylinder and I understand that the moment of inertia's of the aforementioned cans can also...
Two rotating cylinders are held in contact by a force F1. The force is applied through the center of one of the cylinders. One cylinder is the driving cylinder and the other is the driven cylinder .
Does the moment of inertia of the system depends on the force contact force F1? Why?
And...
a=2/3*g*sin(25*(pi/180))=>a=2.8507 m/s^2
vf=vi+at=>vf=0+2.8507*1.50=>vf=4.2760 m/s
So the translational motion of the cylinder is 4.2760 m/s.
4.2760=R*w
w=134.04 rad/s
PE=mgh=>PE=215*9.8*.108=>PE=227.56 J
PE = KE at the end of the roll because of energy conservation.
227.56 =...
If two plates of Aluminum (both are 2" tall, and 1/2" thick) were placed over each other (see photo 1) and then welded together all around the perimeter of the second plate (the blue one) as indicated by the yellow lines in photo 2, then would the moment of inertia be calculated as if the local...
The weight of the rack is supported on an axial bearing as seen in the attached pdf below. I have made an attempt to calculate the torque by taking a look at the chain traction force and the required shaft power to make the plates rotate. For the moment of inertia case i don't know how to treat...
hello guys, I wanted to ask whether I can just consider/think about this as being rotation around a fixed axis in a plane representing it as if it was 'just' a rod. This is mainly so that for the kinetic energy in the second position is where if we think about it in just a plane. Is this...
Hi,
unfortunately, I am completely confused about the task
It is about the task part a
I have now defined the two rotations as follows:
The thin disc rotates around the ##z## axis, red in the picture, and then the rod to which the disc is attached rotates around the ##z_I## axis, in the...
What I did was plug in the outer radius time the force into the torque and then the mass moment of inertia is equal to m*ro^2 so then I plugged in the mass times the radius of gyration squared into I and solved for a but this is not right.
Angular Momentum and Torque are defined about a point. But Moment of Inertia of a body is defined about an axis. There are equations which connect Angular momentum and Torque with Moment of Inertia. How will this be consistent? When I say that the torque of a force acting on a body about a point...
Using the equation above I get Xcm = 0.022 m. I set the origin be at the left of the vertical rod parallel to its centre of mass as in the diagram. But I’m not sure if the equation is correct for 3d.
for the moments of inertia I am using
I = Icm + md^2
= (mr^2)/2 + md^2
where d is the...
I am having trouble to find the moment of inertia of the second rod!
Is it related to the first rod??
At the beginning I thought It's not!
But when took those as constant,the equation had become way much simpler and there is nothing about chaos!
My approach is given below
I placed my Oxy coordinate system at the center of the square, the ##x##-axis pointing rightwards and the ##y##-axis pointing upwards.
I divided the square into thin vertical strips, each of height ##h=2(\frac{L}{\sqrt{2}}-x)##, base ##dx## and mass ##dm=\sigma h...
I have come up with two different approaches, but I'm not sure which one is correct since they give different answers.
We use the following equation to get the total moment of inertia.
##I_o## = moment of inertia of disk about O axis + moment of inertia of road about O axis
Approach 1...
I think the the time given doesn't matter since no torque is acting on the system, but not sure. Therefore, all we need is to determine the angular momentum about the axis passing through O and perpendicular to the plane of disk. This will involve finding the moment of inertia of smaller disk...
What kind of experiment can I design to determine the actual value of the moment of inertia. What should I instruct the sphero to do and what data should I collect?
We solved this problem in class as follows:
Net torque about the center of the pulley taking counterclockwise rotation to be positive = m1gR - m2gR = I_tot α, where I_tot is the moment of inertia of the full system.
My professor said that I_tot = I + m1R^2 + m2R^2, where m1R^2 is the moment...
I am completely stuck on problem 2.45 of Blennow's book Mathematical Models for Physics and Engineering. @Orodruin It says
"We just stated that the moment of inertia tensor ##I_{ij}## satisfies the relation$${\dot{I}}_{ij}\omega_j=\varepsilon_{ijk}\omega_jI_{kl}\omega_l$$Show that this relation...
I used the parallel axis theorem to solve the question but my answer is still wrong. Any ideas where I slipped? I can't seem to figure out the problem?
1) Since the rod is uniform, with mass m and length l, it has a linear mass density of ##\lambda=\frac{m}{l}##, so ##I_{rod_O}=\int_{x=r}^{x=r+l}x^2 \lambda dx=\frac{\lambda}{3}[(r+l)^3-r^3]=\frac{\lambda r^3}{3}[(1+\frac{l}{r})^3-1]=\frac{1}{3}mr^2[3+\frac{3l}{r}+\frac{l^2}{r^2}].##...
Initially, I calculate the moment of inertia of of a square lamina (x-z plane). Thr this square is rotated an angle $\theta$ about a vertex and I need to calculate the new moment of inertia about that vertex.
Can I split the rotated square to two squares in the x-z plane and y-z plane to find...
This was the question
(The line below is probably some translation of upper line in different language)
For disc it was ma^2/2
For ring it was ma^2
For square lamina it was 2ma^2/3
For rods
It was different
Please explain
Thank You🙏
1) By conservation of linear momentum: ##m_1 v_1-m_2v_2=(m+m_1+m_2)v_{cm}\Rightarrow v_{cm}=\frac{m_1}{m+m_1+m_2}v_1-\frac{m_2}{m+m_1+m_2}v_2=\frac{3}{8}\frac{m}{s}##;
2) By conservation of angular momentum: ##-Rm_1v_1-Rm_2v_2=I_{total}\omega=(I_{disk}+m_1R^2+m_2R^2)\omega## so...
Hi!
I would like to calculate (roughly) how much torque is needed bringing the blue plateau in movement. Assume the blue plateau is loaded with 7.5 kg. The radius of the blue circle is 100 mm.
h = d1 + 0.08
d1 = h - 0.08
d2 = h + 0.08
I of the vertical portion
= 1/12 m (l^2 + b^2) + md1^2
= 1/12 m (0.28^2 + 0.04^2) + m(h - 0.08)^2
I of the horizontal portion
= 1/12 m (l^2 + b^2) + md2^2
= 1/12 m (0.28^2 + 0.04^2) + m(h + 0.08)^2
The moment of inertia for the whole T-shape about...
Hello,
It might sound silly, but when I try to calculate the kinetic energy of a rotating rod to form the Langrangian (and in general), why it has both translational and rotational kinetic energy?
Is it because when I consider the moment of Inertia about the centre I need to include the...
I have this moment of inertia problem and is a little confused on the semicircle part and if the rest is really right? I get over 10 if I calculate it in crew CAD but by hand I get 7,568032142. What is right and what am I doing wrong?
I know there are more convenient differential elements that can be chosen to compute the moment of inertia of a disc(like rings).
the mass of the differential element:
$$dm = (M/\pi R^2) (dA) = (M/ \pi R^2) (2\sqrt{R^2 - y^2})(dy)$$
the moment of inertia of a rod through its COM is...
What we know:
The ball is dropped at the tip A with some speed ##v_0## and rebounds with speed ##v##. This collision produces an angular impulse, changing the angular momentum of the bar with the flywheels.
Solution inspired by an answer provided by @TSny in the similar question.
Angular...
I have been given an answer for this but I am struggling to get to that point
$$ANS = 0.430\, kg \cdot m^2$$
So I thought using the moment of inertia of a compound pendulum might work where ##I_{rod} = \frac{ml^2}{12}## and ##I_{disc} = \frac{mR^2}{2}## (##l## is the length of the rod and ##R##...
I = 2/5M R^2 + Md^2
This is analagous to Earth's movement about the Sun. Is the moment of inertia of Earth about the centre of mass of the Earth Sun system = 2/5MR^2 + Md^2, where:
M = Mass of earth,
R = Radius of Earth,
d = distance from Earth to centre of mass of earth-sun system.
We know that impulse is
$$\vec J = \vec F \Delta t = \Delta \vec p$$
Let ##l, m## be the length of single rod and its mass respectively.
Analyzing torques and forces on each rod separately we have:
Rod ##AC##:
$$F\Delta t +N_x\Delta t = mV_{ac,x} \space\space\text{ eq. }(1)$$
$$F\Delta t\cdot...
Hi.
So I was asked the following question whose picture is attached below along with my attempt at the solution.
Now my doubt is, since the question refers to the whole system comprising of these thin rigid body 'mini systems', should the Principle moments of Inertia about the respective axes...
Please, I need help! I need to calculate the moment of inertia of a triangle relatively OY. I have an idea to split my triangle into rods and use Huygens-Steiner theorem, but after discussed this exercise with my friend, I have a question: which of these splits are right (picture 1 and 2)? Or...
Question:
Diagram:
So the common approach to this problem is using polar coordinates.
The definition of infinitesimal rotational inertia at O is ##dI_O=r^2\sigma\, dA##. Therefore the r. inertia of the triangle is
$$I_O=\int_{0}^{\pi/3}\int_{0}^{\sec\theta}r^2r\,drd\theta$$
whose value is...