Moment of inertia Definition and 1000 Threads

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...

We start from the definition$$I_{ij} = \int_V \rho(x_k x_k \delta_{ij} - x_i x_j) dV \iff \dot{I}_{ij} = \int_V \rho (2 x_k \dot{x}_k \delta_{ij} - \dot{x}_i x_j - x_i \dot{x}_j ) dV$$Now since the rigid body rotation satisfies ##\dot{\vec{x}} = \vec{\omega} \times \vec{x} \iff \dot{x}_i =...

Can anyone explain why the moment of inertia for a tubular column in that textbook is like so? (take a look at the attachments). It should be (I = MR^2), as far as I know.

I have done some lab work , and now i have to answer some theoretical questions , but i can not find any data about this on the web or atleast i don't know where to search , i will add some pictures of experiment for you to better understand it. I was wondering can someone share their knowledge...

I never really considered this back when I was taking physics in college but imagine for the sake of thought experiment that you have an extremely and impractically long wrench and it is fixed to the bolt you wish to tighten. Now the longer the lever arm the greater the torque so if you double...

Like I said, objects with the higher acceleration are giving me the lowest values. For a hoop, I got I=0.1*MR^2 For a cylinder, I got I=0.7*MR^2 this seems backwards, no?

Note: the working (taken from iWTSE website) refers to inertia as the symbol ‘J’ (in-case there was any confusion).I found equations of motion for mass m and 2m which were ‘T1 = ma + mg’ and ‘T2 = 2mg – 2ma’, respectively. I know they are connected particles with the same acceleration ‘a’.I have...

Answer is 37.8 g cm^2 new to latex

So i derived the moment of inertia about the axis of symmetry (with height h) and I am confused about the perpendicular axis theorem. The problem ask to find the moment of inertia perpendicular to axis of symmetry So the axis about h, i labelled as z, the two axis that are perpendicular to z, i...

My thought process was to get the mass moment of inertia of the rectangle and then subtract the mass moment of inertia of the quartercircle from it. The MMoI of the rectangle is: (1/3)(0.005*7850*.6*.3)(.3^2)= 0.212 meters The MMoI of the quartercircle is: (1/4)(0.005*7850*¼π 0.3^2)(.3^2) + ? My...

I did in this way: ## I = \int dm \rho^2 ## Dividing the triangle in small rectangles with ##dA = dy x(y) ## where ##x(y) = 2 ctg( \alpha ) (h - y) ## we have : ## dm = \sigma 2 ctg( \alpha ) (h - y) ## Now i have ## \rho^2 = x^2 + (h-y)^2 ## Now I don't know what I can do because it would be...

I'm struggling doing point 5, i have no idea how to solve that question. In point 1 i obtained the following result: ## I=\frac{ML^2}{2}## calculating the integral of dI, the infinitesimal moment of inertia of a small section of the rod of length dr. 2) Through the conservation of angular...

The total moment of inertia is: ##I_{tot} = 2 M_1 R^2 + \frac{1}{2} M_2 R^2## We have ## M_1 = (4 \pi R^2) \sigma ## and ##M_2 = (\pi R^2) \sigma ## , where ## \sigma ## is the density of the disks. We also know that: ## \sigma = \frac{m}{ \pi 5 R^2} ## this leads us to say that: ##I_{tot} =...

The question was: I will also include the solution: So, what is the justification of the first formula [ω=√(C/I)]? I know how to derive simple harmonic equations, this one as I guess is probably similar? But I cannot connect as to how C is used exactly. And the second formula [ω'=ωβ], I...

My answer to part A is correct but for Part B I got an incorrect answer of 0.204J. My working out is sent as an attachment. Part A: Part B: Part C:

When the lamina rotates about A, FA must act on B (because it is the farthest away) perpendicular to AB (so that all of FA contributes to rotation). Same argument is valid for rotation of lamina about B as well. Having noted that, I tried two approaches: Approach 1- If I assume that the...

In order to choose a DC motor according to the article titled: APPYING MOTORS IN LINEAR MOTION APPLICATION by PITTMAN - step 4 : "determine the total reflected inertia (Jt) back from the load to lead screw shaft " . The formula is: Jt = Jscrew + Jload. This calculation relies on the fact...

The formula for moment of inertia is: I=mr^2 A common derivation for this is: 1. F=ma 2. τ=rma 3. τ=rmrα = r^2 mα This is a rotational version of Newton’s second law, where torque replaces force, moment of inertia replaces mass, and angular acceleration replaces tangential acceleration...

Hi, I am building a drone for a school project and I am also looking into how it flies. Recently I have been looking into angular momentum, torque, moment of inertia and angular acceleration. However I am struggling to understand moment of inertia and angular acceleration. If possible please...

Inertia moment of a thin square side 2b about the center of mass... I put the coordinates in the center of the square and came to: Integral of (x²+y²)dm = Integral of (x²+y²)*(dxdy)M But, the interval of the integral is [0,b] to x and y And, since this consider just the integral of one...

Hey guys. Im trying to figure out how to calculate the moment of inertia from a homogeneous ball based on a series of accelerations. The ball is released from the top of an incline plane (3.33 deg) and with a motion sensor, 5 values of acceleration where captured . Together with the radius and...

Apologies if I make anyone frustrated. To start, I've only had up to Calculus II so far but I was curious how to use and evaluate integrals used for moment of inertia. I know that the moment of inertia is basically an object's resistance to rotation, and is the rotational analog of mass. I know...

Why we divide the equation dI = dm*r^2 by two when we calcule the moment of inertia of a cone relative to its symmetry axis?

Hi, I previously posted about the statically indeterminate truck problem. Thank you to everyone who helped me. However, I now realized that isn't the problem I need to solve. I need to know the force of the linear actuators to lift the rear tyres off the ground. Since the tyres will be...

If I take the three masses individually and try to calculate the moment of inertia of the system separately then I=(m*0²)+(m*(l/2)²)+(m*l²) =ml²/4 +ml²=(5/4)ml² But If I try to calculate Moment of Inertia of the system using its Centre of mass then As centre of mass is located at the the...

Given:Thin, homogeneous, curved rod with radius of curvature 𝑅 See figure to the down. Find: The moment of inertia 𝐼𝑥′𝑥 ′ with respect to 𝑥′- the axis passing through the center of mass (point 𝐺). Can someone who can help me ?

So in the above image, I intend to find the moment of inertia of that black rotating object which rotates due to torque which is provided by placing mass on the pulley. But the thing is that this rotating object is kind of like a ball bearing kind of system and even for a small torque it starts...

I tried to find the moment of inertia of 2 rods connected at the corners by adding up their moments of inertia: \frac{1}{3}(\frac{M}{4})a^2 + \frac{1}{3}(\frac{M}{4})a^2 = \frac{1}{6}Ma^2 I then tried to solve for the moment of inertia at the center of mass of the 2 rods using the parallel...

Hi, A well-known part of the formula for calculating the deflection stress is ##I_z=\int \int r^2 dA## Usually a moment of inertia is something related to how difficult is to move an object. In this case is understandable but i don't understand the meaning of the double integral. Using ##r^4##...

Hello, I am a computer science major and Ex-Biology grad student, my knowledge in physics is humble, but I got a little curious when my professor derived the expressions of moment of inertia for different objects. The moment of Inertia of a thin disk is 1/2MR2, but it is the same as the moment...

Hello my dear physicists, I'm trying to model the varied generated (needed)Torque to rotate a washing machine Drum during a Washing Process so i assumed that the Model has as Input the target vilocity and as an Output the new needed torque to rotate the Drum(to be as a input for the motor...

When I solved the problem using the conservation of angular momentum, I have got the correct result (ω = 0.006 rad/s). However, when I tried to find the answer using the conservation of energy the result was incorrect and I do not understand why.

Can someone guide me on how to approach this question? I tried to draw a quick diagram of what I think is happening here Does the question imply that this object is undergoing horizontal circular motion in the shape of a conical pendulum? Thanks for any help!

Here is the problem that I am finding difficult to answer I had tried using conservation of energy to do this question Where I know that the gravitational potential energy at the top of the slope equals to the sum of both the linear and rotational kinetic energy at the bottom of the slope...

I use the moment of inertia I = 1/12ml2 for an axis perpendicular and passing through the center of mass of a rod. In a cube built out of 12 rods I have 8 rods at a perpendicular distance l/2 from the axis through the midpoint of a cube. These 8 rods contribute the moment of inertia I1 =...

Hello, I'm new here so apologies if this is in the wrong place! I was wondering, hypothetically, what sort of canned food would roll the fastest (assuming the ramp remained constant and the can couldn't be altered)? I've been looking this up and I know a solid can would roll faster than an...

moment of inertia= [(1/2)(1.1kg)(0.96)^2+ (1.1kg) (0.75*0.96)^2]= 1.08 kg*m^2 θ=9.8 degrees= 0.17 rad torque= (mass*gravity) * radius * sin(theta) radius= 0.17rad * 0.96m = 0.16m torque= (1.1kg*9.8m/s^2) * 0.16m * sin(0.17rad) = 0.29 N*m torque = inertia * angular acceleration 0.29N*m=...

Homework Statement: Derive the formula for the moment of inertia of a thin spherical shell using spherical coordinates and multiple integrals. Homework Equations: Moment of Intertia is (2MR^2)/3 I = (2MR^2)/3

Homework Statement: So i need to find equations to help me with a bifiler suspension experiment in which i will use a rectangular drop bar as the oscillating object, also any help with the method of this experiment would be greatly appreciated. The end goal is to find the moment of inertia...

I=(1/2)(6kg)(0.4^2 + 0.52^2) = 1.29 kg*m^2 initial: 34 mph= 15.2m/s 15.2m/s = (ω) (0.52m) ω= 29.2 rad/s after: 19 mph = 8.5m/s 8.5m/s = ω(0.52m) ω= 16.3 rad/s acceleration = (16.3 rad/s - 29.2 rad/s) / 5s= -2.58 rad/s^2 torque= |-2.58 rad/s^2 |(1.29 kg*m^2 ) = 3.3 N*m I am confused...

Hello, I tried to put it in an equation, but it didn't really work out. In this situation, the car was about the size of a model, and, while not exact, the radius of each wheel couldn't have been more than like a centimeter. Conversely, the ball was like twice the size of the car and had a...

I= 524kg * 6m^2 = 18.86E3kg*m^2 KE=(1/2) (18.86E3kg*m^2) (0.16 rad/s )^2= 241.5 J torque= 0.25* 57N* (6m)=85.5 N*m

So to start off, what I will do find the center of mass of each of the rods. So for the top rod, COM is at where y= 0.5 L and COM of the rod at the bottom is at x = 0.5 L. From there, how do I proceed in finding the moment of inertia using parallel axis theorem? Do I simply treat: ##I...