Rotational dynamics Definition and 274 Threads

Hi guys, I was just wondering this question: say I have a dumbell which is two plates of equal mass placed on a rod, then is the moment of inertia for this object affected when the distance the plates are placed from the center "R" changes? I've attached an image to illustrate me point Thanks...

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point...

I have solved the problem, but I am not sure if my solution is correct, so I would appreciate if someone would double check it. Thanks (a) Using an inertial reference frame in which the ##x## axis points to the left and the ##y## axis points downwards and computing the torques with respect to...

A thin uniform rod has mass M=0.510 kg and length L=0.470 m. It has a pivot at one end and is at rest on a compressed spring as shown in (A). The sequence below shows that the rod is released from an angle θ1=59 degrees, and moves through its horizontal position at (B) and up to (C) where it...

I am confused about three things: (i) To prove that accelerations are same, in the book it is given: ##mg−T=ma_1## (for the block) ##mg−T=ma_2## (for the cylinder) ##(a_1 = a_2 = a)## And thus the accelerations should be same. I can see that the same forces are acting on the bodies and that...

Hey! In this setup, https://sciencedemonstrations.fas.harvard.edu/presentations/falling-faster-g ,a board rotating about a fixed axis under gravity, The board's angle θ is measured from the horizontal, Angular acceleration is given by α = MgRcosθ / I Then we're asked to find the acceleration of...

Here is a pictorial depiction of the problem From Newton's 2nd law we have $$2T-mg=0\implies T=\frac{mg}{2}$$ Then, considering the torques created by the threads we have $$\vec{\tau}=I\vec{\alpha}=(-r\hat{k}+\frac{l}{2}\hat{i})\times T\hat{j}+(-r\hat{k}-\frac{l}{2}\hat{i})\times...

If we had a wheel rolling without slipping down the inclined plane, kinematically its velocity would be 0 at the contact point to the ground since the rotational and translational components of velocity would cancel out. Speaking of forces, forces acting on body would be static friction and...

[Mentor Note: Two threads merged below and the OP has fixed up their LaTeX in later posts] TL;DR Summary: A billiard ball with zero angular velocity and linear velocity v_0 is released on a horizontal surface with coefficient of friction mu_k. It begins to roll without slipping after...

I am trying to get the torque about the point of contact of the ladder on the ground = mg x l/2, but it is not leaving me anywhere

If there is a net external torque on the system about ##S##, then this net torque vector must have its direction on the ##z##-axis (since the system is only rotating about this axis at all times). This torque generates angular acceleration $$\tau_{S,z}=I_S\alpha_z\tag{1}$$ where ##I_S## is...

First off, I do know how to solve this problem. We use the principle of conservation of angular momentum about the centre of mass of the system which comprises of the loop and the bullet to obtain option B. My doubt is, why do we just not use the principle about the centre of the loop? Where is...

A solution was provided: We take torques about point B. Note that τ = MgL/2 = Iα so α = (3g)/2L. Everything from here is straightforward. I don't understand why in this step, you can take torque about B without accounting for a fictitious force due to the acceleration of the Rod.Thanks for...

Please help to answer this question, I don't know how to start, thanks in advance.

I am trying to solve a problem where we have to find the maximum angle before a cube starts sliding when said cube is placed inside a spinning hollow cylinder (the cylinder is placed horizontally). The radius of the cylinder is 0.4 m, the coefficient of static friction between the cube and the...

According to my current understanding rolling friction rolling friction is the static friction (parallel to the surface on which the object is moving) applied by the frictional surface (rough surface) on the contact point or contact area of the object whose v≠Rw(v is translational velocity and...

The figure illustrates the situation. The radii of the larger and smaller discs are 2R and R, respectively. Their masses are M and 2M, respectively (the largst disc has the smallest mass). Also, m=5/4 M, where m is the mass of the suspended object. The pulley is "massless" (negligible moment...

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...

Suppose the car is moving to the right, so if the wheels roll without slipping, they are rolling clockwise. To get the wheel to slip, a counterclockwise torque would need to be applied to cause the wheel to have some angular acceleration. If the wheel was slipping, then the bottom of the wheel...

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}].##...

a) From impulse-momentum theorem I have: ##J=mv## so ##v=\frac{J}{m}## and since the ball doesn't slip ##v=\Omega b## so ##\Omega=\frac{J}{mb}## and ##\dot{\theta}=\frac{v}{a+b}=\frac{\Omega b}{a+b}##. b) I considered the angular impulse: ##-J(a+b)=I_0 \Omega_0 \Rightarrow...

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🙏

I thought that the force by the pivot A on the pole AB would be the reaction force to the x-component of the gravitational force on AB. This would mean that the force by the pivot would be parallel to the pole, but in my notes from class the force vector seems to be more along the bisector of...

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...

1) ##LT\sin(\frac{\pi}{2}-\theta)-\frac{L}{2}mg\sin\theta=0\Rightarrow T=\frac{mg}{2}\tan\theta##. ##N_{x}-T=0, N_{y}-mg=0\Rightarrow N=\sqrt{N_x ^2+N_y ^2}=mg\sqrt{(\frac{\tan\theta}{2})^2 +1}## 2) ##E_{k_{fin}}=mg\frac{L}{2}[1+\cos\theta]## 3)...

I apologize in advance for any errors in my concepts or assumptions. Feel free to correct me wherever I am wrong. Thanks in advance for the help. There is a vertical shaft which will be operated at around 600 rpm (N) which can be achieved in 2 seconds (or even 4 just an assumption). The shaft...

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...

My doubt is with Method 2 of the given example in KK. I'm unable to understand why the torque around A (where we have chosen a coordinate system at A) becomes zero due to the R x F in z direction with a minus sign {Photo Attached} I have tried to reason out that one way to formulate that term...

My question is this: - Friction exists (for no slipping/pure rolling to occur) - Why is the work done against friction not accounted for in the conservation of energy equation? Thank you!

This looks like a classical setup but I can't find a solution. We can calculate the energy of the system by looking at the work done by the gravity and the spring. But how do we divide the energy between the kinetic energy of the pulley and the rotation of the pulley?

Here's a diagram of what the system looks like: So far I have figured out what the initial angular velocity is, if the system is balanced (no movement): ## \sum F_m = m*\frac{v^2}{R_0}-\frac{Mg}{2}=0 ## ##m \frac{v^2}{R_0}-\frac{Mg}{2}=0 ## divide both sides by m ##\omega_0 =...

Earlier today I realized that, when a strong gust of wind would blow through my area, it would pick up leaves off the ground and typically blow them in circular patterns, and typically the leaves would go in at least several complete circles before coming to rest back down on the ground. Why is...

I am a junior engineer tasked with modeling the dynamics of a small research UAV after landing. The UAV has 3 tires, 1 on the nose landing gear and 2 on the rear landing gear. The rear tires are equipped with disc brake calipers. My coworker has explained that the simplified model (MODEL 1...

I = Icm + mr^2 I = 0.5 mr^2 + mr^2 I= 3/2 mr^2 By COE, mgh = 0.5(3/2 mr^2)(w^2) g(2r) = 3/4(r^2)(w^2) 8g/3 = rw^2 = v^2 / r v = sqrt( 8gr/3) v=0.511m/s ans: v=0.79m/s

Resolving the weight of the cylinder c, i get Mgcosθ (-y) and Mgsinθ (-x) mgsinθ - Fs - T = ma ---(1) (where Fs is frictional force and T is tension) τ = I α (where τ is torque and α is angular acceleration) torque is produced by both tension and frictional force (T-Fs) * r = 0.5 m r^2 α...

The solution states that there's no rotational motion when ##C## is cut (the motion is curvilinear), so we can take torques with respect to the centre of mass of the plate. But, isn't it rotating? I think of it as a pendulum, which describes a circular motion. What's the difference? Wouldn't the...

Hi, I have just joined the forum. Thank you all for being a part of such places so that people like me can get answers to the questions on their minds! --------------------------- I have been trying to understand how a quadcopter yaws. Referring to the figure below which is bird's eye view of...

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...

I've marked correct answers above. Have a look at the solutions: How is the first equation justified? Shouldn't v2 and ωR be of opposite signs? What is v1? And how is it equal to v2? My biggest problem is the source of v1 since the ring is not having vertical displacement, then what is v1?

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...

We have a rod ##AB## of mass ##m##, a force (perpendicular to AB) is applied at ##A##. I want to know how much force will ##B## going to feel? When ##F_1## is applied at ##A## rod will rotate about its COM (which lies at the Center) and hence the point ##B## will also move (a little downwards...

--no explanation as conceptual error--

Basically, I want to know if my assumptions and workings are correct. This is how I see this situation. First, I'm viewing this body as a series of disconnected points, like I have in this animation I made, modeling purely rolling motion. Modeling the body like that worked in that case, and...

I have solved part a using the conservation of energy, getting a (correct) answer of 47.9 km/h, but I am unable to make headway with part b. Based on the flywheel rotating at 237rev/s when the car is moving at 86.5 km/h, I obtained omega = (237*2pi)v/24=62v. Differentiating both sides should...

Where: 1) ##A## is the translational acceleration, ##\Omega## the angular velocity and ##\dot \Omega## the angular acceleration (all relative to the inertial frame attached to the ground ##F##). 2) ##r'##, ##v'## and ##a'## are the position, velocity and acceleration vectors, all relative to...

Firstly I deduced that in this situation the moment of inertia I, is not going to be parallel to w. And I calculated it as a matter of the angle, for the rod and the two point particles attached (with a mass 'm'), and the total moment of Inertia ended up being: I=((R²*sin²α)/2)*(M/6 + m) Being...

A Uniform rod AB of length 7m is undergoing combined motion such that, at some instant, velocities at top most point A is perpendicular to the rod and magnitude is 11 m/s. The mid point/ centre of mass ,say C, has a velocity of 3 m/s and is also perpendicular to the rod. If both the velocities...

is this eqn correct