Rotational Definition and 1000 Threads

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...

So this is what I've attempted: 666 = m*a1 510 = m*a2 a1= ac + 9.8 a2= ac-9.8 666 = m(ac+9.8) 510 = m(ac-9.8) 666 = m*ac + m*9.8 510 = m*ac - m*9.8 156 = 2m(9.8) m = 7.9 kg (which seems very wrong haha) any ideas?? I thought my reasoning was okay, since I considered that at the top of...

We know that Bell States follow the Rotational Invariance property i.e. the probability of results on measurement of two entangled particles do not change if the initial measurement basis (say ##u##) is rotated by an angle θ to a new basis (to say ##v##). Lets take the Bell State ##\psi = \frac...

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...

I watched a video that showed how to calculate the center of gravity of a horizontal bar suspended from two wires, one attached to each end. Each wire was then attached to a vertical wall. The angle each wire made with the wall it was attached to was given. They treated it as an a example of...

I first tried to get the solution by conserving the rotational kinetic energy and got ##\omega'=\frac2{\sqrt5} \omega##. But, it was not the correct answer. Next I tried by conserving the angular momentum and got ##\omega'=\frac 45 \omega##, which is the correct answer. Why is the rotational...

When does an object have rotational energy? Is it only if it rotates around an axis within the object? Does for example a ball attached to a string with a uniform circular movement have rotational kinetic energy?

I'm 3d-printing a part that is 110mm x 40mm (XY-plane). I have to keep it light weight, so I made it thin - 0.6mm (Z direction) and filled the surface with holes to remove as much weight as possible. Then I added a grid of 2,5 mm high "walls" in the XY directions to make it stiffer and a couple...

Near the end of this paper, Sherwood presents a rotational analogue of the centre of mass work equation. The derivation is as follows (##\tau_{i, \text{CM}}## is the torque of the ith force about the centre of mass): $$\sum_i \tau_{i, \text{CM}} = I_{\text{CM}} \alpha$$ $$\int \left( \sum_i...

Very confused at this. https://courses.lumenlearning.com/physics/chapter/10-6-collisions-of-extended-bodies-in-two-dimensions/ "Consider the relatively simple collision shown in Figure 2, in which a disk strikes and adheres to an initially motionless stick nailed at one end to a frictionless...

I found out the time when rotation ceases to be 4 ##v_0## /5*mew*g, where mew=coefficent of friction of surface but I am unable to plot the graph post that time

For a general body, there exists the notion of centre of mass work; that is, computing the work done if all of the forces on the body act through the centre of mass. If we separate the total kinetic energy into that of the CM and that relative to the CM, ##T = T_{CM} + T*##, we can show by...

This isn't really a proper homework question, but something I wondered about myself. To simplify things, we say that pivot point is in the origin of a Cartesian coordinate system, and the angles are constrained to the first quadrant. We see that the weight of the barbell is given as $$F =...

Hello, apologies if this is the wrong forum to post this. I have a question related to fitness. When holding a barbell overhead, sometimes it can spin due to poor timing, flexibility, etc. However, I notice it is significantly harder to counter the barbell's spin when holding it with a narrow...

I am developing my own project called orbital flight simulator myself. I was looking for some IAU rotational model about all planets (Sol solar system) through the Internet and my books but was unable find any information. Does anyone have good books or information (tables and algorithms like...

I know that a hoop should have a higher rotational inertia than a solid disk because its mass is distributed further from the axis of rotation. What I don't understand is how a disk of the same mass and radius can have a higher rotational inertia. If the objects roll freely their axes of...

Friction provides the necessary torque for rolling without slipping.So Rotational Energy must increase.Simultaneously acceleration of centre of mass down the inclined plane is positive so Translational energy also must increase.Overall The Gravitational Potential Energy is getting converted to...

I'm not sure as to why my working is incorrect. When the sign on a_x is postive, i get t = \frac{R\omega_0}{3\mu_kg} which would give the correct value for distance if plugged into the kinematic equation. However, I'm not sure why a_x would be positive though since the friction force is pointing...

I'm inclined to say no, but am by no means certain. The total kinetic energy of a system of particles is $$T = \sum_{i} \frac{1}{2} m_{i}\vec{v_i}\cdot\vec{v_i} = \frac{1}{2} m_{i}(\vec{v_{COM}} + \vec{v_{i}^{'}})^{2} = \frac{1}{2}M\vec{v_{COM}}^{2} + \frac{1}{2}\sum_{i} m_{i}...

I was going over the rolling disk versus rolling hoop problem, in which the hoop has more Kr due to greater I and therefore smaller Kt and v. I know this can be algebraically proved with two unique expressions for V that don't involve omega. The question in class that came up concerns torque. If...

Hello, I'm stuck in this rotational motion problem (advanced high school level). Source: Problems in General Physics- IE Irodov My attempt(s): First I tried using work done by the moment of friction (mgkR) and equated it with change in KE. I got the answer as ## \frac{R (\omega_0)^2}{8 \pi...

I got a confusion about the sings in the angular acceleration. When dealing with system of pulleys, how to define where is the positive and negative direction of the motion and will the choose of positive direction of angular acceleration will effect the positive direction of linear acceleration

Hi there I have been having a go at this question and I'm uncertain if my answer to part b) is valid? The problem is when I plug this into the calculator I get 6.379... revs however this doesn't make sense to me. 2*pi is roughly 6.28 radians so doing 4.061... rads / 6.28 rads = 0.647 revs...

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...

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...

I converted the amount of rotations completed in 5 seconds into radians. 23.4 rot * 2pi = 147 rad I found the angular acceleration of the object in the first 5 seconds it was speeding up. Wf = Wi + at a = 5.881 rad/s^2 I then used the moment of inertia given in the problem to solve for torque. T...

Shouldn't the answer have the same magnitude regardless of sign convention?

A potter's wheel with a 35.9 cm radius rotates with a 2.91 rad/s2 angular acceleration. After 5.37 s, the wheel has rotated through an angle of 77.7 rad. a)What linear distance did a point on the outer edge travel during the 5.37 s? b)What was the initial angular velocity of the wheel? c)What...

What I did: ##\frac{dm}{dA} = \frac{M}{\frac{3\sqrt3 R^2}{2}}## ##dm = \frac{2M}{3\sqrt3 R^2} dA## (1) ##dA=3\sqrt3 rdr## (2) (2) in (1) ##dm = \frac{2M}{3\sqrt3 R^2} 3\sqrt3 rdr## Now in the integral ##I = \int \frac{r^2 2Mrdr}{R^2}## How can I solve the integral interval? I think I...

Hi, this is probably a stupid question, but, does rotational invariance in ##d=2+1## mean to only rotate the spatial coordinates and not the time. I mean bascially I want to show that ## \int d^3 x \epsilon^{\mu\nu\rho}A_{\mu}\partial_{\nu}A_{\rho} ##, yes epsilon the antisymmetric tensor, is...

moment of inertia = (1/3) (2.1kg) (1.2m)^2 = 1.0 kgm^2 center of mass= (0.6i, 0j) magnitude of the gravitational torque=9.8m/s^2*2.1kg*0.6m= 12.34N*m position of the new center of mass now : x direction = cos(20)*0.6m=0.56m y direction= -sin(20) * 0.6m = -0.2m change in gravitational...

For a cylinder rolling down an inclined plane, does the tangential velocity of a point a distance R from the axis of rotation equal the velocity of the center of mass?

'Let’s first see what all of this means in the context of d = 3 + 1 dimensions. If we have rotational invariance then we can’t write down any terms linear in E or B. The first terms that we can write down are instead ...' Why is this? I don't understnad . My thoughts would be pictruing the set...

I have the moment of inertia for the core(initial) and full body(final) but my answer for the moment of inertia for the arms(initial) was incorrect. Arms(initial) moment of inertia:(1/12)(6)(1.7^2)=1.445 this is incorrect for some reason Core(initial) moment of inertia: .9558 Full...

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...

Hi everyone, The equation is one we have been given to calculate the rotational speed of the sun for different latitudes. phi = average latitude. This shouldn't be a problem for me, but for some reason I just can't trust my error calcs. We are given : A = 14.713 ± 0.0491◦/d B = −2.396 ±...

The first doubt that comes to my mind is "I have to determine the acceleration with respect to what?", because the problem doesn't tell. Then, I have some problems when having to plug the data in the formula of acceleration. ##\vec a_B=0## because the origin isn't accelerated, ##\vec{\dot...

Consider a rotating disk with the center at the origin of a stationary Cartesian coordinate system, (x, y). At t = 0, on the circumference of the disk, someone/something shoots a particle with constant velocity components Dvx, Dvy (where the D indicates the rotating disk). Also at time t=0...

I don't have a clue on how to start please help me

Consider two bodies, A and B, of equal mass set at a short distance. Body A is spinning and body B is at rest. Then, through some kind of electromagnetic device, a strong repulsive force is established between them. Will both be displaced at the same speed?

Summary: I want to show by using the rotational energy formula that ##E(2^+) = 92 KeV##? But I got ##E(2^+) = 3096KeV##. Below is what I've done. There must be something I am missing. [Moderator's note: Moved from a technical forum and thus no template.]

Hi all, I found this problem in a new textbook I'm working through. And my energy conservation equation was ## mg\frac {h}{2} = \frac {1}{2} I ω^2 + mg \frac {h}{2}*sin(55) ## My solution was wrong and after checking why I found that they used cos(35) as the angle. The rest was the same. I'm a...

Problem Statement: Finding the rotational inertia Relevant Equations: I=∑m*r^2 A rigid body of 2 massive globes with homogenous mass distribution and a thin rod is connecting the 2 globes. The globes has radius R1 = 0.18 m and R2 =0.28 m and masses m1=193 kg and m2=726 kg. The thin rod has...

There is a fictional planet with two moons. 1) Is it possible for those two moons to orbit that planet on rotational axes that are different from the planet's rotational axis? 2) If so, can the moons themselves also have different rotational axes?

There is a cornering maneuver in rallying called the "Scandinavian flick" or the "pendulum turn". It involves steering away from the corner before actually steering into the corner. This creates a pendulum effect which makes the car turn more sharply into the corner. Sorry for the poor video...

I have a solution, However Cant understand 1 point.Now, This is the solution: ##N_2 l cos\theta + \frac 1 2 F_g l cos\theta - f_2 l sin\theta = 0## ## N_2(1 - \mu tan\theta) + \frac 1 2 F_g = 0## This is the the point that I don't like - yes it is less that 0, but it's even less that...

You can see the effect around minute 1:20 in this video. It seems to me that the un-twist of the elastic band and the rotation of the ball about the line that joins them is what keep the constant the initial zero angular momentum, though I can't tell for sure. The inversion of in the...

I found one answer somewhere else in the internet, It specified there that atoms cannot have rotational and vibrational energies since they don't have a point on them that will allow the atom to be rotated or vibrated. However , that answer did not suffice so I ask the same question here.