Rod Definition and 1000 Threads

I am trying to solve this question by ξ = A*cos(ωt + θ)*sin(kx + Φ) Anyway, the two initial terms of the product helps nothing (i think), what matters is sin(kx + Φ) So, i tried by two ways: First: The stress is essentially zero on the ends, that is, something like cte*∂ξ/∂x (strain) would be...

1) I found: ##x_{CM} = z_{CM} = 0 ## ##y_{CM} = \frac{L}{2} \frac{m_1 - m_2}{M + m_1 + m_2} ## 2) Applying the conservation of linear momentum: ##M v_0 - m_1 v_0 - m_2 v_0 = (M + m_1 + m_2) v_f ## ##v_f = \frac{M - m_1 - m_2 }{M + m_1 + m_2} v_0 ## The velocity should have been found also...

My answer: a. $$H=\frac{dQ}{dt}$$ $$H_{brass}=H_{copper}$$ $$(109\frac{W}{m\cdot K})(0.005m^2)\frac{100^{\circ}C-T_{c}}{0.3m}=(385\frac{W}{m\cdot K})(0.005m^2)\frac{T_{H}-0^{\circ}C}{0.8m}$$ $$T_{H,Cu}=T_{C,brass}=T_{2}$$ $$-1.8166667T_{2}+181.666667^{\circ}C=2.40625T_{2}\Rightarrow...

A homogeneous rod of length l and mass m is free to rotate in a vertical plane around a point A, the constraint is without friction. Initially the rod is stopped in the position of unstable equilibrium, therefore it begins to fall rotating around A and hits, after a rotation of ## \pi ## , a...

A hollow rod closed at the ends A and B, has mass M and length R. The rod is free to rotate on a horizontal frictionless plane around the z axis passing through A and coming out of the sheet. A body can slide without friction inside the cavity point mass m. Initially the rod is stationary and...

The centre of mass of the rod would be at the middle of the rod i.e. at l/2=[50*10^(-2)]/2 The force responsible for torque will be acting downwards = mg The Torque = mg*l/2*sin(30) =mg*l/4 We know that Torque=I*alpha Hence alpha = mg*l/(4*I) Moment of inertia of rod about the end= ml^2/12 +...

The net electric field is ## 2dE \cos\theta ## ## dE = \lambda dx/(4\pi\epsilon (x^2 +r^2)) \\ 2dE \cos\theta = 2r\lambda dx /(4\pi\epsilon (x^2 +r^2)^\frac 3 2) \\ E_{net} = 2\lambda r /(4\pi\epsilon) \int_0^a dx /( (x^2 +r^2)^\frac 3 2) \\ E_{net} = 2\lambda r /(4\pi\epsilon) [\frac x...

I found the correct solution using the equation that relates force and pressure, but I don't REALLY understand what the question is asking and what is actually going on in the machine. I want a better understanding of everything that's going on, not just an answer. Below is a clear diagram and a...

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...

I know that if they had the same density they would have the center of mass at 1,5 m. But now that they don't the center of mass will be shifted towards the part of the rod with higher density. they will have their center of mass where they have equal mass p1*v=p2*v now i don't know how to...

This is my attempt, i am confused at some points a. r = 0; The Electric field is 0 b. At r = a/2.00; I verified the answer and it is non zero, but my understanding is that the net charge should be on the surface of the conductor. Hence the charge q1=5*10^-15 C, should go to the surface of the...

I got the correct answer for the first part but I'm not sure why the answer for (b) is the same for (a). Wouldn't the rings falling off mean that I_f = \frac{1}{12}M_L L^2 only where I_F, M_L, L are the final moment of inertia, mass of the rod and length of the rod as opposed to I_f =...

Summary:: We have a rotating arm, offset from the centre of rotation by a certain length, which is controlled by varying the length of a control rod. Need the angle of the rotating arm in terms of length of the rod. The blue line is a fixed column structure. CE and BD form the rotational...

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...

I think is 25cm but I'm not sure. Help please

Hello, I am new to this forum. I have a question, hope I'll find some answers here. My question is - if there are 2 fishing rod, one lighter (physical weight) than the other. Will the lighter fishing rod cast a lure/bait further than the heavier fishing rod, assuming the rod length...

"When electrified rods are brought near light objects, a similar effect takes place. The rods induce opposite charges on the near surfaces of the objects and similar charges move to the farther side of the object." -from a high school physics book. NCERT Class 12th part 1 to be precise. can...

Please refer to the image

Summary:: This is a question about finding the acceleration of a point in a mechanism Hi, I have a question about the mechanism shown in the attached picture: Question: We are told that \omega = 6 rad/s and the first part is asking me to find the acceleration of point P on the piston when...

Hi, I have a doubt about reaction forces... I've attached a picture that shows two similar situations. The first one shows a rod left with an angle ##\beta## while it is on a smooth surface. The second one shows a rod leaning on a smooth surface and wall. My question is: why is the reaction...

I am not sure which other forces I should consider besides those 3. I cannot consider tensions due to the massless rod on the masses since those will not add up to zero.

Firstly I only consider one of the wheels. This wheel consists of a big wheel (black) with mass M and radius R and inside it a circular region with a negative mass (-m) and radius R/2. (I assume they have same mass density but with opposite signs. I do this because I don't know where the center...

I have to find the inertia tensor of these rods and I don't have the concept that clear... I mean, I know the formulas like: ##I_{xx}=\int y^2 + z^2 dm## ##I_{xy}=\int xy dm## But I don't know what ##x, y, z, dm## stand for. In other words, I don't know what I should replace in the formula...

A rod rotates freely (edit: about an axis perpendicular to its length) in empty space. Working in an inertial coordinate system where the rod rotates around a fixed point, the rod is straight, of length ##2L## in its spinning state, and its mass distribution is symmetric along its length. The...

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...

So I'm a college student currently doing A-Level Physics, I am attempting to build a type 4 double arm barn door tracker and although it has been a complicated project, I believe I've got all the planning done, however, I have a 2Rpm motor, a PWM motor controller, 2 9v batteries, an M8 threaded...

Homework Statement:: Ball of mass mb and velocity vb hits rod of length L , Rod pivots about the center. What is the angular momentum aafter impact? Homework Equations:: I = 1/12 (mR^2) I = mR^2 See the attached figure. I understand the concept of linear and angular momentum separately but I...

My initial thought was to use the conservation of energy law since there're no external forces acting on the system bullet + rod. The rod is in rest, the bullet is moving. Then after the collision, the bullet and the rod are rotating around the pivot together, so the kinetic energy of the bullet...

Let us consider the massless homogeneous helical spring with the infinitesimal wire diameter d and wire length l. We denote the spring radius as R. Now we consider the curvilinear spring section of length dl. We draw radii from the spiral axis to the centers of the end cross-sections of this...

initial total KE= (1/2)(0.6kg)(8m/s)^2 = 19.2J (0.6kg)(8m/s) = (0.6kg+1.8kg)(vf) vf= 2m/s final KE= (1/2)(0.6kg+1.8kg)(2m/s)^2 = 4.8J I tried to use linear speed=angular speed * radius : thus 2m/s= angular speed * (3.3m/2) angular speed= 1.2 rad/s Apparently that is wrong.

we know that the center of instantaneous 0 velocity lies in the interception of 2 perpendicular lines to 2 points, which in this case lies above B. The velocity of any point of the rod can be described relative to the center of instantaneuous 0 velocity ##(Q)## as: $$\vec v_{P/Q}=\vec \omega...

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...

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...

If a rod is on a table (horizontally) and rotating about an axis that passes through one of its ends and vertical to the table, what would be the tension on the opposite end of the rod (the end opposite to the axis) . In this post (Check this post out from Socratic QnA), the limits take while...

I am trying to push the boundaries of special relativity with a self-imposed challenge problem. A common derivation of relativistic kinetic energy involves an object to which a constant force is applied. I want to consider a similar scenario, but instead of a point object we now have a uniform...

∫ λ(x)=0.2 kg/m + 0.061(x/L)^2 kg/m = 0.2(x) + (0.061/3) (x^3) /(1/L^2) mass of rod = 0.2+ (0.061/3) =0.22 kg inertia of rod through nail = (1/3) (mass) (L)^2 inertia of rod through nail = (1/3) (0.22kg) (1m)^2 = 0.073 kg*m^2 torque magnitude = (53N) (0.5m) = 26.5N*m angular...

I begin by drawing the problem. Let the center of the bigger sphere on the left by the origin for the system. Calculating ##x_C = \frac{10 \times 1.1 + 20 \times 2.15}{50+20+10} = \frac{54}{80} = 0.675\; \text{m} \;= \boxed{67.5 \; \text{cm}}\;##. Problem is, doesn't match the answer in the...

A uniform rod AB of length ℓ is free to rotate about a horizontal axis passing through A. The rod is released from rest from the horizontal position. If the rod gets broken at midpoint C when it becomes vertical, then just after breaking of the rod. Choose multiple answeres from the below...

(The answer given in the text says ##\boxed{T_1\; >\; T_2}## but, as I show below, I think it's just the opposite). I begin by putting an image relevant to the problem above. Taking a small particle each of the same mass ##m## at the two positions, the centripetal forces are ##T_1 =...

Alright, to start off: I'm not even sure how this works in the first place. What I do understand is that if they carry current in the opposite direction, using right-hand grip rule, the magnetic field between them will be the same (into the page). Hence using the left-hand rule, I can deduce...

Here is my work done for this problem, along with a diagram of the situation. I'm not worried so much about the arithmetic because our tests are only 50 min long so the problems they give us do not require heavy integration or calculus, but you need to know what goes where in the formula. That...

i don't understand why in the solution of this exercise, the induced electromotive force / current is counterclockwise. Shouldn't it be clockwise? ##emf=-\frac {d \Phi_B} {dt} ## According to the picture the rod is moving upward, the magnetic flux (entering the page) is decreasing due to a...

Homework Statement: A 25.0 cm long metal rod lies in the xy-plane and makes an angle of 36.9 with the positive x-axis and an angle of 53.1 with the positive y-axis. the rod is moving in the +x-direction with a speed of 6.80 m/s. the rod is in a uniform magnetic field B=...

if there's a really good rocket hovering just above a super massive black hole with very low tidal force, and it dips a rod just inside the event horizon, will the rod break? it seems a certainly but the tidal forces are very weak. is it like dipping ur feet into piranha infested water? so when...

Summary: When I tried to find the angular frequency of a rod pendulum, I attempted to find its angular acceleration first, however, I realized that the results are different by using different approaches. i.e. (1) Newton's second law for a system of particles (2) Newton's second law for...

I first calculated induced emf and then calculated torque about O.But what will i do further.I think i need to find current and then fidn emf=IR. Help please.

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...

My solution is on the files. The basic approac was the take moments about A. Then find the reaction at C. Use the reaction at C to find the friction between rod and cylinder. I take it to mean from the question that the friction between the rod and plane is equal to the friction between...

think of a engine. it has a flywheel and a rod connected to it. a string had been totally wrapped around the rod and a mass is hanged from the very end of the rod. the system is in equilibrium. but as the engine starts to rotate, the rod with rotate as well and cause the hanged object to go...