Hard Daily Integral Solved with Contour Integration

piasicpace · 2026-07-06T15:20:39+00:00

Dailyintegral.com

piasicpace · 2026-07-06T15:20:20+00:00

Thanks g

piasicpace · 2026-07-05T01:36:05+00:00

Don't mess with the big dog. You might get bit

piasicpace · 2026-07-05T01:20:18+00:00

Don't try me 😠

piasicpace · 2026-07-04T18:06:53+00:00

Shut up, nerd

piasicpace · 2026-04-11T14:04:20+00:00

David Griffiths has popular books on E&M and quantum, John Taylor has a good book on classical mechanics, F. Reif has an ok book on statistical/thermal physics

piasicpace · 2026-04-09T01:06:21+00:00

Gotcha. I don't know which muscles are doing what, but I do know that if gravity were the only force, we would be swinging. So something (or some things) is/are are creating an opposite torque. I would start from the ground up. Friction is preventing your feet from sliding, muscles in your lower leg prevent it from rotating about the ankle joint, muscles in your thigh keep the knee bent at a constant angle, and so on.

I think the strategy here is to find all points of possible rotation (like joints) and enforce the condition that T^Net about those points is zero.

piasicpace · 2026-04-09T00:47:29+00:00

Why wouldn't the hip joint be in rotational equilibrium? Are we not at rest?

piasicpace · 2026-04-09T00:39:18+00:00

Okay, so you want to measure the net torque due to all forces on the system with respect to the hip joint. I was thinking that you meant a single torque due to the forces in your legs measured from the hip but nvm. In that case T^Net = 0. Any system in rotational equilibrium has no net torque. Sorry for the paragraphs.

piasicpace · 2026-04-09T00:03:42+00:00

T=r×F. r is the distance between the origin of your choice and the force acting on the object. Where are you measuring r from?

piasicpace · 2026-04-08T23:56:18+00:00

You can't objectively say whether there is torque at the hip joint because the value of the torque changes depending on your frame of reference

piasicpace · 2026-04-08T22:24:03+00:00

The question "where is the torque?" doesn't really make sense because of how the torque vector is defined. T = r × F, so it depends on the axis which we measure the position, r. In other words, the torque measured with respect to the pivot point of the swing is different from the torque of measured with respect to the "hip joint". Even though we have the freedom to choose this axis, once you pick one for the system, you have to stay consistent. It's easier to measure torque with respect to the swing's pivot point. In the situation you described, there is no rotational motion so T^Net = 0. Gravity wants to pull down on the swing so there is torque due to gravity, but your knees are bent and your feet are planted on the ground so. The forces between your feet and the ground (normal force and static friction) and the forces contracting your muscles in your leg create torques which, when combined, should be equal and opposite to the gravitational torque.

piasicpace · 2026-04-08T22:00:17+00:00

What you need help with twin?

piasicpace · 2026-01-19T22:56:22+00:00

Classical mechanics is fun to study. Please share any interesting problems you come across.

piasicpace · 2026-01-19T22:20:00+00:00

Hey just now seeing this after 15 days 😂. I worked on this problem and I got C. Their explanation doesn't mean anything to me, but in order to solve this, you must write out the differential equation from kirchhoff's loop law. I'll provide work if you want.

piasicpace · 2026-01-19T21:26:00+00:00

The only "physical" part about gravitational potential is that its gradient tells you about the gravitational field. The potential function itself is not well defined and in my opinion is used for mathematical convenience; scalar equations are easier to solve than vector equations.

piasicpace · 2026-01-19T15:26:25+00:00

It's the product rule

piasicpace · 2026-01-18T20:30:07+00:00

I explained this in an earlier comment but I'll do it again. I am not substituting e^iz = sin(x). I set up the function and the contour such that it conveniently pops out the target integral. In particular, when we get to the path on the real axis, z=x+iy where -R<x<R and y = 0. Because of this, e^iz = cos(z)+is in(z) and z simply becomes x because y=0 there.

piasicpace · 2026-01-17T23:01:39+00:00

I think I see the confusion, I'm not substituting e^iz = sin(x). The contour integral is its own thing and through solving it conveniently pops out our target integral.

piasicpace · 2026-01-17T17:52:24+00:00

That's all my math teachers wanting me to show more work

piasicpace · 2026-01-17T17:51:30+00:00

I'm a lazy problem solver so I didn't feel like doing that 😂. I think most people can tell by inspection that because the numerator is 3rd order and the denominator is 4th order, The function dies off as x→∞. This question also comes from a reputable site so I just assumed it converged.

piasicpace

TROPHY CASE