Recent content by solarcat

No, but you asked "How far does the mass centre of the rod rise?" If you consider the change in height of the mass center of the rod, instead of the end of the rod as I did previously, you have to account for the fact that the mass center is initially at height R/2 = 3 m. Anyway, the problem...

If you calculate the potential energy based on the mass center, you'd have to have some initial potential energy, right? Initial Energy = 1/2 (2808)( 0.662 √h)2 + (94 kg) (9.8 m/s/s) (3 m) = 615.295 h + 2763.6 I'm having trouble calculating the final height. I drew this diagram, and I used 6...

1. Descent of slide - Use conservation of energy to find speed at bottom of slide mpersongh = 1/2 mpersonv2 gh = 1/2 v2 2gh = v2 v = √2gh 2. Seizing the pole - Use conservation of angular momentum Moment of inertia of person = (70 kg) (6 m)2 = 2520 Moment of inertia of rod + person = 2808, as...

First I found the moment of inertia of the rod + rider to be I = (70 kg * 62 m2+ 1/3 * 24 kg * 62 m2)= 2808 Then I found initial angular momentum of the pole and rider just as the rider grabs the pole: L = Iw = 2808 v0/ 6 m = 468 v0 Then I found the final angular momentum of the pole and rider...

OK, I see what I did there. But I still don't understand the difference between calculating torque using the radius or the horizontal and vertical components. If I look at the components, it looks like the horizontal forces Fb and Tcos 30 don't produce any torque. The sum of torques would be 4 m...

Okay, so using the side lengths of the rectangle, I can see the angle between the gates bisector is tan theta = 2/4; theta = 26.6 degrees. From the geometry of the rectangle the angle between the bisector and the tension should be 52.5. 1400=T*sqrt 20 * sin 26.6 deg T = 375 N? So in that case...

OK, so I don't know what to do next though. You said earlier that I wasn't taking the vector part into account, but can't you also use the equation torque = Fd sin theta? I don't understand why this doesn't work, and I don't understand why the radius isn't the square root of 20.

I thought the free body diagram might look something like this

I thought that the upward/vertical torque due to the cord might be T*sin 30 deg * sqrt (4^2 + 2^2). Then this would have to equal the weight of the gate times the distance (700 N * 2 m). This gives a tension force of 626 Newtons but it's not the right answer. I don't even know what the lever arm...

A net torque is required to make the pull rotate, so this can only occur if the tension forces differ. I know mass 1 will accelerate upwards and mass 2 will accelerate downwards, so T1>mAg and T2<mBg. But I don't know where to go from there.

This is what I did to find the work required to get the spacecraft to height h: mv^2 / (re + h) = GmMe/(re+h)^2 (v is velocity at height h) Cancel m: v^2/(re + h) = GMe/(re+h)^2 Multiply (re+h) by both sides; v^2 = GMe/(re+h) h is negligible: v^2 = GMe/re Final KE = 1/2 * m * GMe/re Also...

Homework Statement Say you're launching a spacecraft of mass m from the surface of the Earth (mass Me and radius re) to a low height h (h is much smaller than r, so h is essentially negligible). How much work is required to move the spaceship from its low orbit to a great distance from earth...

OK, fair enough. But I have also derived the equation amax = k/m(H-L) - g. This is because when the cord starts stretching, the net force on the jumper ma = Ft-mg. a is maximized when Ft (tension) is greatest, which would be when the cord is stretched the most, which would be at the bottom. Then...

Homework Statement A person is bungee jumping from the top of a cliff with height H. The un-stretched length of the bungee rope is L. The person comes to a stop just before hitting the ground. The length of the cord is equal to H(amax-g)/(amax+g), where amax is the maximum acceleration upward...