Recent content by Dennydont

These equations are rather useful, but I just want to know if Pafter = (E', p', 0, 0). If so I can substitute E' as E+m and p' as p, using conservation of energy and momentum laws. Thanks to the third equation you've given me, I can say that p = √(E2-m2) and p' = √(E2-m2) and solve for m' since...

Great, do I have all the tools to solve this then? Is Pafter = (E', p', 0, 0) correct? How can I find the speed then?

Homework Statement A photon with energy E collides with stationary mass m. They form a single particle together, what is this new particle's mass and what is its speed? Homework Equations Energy-momentum 4-vector P=(E, px, py, pz) Possibly P2=m2 The Attempt at a Solution Using 4- momenta, the...

If you had twin 1 on the earth, and twin 2 fly to a star and back at a speed of v with the Earth and star separated by a distance L, twin 1 sends out flashes at intervals of t seconds (measured in his frame). Taking into consideration the numbers of redshifted and blueshifted flashes that the...

Homework Statement Two trains of proper length L move toward each other in opposite directions on parallel tracks. They both move at speed v with respect to the ground. Both trains have clocks at the front and back, and these clocks are synchronized as usual in the frame of the train they are...

ml^2 = I? Multiplied by 2 since the inertia for one of the bars is the same as the other? How on Earth do I solve this differential equation for period T?

I always have issues with the moment of inertia. But okay the expression should look like: d2θ/dt2*I = -kθ and I should rearrange everything to find dt which happens to be the period? I don't quite understand the inertia but perhaps it is the sum of the contributions from the two masses? So m =...

x = 0.5, that's the distance between the two m's. Okay, but is that the right equation for two-point masses? I = (m2/2m)x^2 since m and M are the same? α = d2θ/dt2 something like that?

Good question... I don't know. The drag force equation?

I'm not sure if this is correct, but could I relate spring constant to torque? τ=-kθ And I realize now that ml^2 isn't the inertia for two-point masses? That should be: I = (mM/m+M)x^2 If I relate it to τ= Iα then: -kθ = Iα Hmmm, can I say that α = ω/t where t is the period and just try to find...

To the velocity, but it is also opposite to the velocity?

Okay! That's what I was thinking, I just needed some confirmation just to be super safe. And finally got the right answer! Thank you so much for your help! You and ehild. :smile:

Hmmmm, okay! So here are the variables I currently have: d (length of rod) = 1.4m r = 0.2m m = 3.9kg I = 1/3 m d2+ ( 1/2 mr2 + md2 ) = 10.27kg*m2 T = 2π√(I/MgL) M (combined mass) = 7.8kg I = 10.27kg*m2 g = 9.81m/s2 L = Wait how do I get this using the centre mass of 1.05m?

Well there is the equation for frictional force: Ff=μFn

Okay, so using that equation I get 1.05m for the centre of mass. How does that work into the problem? Can I use that to find L in T = 2π√(I/mgL)?