Recent content by Np14

Doing it symbolically makes it more confusing for me, so I will use numbers. Torque on cylinders: τ = Iα = ΣFrperp. τ = 45α = ΣF(0.5) ΣF = 90α Net force on hanging mass: ΣF = Fg - Ft = 20(9.8) - 20(a) = 196 - 20(αr) = 196 - 20(α(0.5)) Combine: 90α = 196 - 20(α(0.5)) 100α =...

Doing it symbolically makes it more confusing for me, so I will use numbers. Torque on cylinders: τ = Iα = ΣFrperp. τ = 45α = ΣF(0.5) ΣF = 90α Net force on hanging mass: ΣF = Fg - Ft = 20(9.8) - 20(a) = 196 - 20(αr) = 196 - 20(α(0.5)) Combine: 90α = 196 - 20(α(0.5)) 100α =...

I meant the mass of the two cylinders combined, which I thought constituted the entire system, but now I realize I didn't factor in the hanging weight, m1. Regardless I don't think that solving for the 90α (or the mass of the entire system) is relevant to the problem.

I did not use that equation, but I got the wrong answer, so it is most likely necessary to solve the problem. I'm pretty sure I did, however, differentiate between the two variables.

Yes I understand that. aT = rα

PART B ONLY: The cylinder undergoes torque when the mass m2 is removed: τNET = Iα = FNETr = 45α = FT(0.5) FT = 90α, therefore msystem = 90 kg After this step, I am not sure what to do. τ = ΣF = Fg - Ft = ma = (20 kg)(9.8 m/s2) - (90α) = 196 -...

EDIT: EDIT: My bad, there should only be one F for the first expression (didn't use two in my calculations anyways) I see what you mean. The length should be some constant times L, I just don't know what it should be. Using the first way, sin30 x 0.6 = 0.3 I guess I just typed the...

The system is in rotational equilibrium and therefore experiences no net torque, meaning all individual torques must add to zero. τNET = 0 = FFTsin(θ)L - FgL - Fg(L/2) τNET = 0 = FTsin30°(0.6?) - (0.5)(9.8)(0.6) - (2.0)(0.6/2) My only problem (I think) is figuring out what the length L is for...

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Thank you, now it makes so much more sense. Ft = 0.5(3.7 - (-4.85)) Ft = 4.28 Ns

Ft = m(v - vo) Ft = (0.5)(3.7 -4.85) Ft = - 0.575 Ns, which is incorrect I misspoke. The ball falls due to gravity and at the instant of collision changes velocity.

I am not sure what you mean by asking what the new kinetic energy is. It would just be K = Ug, right?? 1/2mv2 = magh 1/2(0.5)(VB)2 = (0.5)(9.8)(0.7) VB = 3.7 m/s I understand your point that velocity is a vector, and the answer would be a nonzero ±2 (depends on which direction is specified as...

Homework Statement A fireworks rocket is moving at a speed of 45.0 m/s. The rocket suddenly breaks into two pieces of equal mass, which fly off with velocities v1 and v2. What are the magnitudes of v1 and v2? Homework Equations Conservation of Momentum m1v1 + m2v2 = m1vo1 + m2vo2 The...

Okay so I tried to find second impulse but it did not work out... KA = Ug,B + KB 1/2(0.5)(4.85)2 = (0.5)(9.8)(0.7) + 1/2(0.5)(VB)2 VB = 5.55 m/s ImpulseB = (0.5)(5.55 - 4.85) ImpulseB = 0.35 N(s) ImpulseA + ImpulseB = 2.42 + 0.35 = 2.77 N(s)

What do you mean by impulse to stop the ball? Please elaborate.