Recent content by frostedpoptar

Thanks so much! After looking at what you did I worked it out on my own and it all came out perfectly! I think this was the only problem I had a problem on in preparation for the final so if I get a 100 it goes out to you! Thank you a million times! :)

OK, I went through your entire method and the answer I got was .43 s. The correct answer however is .56 s. Did I do something wrong, or do you get the same answer as well? Thanks for the continued support.

Oh nevermind I see what you did. You just didnt substitute the 2 for m.

Just one thing I'm confused about in your explanation now In "T1-10=Ia/r^2=(1/2Mr^2*a)/r^2=(ma/2)" how did you get ma/2? When I simplify this I get T1-10 = ((1/2)(M)(r^2)(a))/(r^2) T1 - 10 = ((1/2)(2)(r^2)(a))/(r^2) 1/2 cancels the 2 T1 - 10 = ((r^2)(a))/(r^2) r^2's cancel T1-10 =...

Houdini, thanks so much for the reply. I see what you're getting at. It's probably just me but I can't seem to apply it. Torque cw = T * .25m Torque ccw = mg - T * .25m? I'm probably making the same mistake you mentioned in your first paragraph though. Can you please elaborate? It's greatly...

Then how do I go about doing this?

I also tried to do F = ma and in turn, 4.7 = 1.5a I then substituted that a into kinematics and got something like .43 s. Still wrong.

Homework Statement problem #62 Homework Equations Torque = F * radius Tangential accel = radius * alpha w = mg Kinematic equations for linear motion The Attempt at a Solution I tried to find the net torque and go from there. For clockwise torque I got T = 10 * .25 = 2.5 Nm...

I got it! its 5.9 when you factor in the .5 year of acceleration. Thanks so much guys.

I got 5.3589 years as the time it takes from the moment the spacecraft stops accelerating to the moment it reaches the star..

Gotcha. I plugged in the new value (15,768,000 s) and got these values for v(f) and delta x: v(f) = 231,789,600 m/s delta x = 1.827429206 x 10^15

I also plugged that into v(f)^2 = v(i)^2 + 2a(delta x) and got delta x as 1.26904806 x 10^13 meters.

I just tried to work it out partially. I used v(f) = v(i) + a*t As v(f) = 14.7(1314000) so I got v(f) as 19,315,800 m/s

G01, I was attempting to do that yesterday and it didn't work out. I erased it from my whiteboard but I will attempt to do it again and post my results. gneill, You're completely write, I left some things out in the main post, I just edited it. Thanks for noticing. I'll attempt to work it...

Homework Statement People hoping to travel to other worlds are faced with huge challenges. One of the biggest is the time required for a journey. The nearest star is 4.1 x 10^16 m away. Suppose you had a spacecraft that could accelerate at 1.5g for half a year, then continue at a constant...