Recent content by benoconnell22

Homework Statement Suppose the motion of a spring has natural frequency 1/2 and is undamped. If the weight attached is 32lb, write a differential equation describing the motion. Homework Equations my''+ky=F_ocosωt 32y"+8y=? ω_o= (k/m)^.5 The Attempt at a Solution → .5=(k/32)^.5 → k=8...

I'm sorry I meant .934 m/s^2 I was practicing this problem with different quantities and accidentally wrote the answer one of the others.

I ended up getting a=.901m/s^2

Thank you Sammy and Tim. You guys are very helpful, thank you for pushing me and allowing me to think. Even though I didn't come up with the answer on my own, I stressed to think about this problem and was incorrect on a few concepts but you guys allowed me to learn more about moments of...

Alright guys this still isn't right. The answers are all off, I plugged in the respective values and continue to get the wrong answer. Solve for Fp with Fp(2r)=(.5(2m)(2r)^2+2m(2r)^2)a/2r then plug into F-Fp=ma and still get the wrong answer. Still get Fp=6a F-Fp=ma a=.551

May I ask your reasoning why you multiplied Fp by 2R? Is that because there are two cylinders?

A plank with a mass M = 5.60 kg rides on top of two identical, solid, cylindrical rollers that have R = 4.40 cm and m = 2.00 kg. The plank is pulled by a constant horizontal force of magnitude 6.40 N applied to the end of the plank and perpendicular to the axes of the cylinders (which are...

ok, so I decided to do the parallel axis theorem and got the wrong answer. t=Fpr=(.5(2m)(r^2)+2mr^2)a/r I have 2m because I am treating it as one cylinder with twice the mass. Once I solved for Fp I plugged that into my F-Fp=Ma and got the wrong answer again.

How else can you do this without having Fg in the equation?

I got a=.877 but there is a flaw some where in my calculating because for all my answers, it say's I'm 10% within the expected value and to check my calculations. What do you think?

I believe I figured it out, ok so I took F-Fp=ma T=FpR and T=Iα For figuring the correct I value, I did the parallel axis theorem, I=I_cm+MD^2 where I_cm=.5MR^2 and MD^2=4MR^2 so my combined I value is .5MR^2+4MR^2=9/2MR^2=I The I took my I value and multiplied it by α=a/R...

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Alright so would I do (Fp-Fg)(R)=(.5(MR^2)+M(2R^2))a/R for the parallel axis theorem then? T=RxF which is equal to the moment of inertia (using the parallel axis theorem) times a/r. Then solve for a but then if i plug that into my F=ma equation i still have three variables :(

Ok so add the moment of inertia of the center of mass to the moment of inertia from the ground? Would the moment of inertia from the ground be .5M2R^2

You're right, 2(Fp-Fg)/M=a So 6.4-Fp=M(2(Fp-Fg)/M) I'm missing something huge because I continue to get the wrong answer. I get something that's close but off by a bit.