Recent content by kenway

Sorry it's kg*m2. But how does that affect my equation? Torque is measured in N*m... Do I have to multiply 0.30 by 9.8m/s2 because of gravity to put it in Newtons?

So .72rev/s = 1.44π(rad)/s or 4.52rad/s So then I put that into my equations: α=Δω/Δt =4.52rad/sec / 12s =0.38rad/sec2 ∑τ=Iα = (0.30kg/m2)(0.38rad/s2) = 0.79Nm And it doesn't state which direction the wheel was spun, so I am assuming the negative/positive is irrelevant

I honestly don't know how to find the friction, much less frictional torque, when given things like this. 1. Homework Statement "If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.72rev/s, friction in the bearings causes the wheel to stop...

I used the y comp. of the ramp (that's what I used in the first equation, 5.14m), but I can see where that could be wrong. The length of the ramp (hyp of the triangle) would then be 12.16 m. Then to find the time: sqrt(12.16/2.36/2) = t 3.21 s = t v=at v = 2.36(3.21) v = 7.58 m/s

Okay, this makes more sense now! Thank you both for all your help. I feel much more confident about this. I appreciate it very much!

Going up, the block faced friction and it's own weight, but it's initial velocity overcame that. Going down, it doesn't have the velocity, but the friction is now facing up, away from the direction the block is moving, and it's "weight" still faces down, which pushes it down the ramp, so to...

So that would be the x-comp. of the weight? I found: mgsinθ = x But that was the x-comp. of the block's weight, not gravity. If that's not the correct answer, then I did the first part wrong

Oh! So then I find the comp. that runs along the incline. So: sin25=9.8/hyp 9.8/sin25=hyp 23.19 The hyp runs along the incline, but this seems like a high number. If it's supposed to be, then I will go with it, but when I plugged it into the equations I did in the last post I got 20.59 m/s...

Gravity causes it to slide, friction opposes it. So do I do it like this?: a = (9.8-5.33)/3 Gravity - friction (found above) / mass a = 1.49 m/s2 Then I would find the time it takes to get to the bottom? So: y = .5at2 5.14 = .5(1.49)t2 sqrt(5.14/.0745 = t 2.63 s = t Which I can then use in...

So how do I calculate the second answer of the question: What speed does it have when it slides back down to its starting point? I honestly don't even know where to start.

As I described in the above paragraph, I tried again by plugging in 12 for the 0 (that you have highlighted) and got 12.16 m. If that is the answer, then I can use this equation to find the height: 12.16sin25=y 5.14 m = y I think this is the answer

They're both negative? So I would set it up as: a = (-12.42 + (-5.33))/3 a = -5.92 m/s2 I wouldn't subtract the friction, because the two forces are adding on each other, so to say. Then I can find time: v = at 12/-5.92=t 2.03 s = t Then I can plug all my variables into this kinematic...

Homework Statement A 3 kg wood block is launched up an inclined wooden ramp that has an angle of 25° to the horizontal. the Block's initial speed is 12 m/s and the kinetic coefficient of friction is 0.2. What is the vertical height the block reaches above it's starting point? What speed does it...