Recent content by cpgp

Thanks, I understand now.

A cylinder of radius R spins with angular velocity w_0 . When the cylinder is gently laid on a plane, it skids for a short time and eventually rolls without slipping. What is the final angular velocity, w_f? The solution follows from angular momentum conservation. $$L_i = I \omega_0 = L_f =...

Yes, x is l(v0-V)/(v0+2V).

The total time for those two transits is 2l/(v0+V). The distance traveled is l(v0-V)/(v0+V).

Assuming the ball is released at r = 0, it first collides with the moving wall at t = 2l/(v0+V).

I thought that might be the way to calculate the force in terms of x.

I have been able to solve part a correctly, but am having some trouble with part b. I started out calculating values of v and x at various points in time when the ball collides with either wall. I found: t = 0, v = v0, x = l t = (l/v0), v = v0, x = l(v0-v)/v0 t = 2l/(v0+V), v = v0 + 2V, x =...

Thanks a lot, that has given me the answer!

That makes sense, but I still only get alpha = 52.4 rad/s^2.

I have shown that in my post above

I've typed up my work a little better. Relation derived in part a: $$\omega = \frac{237\times 2\pi}{24}\times v$$ $$\omega = 62v$$ For part b: $$\Sigma F = ma$$ $$ma = \frac \tau r + mg\sin(5)$$ $$ma = \frac {I\alpha} r +mg\sin5$$ $$ma = \frac {Ia} {r^2} +mg\sin5$$ $$a(m-\frac I {r^2}) =...

R is the radius of the flywheel.

So, then it should be given by ma = I*alpha/r + mg sin5? This gives me alpha=(g/r)sin5/(1-I/(mr^2)). Substituting the values of I=2.83, r=0.54, m=19.4, I get alpha=3.17 rad/s^2, but even that is not the right answer.

I have solved part a using the conservation of energy, getting a (correct) answer of 47.9 km/h, but I am unable to make headway with part b. Based on the flywheel rotating at 237rev/s when the car is moving at 86.5 km/h, I obtained omega = (237*2pi)v/24=62v. Differentiating both sides should...