Recent content by Phys student

So in the end we have: K=\frac{1}{2}m\dot{ξ}^2(\sin^2{θ}+\cos^2{θ}(1-\frac{m}{M})^2)+\frac{2}{10}m\dot{ξ}^2 +\frac{m^2\cos^2{θ}\dot{ξ}^2}{2M} V=-mgξ\sin(θ) \dot{E}=m\ddot{ξ}(\sin^2{θ}+\cos^2{θ}(1-\frac{m}{M})^2)+\frac{4}{10}m\ddot{ξ}+\frac{m^2\cos^2{θ}\ddot{ξ}}{M}-mg\sin{θ}=0 When we solve...

What about the rotational kinetic energy? What v do you use?

The x component of the ball's velocity with respect to the ground is: v_x= \dot{ξ}\cos{θ}-\dot{η}=\dot{ξ}\cos{θ}(1-\frac{m}{M}) And the y component is: v_y =\dot{ξ}\sin{θ} Right?

Is the kinetic energy of the ball the sum of the kinetic energies of both coordinates? (ball moving with respect to the incline and the ground) \frac{1}{2}m\dotξ^2 +\frac{1}{2}m\dotη^2

So the expression should have been: \frac{1}{2}m({{\vec{\dotξ}}}+{{\vec{\dotη}}})^2 ?

It is along the slope, How did the kinetic energy expression make you assume it's relative to the ground?

Homework Statement [/B] A uniform solid ball of mass m rolls without slipping down a right angled wedge of mass M and angle θ from the horizontal, which itself can slide without friction on a horizontal floor. Find the acceleration of the ball relative to the wedge. 2. The attempt at a...

I see what you mean now. If we substitute x= \frac{L}{2} we get: T_m = \frac{Mω}{3gλμ} Now, when we substitute x=0 or x=L we get: T_M = \frac{2Mω}{3gλμ} So: \frac{T_m}{T_M} = \frac{1}{2} Can this result be derived intuitively, without calculus? Anyway, thank you very much...

WolframAlpha link for derivative: http://goo.gl/M7Psfe I guess it could be possible that there's only one extremum but it would mean that we don't have both a maximum and minimum time.

We have: α=\frac{μgλ(L^2 +2(x^2-Lx))}{\frac{1}{6}ML^2 +2M(\frac{L}{2} -x)^2} T= \frac{ω}{α} = \frac{ω(\frac{1}{6}ML^2 +2M(\frac{L}{2} -x)^2)}{μgλ(L^2 +2(x^2-Lx))} When we take the derivative we get: \frac{d}{dx} \frac{ω(\frac{1}{6}ML^2 +2M(\frac{L}{2} -x)^2)}{μgλ(L^2 +2(x^2-Lx))} =...

I used the parallel axis theorem. But I think I made a mistake, because the theorem applies to the moment of inertia around the center of mass so it should have been: I=\frac{1}{12}ML^2 +M(\frac{L}{2} -x)^2

Homework Statement A uniform rod of mass M and length L can rotate around point P which is at position x from one end of the rod as shown in the figure. The rod is gently placed on a rough horizontal surface that has a friction coefficient μ. and at t=0 starts rotating with angular velocity...

You want to find a relation between the velocity of car A and car B, and you're given a relation between the radii of their circular paths and a relation between their centripetal accelerations. Try to write down some equations.

So the answer is 1.7A because after changing to fuse C a larger maximal current is possible where fuse A would burn and the current (1.7A) would be redirected to fuse C only.

Then fuse A would burn and the current would be redirected to fuse C only?