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...
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
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.