haruspex said:
The right hand side is wrong. It treats the ball as rotating as a whole about the centre of curvature of the trough, and ignores the rotation involved in its rolling.
Thanks a lot! At last, I understand why I was wrong! The net torque equation I wrote down just neglects the rotation involved in its rolling. However, that means I can't use the net torque equation about the center of curvature of the trough because that's meaningless. I can't use the net torque equation unless all points on the object has the same angular velocity about the pivot, right? Is this correct?
I guess I still can use the net torque equation because that's the derivative of kinetic energy with respect to θ.
Well, I still have some problem on it. Here's my reasoning.
Its total kinetic energy is
E_{k,total}=\sum_{i=1}^{n}\frac{1}{2}m_{i}v_{i}^{2}=\sum_{i=1}^{n}\frac{1}{2}m_{i}(v_{ic}+v_{c})^{2}=\frac{1}{2}Mv^{2}+\frac{1}{2}I(\frac{d\theta }{dt})^{2}
where dθ/dt is the rotation rate of the ball about its center of mass. That is,
\frac{d\theta }{dt}\neq \frac{d\Omega }{dt}
However, we still can find their relation through the instantaneous speed of center of mass of the ball:
v=4R(\frac{d\Omega }{dt})=R(\frac{d\theta }{dt})
So, I can say its total kinetic energy is:
E_{k,total}=\frac{1}{2}M(4R\frac{d\Omega }{dt})^{2}+\frac{1}{2}(\frac{2}{5}MR^{2})(\mathbf{4}\frac{d\Omega }{dt})^{2}
\therefore \: E_{k,total}=\frac{1}{2}[M(4R)^{2}+(\frac{2}{5}MR^{2})(\mathbf{4})^{2}](\frac{d\Omega }{dt})^{2}=\frac{1}{2}I_{eff}(\frac{d\Omega }{dt})^{2}
\therefore \: \frac{dE_{k,total}}{dt}=I_{eff}(\frac{d\Omega }{dt})(\frac{\frac{d\Omega }{dt}}{d\Omega })=I_{eff}(\frac{d^{2}\Omega }{dt^{2}})=I_{eff}\alpha =\tau _{net}
Now I can see why the right-hand side of my net torque equation was wrong, the true effective moment of inertia is
I_{eff}=M(4R)^{2}+(\frac{2}{5}MR^{2})(\mathbf{4})^{2}
The factor
4 is the key point...
But when I plug it in the net torque equation, I found I'm still wrong.
[-Mg\times 4Rsin(\Omega )]+[\frac{2}{7}Mgsin(\Omega )\times 5R]=[m(4R)^{2}+(\frac{2}{5}mR^{2})(\mathbf{4})^{2}](\frac{d^{2}\Omega }{dt^{2}})
\therefore \: -\frac{18}{7}MgRsin(\Omega )=\frac{112}{5}MR^{2}(\frac{d^{2}\Omega }{dt^{2}})
\therefore \: -\frac{45g}{392R}\Omega \approx (\frac{d^{2}\Omega }{dt^{2}})
\therefore \: T\approx 2\pi \sqrt{\frac{392R}{45g}}=2\pi \sqrt{\frac{43.\bar{5}R}{5g}}
That means my static frictional force is wrong, but why?...
Thanks a lot!edit:
If I calculate the "
right static frictional force", then ..
[-Mg\times 4Rsin(\Omega )]+[+f_{s}\times 5R]=(\frac{112}{5}MR^{2})\frac{d^{2}\Omega }{dt^{2}}
\because \: T=2\pi \sqrt{\frac{28R}{5g}}\: \therefore \: -\frac{5g}{28R}\Omega \approx \frac{d^{2}\Omega }{dt^{2}}
\therefore \: \frac{-20Mg\Omega +25f_{s}}{112MR}=-\frac{5g}{28R}\Omega
\therefore \: f_{s}=0
That is to say, the net torque about the center of mass of the ball is zero so that its acceleration should be zero. Why?...