SOLUTION
The top portion of the rope moves with same speed as the hanging portion. Hence \(z\) is used to describe the motion as the generalized coordinate. From the above\begin{align*} U & =-\left ( \frac{1}{2}z\right ) \left ( \frac{z}{l}\right ) mg=-\frac{1}{2}\left ( \frac{z^{2}}{l}\right ) mg\\ T & =\frac{1}{2}\left ( \frac{z}{l}\right ) m\dot{z}^{2}+\frac{1}{2}\left ( \frac{l-z}{l}\right ) m\dot{z}^{2}=\frac{1}{2}m\dot{z}^{2} \end{align*}
In finding \(U\) we used \(\frac{1}{2}\) since the center of mass of the hanging part is half way over the length. So the potential energy is taken from the center of mass. In the above, \(\dot{z}\) is used for both parts of the rope, since both parts move with same speed. Applying Lagrangian equations gives\begin{align*} L & =T-U\\ & =\frac{1}{2}m\dot{z}^{2}+\frac{1}{2}\left ( \frac{z^{2}}{l}\right ) mg \end{align*}
Hence\begin{align*} \frac{\partial L}{\partial z} & =\frac{z}{l}mg\\ \frac{d}{dt}\frac{\partial L}{\partial \dot{z}} & =m\ddot{z} \end{align*}
And therefore\begin{align*} \frac{d}{dt}\frac{\partial L}{\partial \dot{z}}-\frac{\partial L}{\partial z} & =0\\ m\ddot{z}-\frac{z}{l}mg & =0\\ \ddot{z} & =\frac{z}{l}g \end{align*}
When \(z=0\) then the acceleration is zero as expected. When \(z=\frac{l}{2}\) then \(\ddot{z}=\frac{1}{2}g\) and when \(z=l\) then \(\ddot{z}=g\) as expected since in this case the rope will all be falling down on its own weight due to gravity and should have \(g\) as the acceleration.
SOLUTION
There are two coordinates \(r,\theta \) (polar) and one constraint \begin{equation} f\left ( r,\theta \right ) =r-R=0 \tag{1} \end{equation} Now we set up the equations of motion for \(m\)\begin{align*} T & =\frac{1}{2}m\left ( \dot{r}^{2}+r^{2}\dot{\theta }^{2}\right ) \\ U & =mgr\sin \theta \\ L & =T-U\\ & =\frac{1}{2}m\left ( \dot{r}^{2}+r^{2}\dot{\theta }^{2}\right ) -mgr\sin \theta \end{align*}
Hence the Euler-Lagrangian equations are\begin{align} \frac{d}{dt}\frac{\partial L}{\partial \dot{r}}-\frac{\partial L}{\partial r}+\lambda \frac{\partial f}{\partial r} & =0\tag{2}\\ \frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }}-\frac{\partial L}{\partial \theta }+\lambda \frac{\partial f}{\partial \theta } & =0 \tag{3} \end{align}
But \begin{align*} \frac{d}{dt}\frac{\partial L}{\partial \dot{r}} & =m\ddot{r}\\ \frac{\partial L}{\partial \dot{\theta }} & =mr^{2}\dot{\theta }\\ \frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{\theta }}\right ) & =m\left ( 2r\dot{r}\dot{\theta }+r^{2}\ddot{\theta }\right ) \\ \frac{\partial L}{\partial r} & =mr\dot{\theta }^{2}-mg\sin \theta \\ \frac{\partial L}{\partial \theta } & =-mgr\cos \theta \\ \frac{\partial f}{\partial r} & =1\\ \frac{\partial f}{\partial \theta } & =0 \end{align*}
Hence (2) becomes\begin{equation} m\ddot{r}-mr\dot{\theta }^{2}+mg\sin \theta +\lambda =0 \tag{4} \end{equation} And (3) becomes\begin{align} m\left ( 2r\dot{r}\dot{\theta }+r^{2}\ddot{\theta }\right ) +mgr\cos \theta & =0\nonumber \\ r\ddot{\theta }+2\dot{r}\dot{\theta }+g\cos \theta & =0 \tag{5} \end{align}
We now need to solve (1,4,5) for \(\lambda \). Now we have to apply the constrain that \(r=R\) in the above to be able to solve (4,5) equations. Therefore, (4,5) becomes\begin{align} -mR\dot{\theta }^{2}+mg\cos \theta +\lambda & =0\tag{4A}\\ R\ddot{\theta }+g\cos \theta & =0 \tag{5A} \end{align}
Where (4A,5A) were obtained from (4,5) by replacing \(r=R\) and \(\dot{r}=0\) and \(\ddot{r}=0\) since we are using that \(r=R\) which is constant (the radius).
From (5A) we see that this can be integrated giving\begin{equation} R\dot{\theta }^{2}+2g\sin \theta +c=0 \tag{6} \end{equation} Where \(c\) is constant. Since if we differentiate the above with time, we obtain\begin{align*} 2R\dot{\theta }\ddot{\theta }+2g\dot{\theta }\cos \theta & =0\\ R\ddot{\theta }+g\cos \theta & =0 \end{align*}
Which is the same as (5A). Therefore from (6) we find \(\dot{\theta }^{2}\) to use in (4A). Hence from (6)\[ \dot{\theta }^{2}=-2\frac{g}{R}\sin \theta +c \] To find \(c\) we use initial conditions. At \(t=0\), \(\theta =90^{0}\) and \(\dot{\theta }\left ( 0\right ) =0\) hence \[ c=2\frac{g}{R}\] Therefore \begin{align*} \dot{\theta }^{2} & =-2\frac{g}{R}\sin \theta +2\frac{g}{R}\\ & =2\frac{g}{R}\left ( 1-\sin \theta \right ) \end{align*}
Plugging the above into (4A) in order to find \(\lambda \) gives\begin{align*} -mR\left ( 2\frac{g}{R}\left ( 1-\sin \theta \right ) \right ) +mg\sin \theta +\lambda & =0\\ \lambda & =m\left ( 2g\left ( 1-\sin \theta \right ) \right ) -mg\sin \theta \\ \lambda & =2mg-2mg\sin \theta -mg\sin \theta \\ & =mg\left ( 2-3\sin \theta \right ) \end{align*}
Now that we found \(\lambda \,,\)we can find the constraint force in the radial direction\begin{align*} N & =\lambda \frac{\partial f}{\partial r}\\ & =mg\left ( 2-3\sin \theta \right ) \end{align*}
The particle will leave when \(N=0\) which will happen when\begin{align*} 2-3\sin \theta & =0\\ \theta & =\sin ^{-1}\left ( \frac{2}{3}\right ) \\ & =41.8^{0} \end{align*}
Therefore, the angle from the vertical is \[ 90-41.8=48.2^{0}\]
SOLUTION
From the above, we see that the center of mass has height above the ground level after rotation of\[ h=a-\left ( a-b\right ) \cos \theta \] Taking the ground state as the floor, the potential energy in this state is\begin{align*} U & =mgh\\ & =mg\left ( a-\left ( a-b\right ) \cos \theta \right ) \end{align*}
And the kinetic energy\[ T=\frac{1}{2}I\dot{\theta }^{2}\] Hence the Lagrangian is \begin{align*} L & =T-U\\ & =\frac{1}{2}I\dot{\theta }^{2}-mg\left ( a-\left ( a-b\right ) \cos \theta \right ) \end{align*}
Therefore the equation of motion is\begin{align*} \frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }}-\frac{\partial L}{\partial \theta } & =0\\ I\ddot{\theta }-\frac{\partial }{\partial \theta }\left ( \frac{1}{2}I\dot{\theta }^{2}-mg\left ( a-\left ( a-b\right ) \cos \theta \right ) \right ) & =0\\ I\ddot{\theta }+\frac{\partial }{\partial \theta }mg\left ( a-\left ( a-b\right ) \cos \theta \right ) & =0\\ I\ddot{\theta }-\frac{\partial }{\partial \theta }mg\left ( a-b\right ) \cos \theta & =0\\ I\ddot{\theta }+mg\left ( a-b\right ) \sin \theta & =0 \end{align*}
For small \(\theta ,\sin \theta \simeq \theta \), hence the above becomes\[ \ddot{\theta }+\frac{mg\left ( a-b\right ) }{I}\theta =0 \] Therefore the natural angular frequency is\[ \fbox{$\omega _n=\sqrt{\frac{mg\left ( a-b\right ) }{I}}$}\] When \(a=b\) then \(\omega _{n}=0\) and the mass do not oscillate but remain at the new positions. When \(b>a\) then \(\omega _{n}\) is complex valued. This is not possible, as the natural frequency must be real. So center of mass can not be in the upper half.
SOLUTION
Part (1): There are two coordinates are \(\theta ,\phi ,\) but due to dependency between them (no slip) then this reduces the degree of freedom by one, and there is one generalized coordinate \(\theta \). The constraints of no slip means\[ f\left ( \theta ,\phi \right ) =\left ( R-r\right ) \theta -r\phi =0 \] Which means the center of the small disk move in speed the same as the point of the disk that moves on the edge of the larger cylinder as shown in the figure above.\begin{align*} T & =\frac{1}{2}I\dot{\phi }^{2}+\frac{1}{2}m\left ( \left ( R-r\right ) \dot{\theta }\right ) ^{2}\\ U & =mgh=mg\left ( R-\left ( R-r\right ) \cos \theta \right ) \end{align*}
Using \(I=\frac{2}{5}mr^{2}\) and using \(\dot{\phi }=\frac{\left ( R-r\right ) }{r}\dot{\theta }\) from the constraint conditions, then \(T\) becomes\begin{align*} T & =\frac{1}{2}\left ( \frac{2}{5}mr^{2}\right ) \left ( \frac{\left ( R-r\right ) }{r}\dot{\theta }\right ) ^{2}+\frac{1}{2}m\left ( \left ( R-r\right ) \dot{\theta }\right ) ^{2}\\ & =\frac{1}{5}m\left ( R-r\right ) ^{2}\dot{\theta }^{2}+\frac{1}{2}m\left ( R-r\right ) ^{2}\dot{\theta }^{2}\\ & =\frac{7}{10}m\left ( R-r\right ) ^{2}\dot{\theta }^{2} \end{align*}
Hence\begin{align*} L & =T-U\\ & =\frac{7}{10}m\left ( R-r\right ) ^{2}\dot{\theta }^{2}-mg\left ( R-\left ( R-r\right ) \cos \theta \right ) \end{align*}
And\begin{align*} \frac{\partial L}{\partial \theta } & =-mg\left ( R-r\right ) \sin \theta \\ \frac{\partial L}{\partial \dot{\theta }} & =\frac{7}{5}m\left ( R-r\right ) ^{2}\dot{\theta } \end{align*}
Therefore the equation of motion is\begin{align*} \frac{d}{dt}\frac{\partial L}{\partial \dot{\theta }}-\frac{\partial L}{\partial \theta } & =0\\ \frac{7}{5}m\left ( R-r\right ) ^{2}\ddot{\theta }+mg\left ( R-r\right ) \sin \theta & =0\\ \ddot{\theta }+\frac{g}{\frac{7}{5}\left ( R-r\right ) }\sin \theta & =0 \end{align*}
For small angle\[ \ddot{\theta }+\frac{5g}{7\left ( R-r\right ) }\theta =0 \] The frequency of oscillation is \[ \omega _{n}=\sqrt{\frac{5g}{7\left ( R-r\right ) }}\] Using \(\omega _{n}=\frac{2\pi }{T}\) then the period of oscillation is\[ T=\frac{2\pi }{\sqrt{\frac{5g}{7\left ( R-r\right ) }}}=2\pi \sqrt{\frac{7\left ( R-r\right ) }{5g}}\]
Part (2):
There are now two generalized coordinates, \(\theta \) and \(z.\) The sphere now rotates in 2 angular motions, \(\dot{\phi }\) which is the same as it did in part 1, and in addition, it rotate with angular motion, \(\dot{\alpha }\) which is rolling down the \(z\) axis. The new constraint is that \begin{equation} f_{1}\left ( \alpha ,z\right ) =z-r\alpha =0\tag{1} \end{equation} So that no slip occurs in the \(z\) direction. This is in additional of the original no slip condition which is\begin{equation} f_{2}\left ( \theta ,\phi \right ) =\left ( R-r\right ) \theta -r\phi =0\tag{2} \end{equation} The following diagram illustrates this
Now there are translation kinetic energy in the \(z\) direction as well as new rotational kinetic energy due to spin \(\alpha \). Therefore\begin{align*} T & =\overset{\text{part(1)}}{\overbrace{\frac{1}{2}I\dot{\phi }^{2}+\frac{1}{2}m\left ( \left ( R-r\right ) \dot{\theta }\right ) ^{2}}}+\overset{\text{due to moving in z}}{\overbrace{\frac{1}{2}m\dot{z}^{2}+\frac{1}{2}I\dot{\alpha }^{2}}}\\ U & =mgh=mg\left ( R-\left ( R-r\right ) \cos \theta \right ) \end{align*}
Notice that the potential energy do not change, since it depends only on the height above the ground. Using \(I=\frac{2}{5}mr^{2}\) and from constraints (1,2) then \(T\) becomes\begin{align*} T & =\frac{1}{2}\left ( \frac{2}{5}mr^{2}\right ) \overset{\dot{\phi }^{2}}{\overbrace{\left ( \frac{\left ( R-r\right ) }{r}\dot{\theta }\right ) ^{2}}}+\frac{1}{2}m\left ( \left ( R-r\right ) \dot{\theta }\right ) ^{2}+\frac{1}{2}m\dot{z}^{2}+\frac{1}{2}\left ( \frac{2}{5}mr^{2}\right ) \overset{\dot{\alpha }^{2}}{\overbrace{\left ( \frac{\dot{z}}{r}\right ) ^{2}}}\\ & =\left ( \frac{1}{5}mr^{2}\right ) \frac{\left ( R-r\right ) }{r^{2}}\dot{\theta }^{2}+\frac{1}{2}m\left ( R-r\right ) ^{2}\dot{\theta }^{2}+\frac{1}{2}m\dot{z}^{2}+\left ( \frac{1}{5}mr^{2}\right ) \frac{\dot{z}^{2}}{r^{2}}\\ & =\frac{7}{10}m\left ( R-r\right ) \dot{\theta }^{2}+\frac{7}{10}m\dot{z}^{2} \end{align*}
Hence the Lagrangian is\begin{align*} L & =T-U\\ & =\frac{7}{10}m\left ( R-r\right ) \dot{\theta }^{2}+\frac{7}{10}m\dot{z}^{2}-mg\left ( R-\left ( R-r\right ) \cos \theta \right ) \end{align*}
This part only now asks for motion in \(z\) direction. Hence\begin{align*} \frac{\partial L}{\partial z} & =0\\ \frac{\partial L}{\partial \dot{z}} & =\frac{7}{5}m\dot{z} \end{align*}
Since \(\frac{\partial L}{\partial z}=0\) then \[ \frac{d}{dt}\frac{\partial L}{\partial \dot{z}}=0 \] Hence \(\frac{\partial L}{\partial \dot{z}}\) is the integral of motion. Or \[ \frac{7}{5}m\ddot{z}=0 \] or\begin{align*} \ddot{z} & =0\\ \dot{z} & =c \end{align*}
Where \(c\) is constant. This means the sphere rolls down the \(z\) axis at constant speed.
SOLUTION
This is first solved using energy method, then solved using Newton method.
Energy method
Constraint is \(f\left ( y,\theta \right ) =y-a\theta =0\). Hence \(\dot{\theta }=\frac{\dot{y}}{a}\)\begin{align*} U & =-mgy\\ T & =\frac{1}{2}I\dot{\theta }^{2}+\frac{1}{2}m\dot{y}^{2}\\ & =\frac{1}{2}I\left ( \frac{\dot{y}}{a}\right ) ^{2}+\frac{1}{2}m\dot{y}^{2}\\ & =\frac{1}{2}\left ( \frac{1}{2}ma^{2}\right ) \left ( \frac{\dot{y}}{a}\right ) ^{2}+\frac{1}{2}m\dot{y}^{2}\\ & =\frac{1}{4}m\dot{y}^{2}+\frac{1}{2}m\dot{y}^{2}\\ & =\frac{3}{4}m\dot{y}^{2} \end{align*}
Hence \begin{align*} L & =T-U\\ & =\frac{3}{4}m\dot{y}^{2}+mgy \end{align*}
Therefore\begin{align*} \frac{\partial L}{\partial y} & =mg\\ \frac{\partial L}{\partial \dot{y}} & =\frac{3}{2}m\dot{y}\\ \frac{d}{dt}\frac{\partial L}{\partial \dot{y}} & =\frac{3}{2}m\ddot{y} \end{align*}
And the equation of motion becomes\begin{align*} \frac{d}{dt}\frac{\partial L}{\partial \dot{y}}-\frac{\partial L}{\partial y} & =0\\ \frac{3}{2}m\ddot{y}-mg & =0\\ \ddot{y} & =\frac{2}{3}g \end{align*}
Newton method
Using Newton method, this can be solved as follows. The linear equation of motion is (positive is taken downwards)\begin{align} F & =m\ddot{y}\nonumber \\ -T+mg & =m\ddot{y}\tag{1} \end{align}
And the angular equation of motion is given by\begin{equation} Ta=I\ddot{\theta }\tag{2} \end{equation} Due to constraint \(f\left ( y,\theta \right ) =y-a\theta =0\), then \[ \frac{\ddot{y}}{a}=\ddot{\theta }\] Using the above in (2) gives\begin{align} Ta & =I\frac{\ddot{y}}{a}\nonumber \\ T & =I\frac{\ddot{y}}{a^{2}}\tag{3} \end{align}
Replacing \(T\) in (1) with the \(T\) found in (3) results in\begin{align*} m\ddot{y} & =-I\frac{\ddot{y}}{a^{2}}+mg\\ \ddot{y}\left ( m+\frac{I}{a^{2}}\right ) & =mg\\ \ddot{y} & =\frac{mg}{m+\frac{I}{a^{2}}} \end{align*}
But \(I=\frac{1}{2}ma^{2}\) then the above becomes\begin{align*} \ddot{y} & =\frac{mg}{m+\frac{\frac{1}{2}ma^{2}}{a^{2}}}\\ & =\frac{g}{1+\frac{1}{2}}\\ & =\frac{2}{3}g \end{align*}
Which is the same (as would be expected) using the energy method
