Solved Problems

\[\tau =I\alpha =I\frac{d\omega }{dt}\ \Longrightarrow \frac{d\omega }{dt}=\frac{\tau }{I}\] 
\[\omega =\frac{1}{I}\underbrace{\int^t_0{\tau \left(t\right)dt}}_{area}\] 
Instead of evaluate the above integral, compute the area under the $\tau -t$ diagram
\[\omega =\frac{1}{I}\left(area\right)=\frac{1}{0.5\times 2\times {\left(0.06\right)}^2}\left(\frac{1}{2}\left(0.2\right)\left(8\right)-\frac{1}{2}\left(-0.1\right)\left(4\right)\right)=170\ \frac{\mathrm{rad}}{\mathrm{s}}\] 
 

(a) Take the pivot about the left end. Identify all of the forces that act on the beam. Since we want the beam to be in static equilibrium after the child reaches the desired point so apply this condition to the beam:
\[\Sigma \vec F_y=0\to \ mg+wg=F_1+F_2\] 
Where $F_1$ and $F_2$ are the normal forces acted by the posts on the beam.

All of the above forces create the following torque about left post: 
\[\Sigma \vec \tau=\vec r \times \vec F=0 \Rightarrow F_1D=Mg\,\frac{L}{2}+wg\left(L-x\right)\] 
Since $F_2\ge 0$ and $F_1D\le \left(Mg+wg\right)D$ then
\[Mg\,\frac{L}{2}+wg\left(L-x\right)\le \left(M+w\right)gD\Rightarrow \ -\frac{mg}{wg}\left(D-\frac{L}{2}\right)+\left(L-D\right)\le x\ \] 
\[\Longrightarrow x\ge 1\ \mathrm{m}\] 
(b) Since now the beam is attached to the left post, the force on the left end of the beam can point down. Hence, the child can easily be all the way to the right.
$F_1D\cong Mg\frac{L}{2}+wgL\Rightarrow \ F_1\approx Mg\ \left(\frac{1}{2}\frac{L}{D}\right)+wg\frac{L}{D}$
Let $\mathrm{\Delta }S$ be the compression of post 2. Hence $\mathrm{\Delta }S_{child\ on}\approx \left(\frac{F_1}{A}\right)\frac{S}{Y}$.
With the child off, $F^{'}_1\approx Mg\ \left(\frac{L}{2D}\right)$ and $\mathrm{\Delta }S_{child\ off}\approx \left(\frac{F^{'}_1}{A}\right)\frac{S}{Y}$
\[\therefore \mathrm{\Delta }S_{child\ on}-\mathrm{\Delta }S_{child\ off}\approx \frac{\left(F_1-F^{'}_1\right)}{A}\frac{S}{Y}=\left(\frac{wg\frac{L}{D}}{A}\right)\frac{S}{Y}=\left(\frac{327}{\frac{1}{2}}\right)\frac{10}{{10}^{10}}\approx 7\times {10}^{-7}\mathrm{m}\] 
 

The angular speed in the rotational motions with period $T$ and frequency $f$ is defined by 
\[\omega =\frac{2\pi }{T}=2\pi f\] 
Where $T$ is in seconds and $f$ in Hertz thus $\omega $ has unit $\mathrm{rad/s}$. So only convert $\mathrm{rev/s}$ to $\mathrm{rad/s}$.
\[1\ rev=2\pi \ \mathrm{rad}\] 
\[\omega =8.3\frac{\mathrm{rev}}{\mathrm{s}}\left(2\pi \frac{\mathrm{rad}}{\mathrm{1\ rev}}\right)=52\frac{\mathrm{rad}}{\mathrm{s}}\] 
 

The torque about an axis or point is defined as the cross product of moment arm distance-vector $\vec{r}$ and force $\vec{F}$ which produce rotation. 
\[\vec{\tau }=\vec{r}\times \vec{F}\] 
Note: the moment arm is the minimum distance between the pivot point and the line of action (the line along which the forces act)
First, find the magnitude of the torques due to the forces $F_1$ and $F_2$ about the pivot point $A$. 
\[\left|{\vec \tau }_1 \right|=r_1F_1\,{\sin \theta \ }=\left(\frac{L}{2}\right)F_1\,{\sin 90{}^\circ \ }=50\ \mathrm{\ N}\] 
Where $\theta $ is the angle between the lever arm and force vector $\vec F$.
\[\left|{\vec \tau }_2\right|=r_2F_2\,{\sin \theta \ }=LF_2\,{\sin 60{}^\circ \ }=2\times 100\times \frac{\sqrt{3}}{2}=100\sqrt{3}\] 
The torque is a vector quantity so use the right-hand rule to find its direction. Curl your fingers of the right hand from the direction of $\vec r$ into the direction of $\vec F$. Then your right thumb points in the direction of torque $\vec \tau$ (or the direction of curling the right fingers shows the direction of torque as clockwise or counterclockwise).  Let us consider the positive direction to be out of the page (counterclockwise) and vice versa. So \[{\vec \tau }_1=-50\ \mathrm{N} \ \ and \ \vec \tau_2=+100\sqrt{3}\ \mathrm{N}\] 
The total torque exerted on the pivot point $A$ is the sum of them
\[\mathrm{\Sigma }{\vec \tau }_A=+100\sqrt{3}-50=+123.2\ \mathrm{N}\] 
Thus the total torque is counter clockwise. 
 

(a) Note: parallel axis theorem tells us that the moment of inertia is minimal when the rotation axis passes through the center-of-mass(CM) and increases as the rotation axis are moved further from the (CM). i.e. $I=I_{cm}+mh^2$, where $h$ is the distance from the CM to the parallel axis of rotation.
So, $I=\frac{1}{2}mr^2+mL^2$
(b) The radial part of $mg$ will be balanced by the tension on the cord (so there is no radial acceleration).

The tangential part of it provides the torque on the pendulum. 
\begin{gather*} \vec{\tau }=I\alpha =I\frac{d^2\theta }{dt^2}\\\\  \Rightarrow \quad -mgL\,{\sin \theta}=I\frac{d^2\theta}{dt^2} \\\\ \therefore \frac{d^2\theta}{dt^2}+\frac{mgL}{I}{\sin \theta}=0\end{gather*} Thus, the small angle approximation gives us \[\frac{d^2\theta}{dt^2}+\frac{mgL}{I}\theta \approx 0\] 
Note: In above the minus sign is there because the torque opposes the angular displacement from equilibrium.

By considering $\frac {mgL}{I}$ as the squared angular momentum $\omega^2$, the above equation have similar form of an SHM equation, therefore
\begin{gather*}{\omega}^2=\frac{mgL}{I}=\frac{mgL}{mL^2+\frac{1}{2}mr^2} \\\\ f=\frac{\omega}{2\pi} \end{gather*}

Practice more problems on simple harmonic motions (SHM). 

(c) To find the extremes of a function, we must take the derivative of it.If $\omega $ is an extreme, so ${\omega }^2$ is also, therefore we can examine $d{\omega }^2/dL$ rather $d\omega /dL$
\begin{gather*} \frac{d{\omega }^2}{dL}=1-\frac{r^2}{L^2}\frac{1}{1+r^22L^2}=0\\\\ \Longrightarrow \quad L=R/\sqrt{2}\end{gather*}
It is less than $R$!

More related article:
Simple pendulum problems 

(a) Recall that the angular momentum of a rotational object is given by $\vec{L}=\vec{r}\times \vec{p}$. So the magnitude of the angular momentum about CM for this case is 
\[\left|\vec{L}\right|=Mv\left(\frac{d}{2}\right)+Mv\left(\frac{d}{2}\right)=dMv\] 
By the Right hand rule, the direction of it is out of page.
(b) The rotational energy of a rotating system defines as $K_r=\frac{1}{2}I{\omega }^2$, where $\omega $ is angular velocity $\vec{v}=\vec{\omega }\times \vec{r}$ and $I=\Sigma m_ir_i^2$ is the moment of inertia of the system.
In this problem, $\vec \omega \bot \vec r\ \Rightarrow \omega =\frac{v}{r}$
\[K_r=\frac{1}{2}\left(Mr^2\right){\left(\frac{v}{r}\right)}^2=\frac{1}{2}Mv^2\] 
\[K_{r,tot}=2K_r=2\left(\frac{1}{2}Mv^2\right)=Mv^2\] 
(c) Because there is no external torque acting on the system (in space effects due to friction and air resistance are negligible), so the total angular momentum is constant i.e. 
\[\vec {\tau }_{net\ ext}=\frac{d{\vec{L}}_{sys}}{dt}=0\Rightarrow \ L_i=L_f=dMv\] 
(d) $L_f=dMv=2\left(Mv_f\frac{d}{4}\right)\Rightarrow v_f=2v$
(e) $K^{'}_{r,tot}=2\frac{1}{2}\left(I_f{\omega }^2_f\right)=M{\left(\frac{d}{4}\right)}^2{\left(\frac{v_f}{\frac{d}{2}}\right)}^2=Mv^2_f=M{\left(2v\right)}^2=4Mv^2$
\[\Rightarrow \ K^{'}_{r,tot}=4K_{r,tot}\] 
(f) Use the work - energy theorem $\Delta K=W$
\[K_f-K_i=W\Rightarrow 4Mv^2-Mv^2=W\] 
\[\therefore W=3Mv^2\] 
 

(a) Draw a free body diagram as below and apply Newton's 2${}^{nd}$ law to it:
\[\Sigma \vec F_y=ma_{lin}=0\] 
\[T-mg=0\Rightarrow T=mg\] 
Where $a_{lin}$ is the acceleration of the center of mass of the cylinder that is stated does not move.
(b) The free body diagram shows that the tension of the string causes the rotation of the cylinder ($mg$ acts at the center of mass, so it does not provide a torque about the center of mass) so Newton's 2${}^{nd}$ law for rotating objects states: 
\[\Sigma \tau =I\alpha \Rightarrow \ TR=I\alpha \Rightarrow \alpha =\frac{TR}{I}=\frac{TR}{\frac{1}{2}mR^2}=\frac{2T}{mR}\] 
(c) The angular acceleration is related to the tangential acceleration of the cylinder via
\[a=\alpha R\Rightarrow a=\frac{2T}{m}\] 
 

By definition, the torque acts on an object is $\vec{\tau }=\vec{r}\times \vec{F}$. There are two solutions, using a determinant method or direct product method. 
Determinant method:
\begin{align*} \vec{\tau }&=\vec{r}\times \vec{F}\\\\&=\left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 0 \\ 3 & -2 & 0 \end{array} \right|\\\\&=\hat{i}\left(2\left(0\right)-\left(0\right)\left(-2\right)\right)-\hat{j}\left(1\left(0\right)-\left(0\right)\left(3\right)\right)\\\\&+\hat{k}\left(\left(1\right)\left(-2\right)-\left(2\right)\left(3\right)\right)\\\\&=-8\hat{k}\quad \mathrm{(N.m)}\end{align*}
Direct product:
\begin{align*} \vec{\tau }=\vec{r}\times \vec{F} &=\left(1\ \hat{i}+2\hat{j}\right)\times \left(3\hat{i}-2\hat{j}\right)\\\\ &=\left(1\right)\left(3\right)\left(\underbrace{\hat{i}\times \hat{i}}_{0}\right)+\left(1\right)\left(-2\right)\left(\underbrace{\hat{i}\times \hat{j}}_{\hat{k}}\right)\\\\&+\left(2\right)\left(3\right)\left(\underbrace{\hat{j}\times \hat{i}}_{-\hat{k}}\right)+\left(2\right)\left(-2\right)\left(\underbrace{\hat{j}\times \hat{j}}_{0}\right)\\\\&= -2\hat{k}+6\left(-\hat{k}\right)=-8\hat{k}\quad {\rm N.m}\end{align*}
Note:  an easy way to remember the cross product of unit vectors are shown in the figure below. 
By going around this figure in the clockwise direction, take the positive cross product and vice versa. That is 
\[\hat{i}\times \hat{j}=\hat{k}\ \ and\ \ \hat{j}\times \hat{i}=-\hat{k}\] 
\[\hat{k}\times \hat{i}=\hat{j}\ \ \ and\ \ \hat{i}\times \hat{k}=-\hat{j}\] 
And so on. The cross product of the same unit vectors with each other is always zero. 
\[\hat{i}\times \hat{i}=\hat{j}\times \hat{j}=\hat{k}\times \hat{k}=0\] 

(a) Use Newton's 2${}^{nd}$ law for linear and rotational motion as follows
\[\mathrm{\Sigma }\vec F_y=Ma\Rightarrow T\hat{j}+Mg\left(-\hat{j}\right)=Ma\left(-\hat{j}\right)\Rightarrow T=M(g-a)\] 
Where $a$ is the linear acceleration of the man. Recall that in the rotational motion linear acceleration (if the string around the pulley does not slip) is related to the angular acceleration by $a=\alpha r$. 
The tension of the rope acts a torque on the cylinder about the horizontal axis passing through the central axis of the cylinder. So $\tau =I\alpha $, where $I$ is the moment of inertia and $\alpha $ is the angular acceleration. By definition of the torque $\vec{\tau }=\vec{r}\times \vec{F}$ we obtain
\[\tau =I\alpha \to rT=I\alpha \Rightarrow \ \alpha =\frac{rT}{I_{cylinder}}\] 
Now substitute the tension on the rope into the above relation: 
\[\alpha =\frac{rM\left(g-a\right)}{I_{cyli}}=\frac{rM\left(g-\alpha r\right)}{I_{cyli}}\to \alpha \left(I_{cyli}+Mr^2\right)=rMg\] 
\[\Rightarrow \alpha =\frac{Mgr}{I_{cyli}+Mr^2}=\frac{Mgr}{\left(2M\right)r^2+Mr^2}=\frac{g}{3r}=6.5\ {\mathrm{s}}^{\mathrm{-}\mathrm{2}}\] 
Where we have use the $I_{cylinder}=mr^2$ about central axis. 
(b) Use the following kinematic equation to find his speed.
\[v^2-v^2_0=2a\mathrm{\Delta }y\] 
Since the man start at rest, $v_0=0$. If we choose the starting point as the base, the man has reached $-y$ below the base. Therefore, 
\[v=\sqrt{2\underbrace{a}_{\mathrm{is\ negative}}\mathrm{\Delta }y}=\sqrt{2\left(-\alpha r\right)\mathrm{\Delta }y}=\sqrt{2(-0.5\times 6.5)(-20)}=11.4\ \frac{\mathrm{m}}{\mathrm{s}}\] 
 

In this problem we have a rotational motion. So use the conservation of mechanical energy between initial and final points to find the velocity of the center of mass of the hoop at the bottom of the incline plane. 
\[E_{top}=E_{bottom}\] 
\[mgh=\underbrace{\frac{1}{2}mv^2_c}_{ \begin{array}{c} trans.\ \ kinetic \\ energy \end{array} }+\underbrace{\frac{1}{2}I{\omega }^2}_{ \begin{array}{c}rotational\ kinetic \\ energy \end{array} }\] 
We know that in the rotational motions, the angular and tangential velocities are related together via $v=r\omega $. 
By substituting the given values, we obtain
\[mgh=\frac{1}{2}mv^2_c+\frac{1}{2}\left(mr^2\right){\left(\frac{v_c}{r}\right)}^2\Rightarrow v_c=\sqrt{gh}=\sqrt{g(L\,{\sin 20{}^\circ \ })}\] 
From the geometry, we have substituted $L\,{\sin 20{}^\circ \ }$ for height of the incline plane $h$.
Now using the kinematic equation $v^2-v^2_0=2ax$, find the acceleration of the CM and then use the equation $y=\frac{1}{2}at^2+v_0t$ to determine the required time.
\[v^2-\underbrace{v^2_0}_{0}=2a\underbrace{x}_{L}\Rightarrow \ a=\frac{gL}{2L}{\sin 20{}^\circ \ }\] 
If we suppose the axis parallel to the incline plane as the $y$ axis then $y=L$
\[y=\frac{1}{2}at^2+\underbrace{v_0}_{0}t\to L=\frac{1}{2}at^2\] 
\begin{align*} \Rightarrow t & =\sqrt{\frac{2L}{a}}=\sqrt{\frac{2L}{\left(\frac{gL}{2L}\,{\sin 20{}^\circ \ }\right)}}=\sqrt{\frac{4L}{g{\sin 20{}^\circ \ }}}\\& =\sqrt{\frac{4\left(12\right)}{9.8\,{\sin 20{}^\circ \ }}}=3.78\ \mathrm{s}\end{align*}

(a) By using the conservation of mechanical energy, we have
\begin{gather*} E_i=E_f \\\\ \frac 12 Mv_0^2 + \frac 12 I\omega^2 =\frac 12 Mv_f^2 +\frac 12 I \omega^2 +Mgh \end{gather*} 
The required condition for rotation without slipping is $v=r\omega $, where $v$ is the tangential velocity of the ball and $r$ is its radius. So \begin{gather*} \frac 12 Mv_0^2+\frac 12 \left(\frac 25 Mr^2 \right) \left(\frac {v_0}r \right)^2 = \frac{1}{2}Mv^2_f+\frac 12 \left(\frac 25 Mr^2\right){\left(\frac{v_f}{r}\right)}^2+Mgh \\\\ \frac{7}{10}Mv^2_0=\frac{7}{10}Mv^2_f+Mgh \\\\ \Rightarrow v_f=\sqrt{v^2_0-\frac{10}{7}gh} \end{gather*} Substituting the numerical values into that, we get \[ v_f=\sqrt{{30}^2-\frac{10}{7}(9.8\times 20)}=25\,\rm m/s \] 
(b) After the ball leaves the cliff, it moves along a projectile motion path (with $\alpha =0^\circ $). In the free-fall stage, use the following projectile motion formula \[y=-\frac{1}{2}gt^2+\underbrace{v_0\,{\sin \alpha\ }}_{v_{0y}}t+y_0\] 
Here $v_0$ is the initial velocity at the moment of free fall but at the end of the cliff, we have $v_0=v_f$. If above the cliff is chosen as the base then the hitting coordinate is $y=-20\ \mathrm{m}$. 
\[-20=-\frac{1}{2}(9.8)t^2+0t+0 \Rightarrow t=2.02\,\rm s\] 
(c) In the projectile motion, for finding the traveled horizontal distance, we must use the following kinematic equation
\[x=\underbrace{v_0\,{\cos \alpha \ }}_{v_{0x}}t\Rightarrow x=24.9\times {\cos 0{}^\circ \ }\times 2.02=50\,\rm m\] 
(d) In the projectile motion, the projectile has no horizontal acceleration so the only acceleration is the downward free fall acceleration $-g$. Therefore, at any moment first, find the components of the velocity as follows 
\[v_x=v_{0x}=v_0\,{\cos \alpha \ }=25\, \rm{m/s}\] 
\[v_y=\underbrace{v_{0y}}_{v_0\,{\sin \alpha \ }}-gt=-9.8\times 2.02=-20\ \rm{m/s}\] 
Now use the following expression to determine the magnitude of the velocity 
\[v=\sqrt{v^2_x+v^2_y}=\sqrt{{25}^2+{\left(-20\right)}^2}=32\ \mathrm{m/s}\] 

(a) There are no external torques that act about the vertical axis passing through the hole since the surface is frictionless. So the total angular momentum of the system is conserved. By definition, the angular momentum is $\vec{L}=\vec{r}\times \vec{P}$. Therefore, we have
\[L_i=L_f\Rightarrow rmv=\left(\frac{r}{2}\right)mv^{'}\Rightarrow v^{'}=2v\] 
(b) The tension in the string provides the radial acceleration of the block. So the initial tension is 
\[T=\frac{mv^2}{r}\] 
After the string reaches the new radius $r/2$, the tension in it is 
\begin{gather*}T'=\frac{m{v'}^{2}}{r'}=\frac{m{\left(2v\right)}^2}{\frac{r}{2}}=8\frac{mv^2}{r}\\\\ \Rightarrow T'=8T\end{gather*}

First, draw a sketch and show the forces that act on the disk. Consider the direction of the motion of the disk to be the positive direction. Apply Newton's second law in component form for the $x$ axis.

\[\Sigma F_x=ma_{cm}\Rightarrow \ -f_k-mg\,{\sin \theta \ }=ma_{cm}\] 
Now apply Newton's second law for the rotational motion about a horizontal axis through the center of mass and perpendicular to ${\vec{v}}_{cm}$. 
\[\Sigma {\tau }_i=I_{cm}\alpha \] 
\[f_kR=I_{cm}\alpha \] 
Other forces ($N, mg\,{\sin \theta \ }$) do not exert torques on the ball since they act on the center of mass or axis of rotation. 
Relate $\alpha $ and $a_{cm}$ using the nonslip condition that is $a_{cm}=\alpha R$. Combining these equations, gives 
\[\left\{ \begin{array}{rcl}f_k & = &-mg\,{\sin \theta \ }-ma_{cm} \\ f_kR & = & I_{cm}\alpha \\ \alpha & = & \frac{a_{cm}}{R} \end{array}\right.\] 
\[\Rightarrow \ -\left(mg\,{\sin \theta \ }+ma_{cm}\right)R=I_{cm}\left(\frac{a_{cm}}{R}\right)\] 
Solving for $a_{cm}$, we obtain
\[\Rightarrow \ a_{cm}=-\frac{mg{\sin \theta \ }}{\frac{I_{cm}}{R}+mR}\] 
With $a_{cm}$ given by the above equation, use the following kinematic equation to find the time required for the disk to come to a stop ($v=0$).
\[v=v_0+at\Rightarrow 0=v_0+a_{cm}t\Rightarrow t=-\frac{v_0}{a_{cm}}\] 
\[\Rightarrow t=\frac{\left(\frac{I_{cm}}{R}+mR\right)v_0}{mg{\sin \theta \ }}\] 
The moment of inertia for disk is $I_{cm}=\frac{1}{2}mR^2$, by substituting it into above, we get
\[t=\frac{3v_0}{2\left(g{\sin \theta \ }\right)}=\frac{3\times 3}{2\times 9.8\times {\sin 20{}^\circ \ }}=1.34\ \mathrm{s}\] 

Apply Newton's second law to each of the masses and pulley in the figure. Since the pulley has the mass (moment of inertia), so the tension on the string is different on each side of the pulley. 
\[\Sigma \vec F_1=m_1a\to m_1g-T_1=m_1a\] 
\[\Sigma \vec F_2=m_2a\to T_2-m_2g\,{\sin \theta \ }=m_2a\] 
Now apply Newton's second law for rotational motion of the pulley as follows
\[\Sigma {\tau }_{ext}=I\alpha \to \ R\left(T_1-T_2\right)=I\left(\frac{a}{R}\right)\] 
Where the angular acceleration $\alpha $ is related the linear (center of mass) acceleration by $\alpha =a/R$. Substituting the tensions $T_1$ and $T_2$ into the above equation, we obtain
\[R\left(\left(m_1g-m_1a\right)-\left(m_2a+m_2g{\sin \theta \ }\right)\right)=I\left(\frac{a}{R}\right)\] 
\[g\left(m_1-m_2{\sin \theta \ }\right)-a\left(m_1+m_2\right)=\frac{Ia}{R^2}\] 
\[a=g\frac{m_1-m_2{\sin \theta \ }}{m_1+m_2+\frac{I}{R^2}}\] 

(a) The angular velocity $\omega $ of a rolling object is related to its tangential velocity by $v=R\omega $. Thus 
\[v_{cm}=R\omega \to \omega =\frac{v_{cm}}{R}=\frac{5}{0.5}=10\ \mathrm{rad/s}\] 
(b) A rolling object has two kinetic energy, translational and rotational. Therefore,
\[K_{tot}=K_{tran}+K_{rot}=\frac{1}{2}Mv^2_{cm}+\frac{1}{2}I_{cm}{\omega }^2=\frac{1}{2}Mv^2_{cm}+\frac{1}{2}\left(\frac{1}{2}MR^2\right){\omega }^2\] 
\[\Rightarrow K_{tot}=\frac{1}{2}\left(10\right){\left(5\right)}^2+\frac{1}{4}\left(10\right){\left(0.5\right)}^2{\left(10\right)}^2=188\ \mathrm{J}\] 
(c) Use the conservation of the mechanical energy as $E_i=E_f$. 
\[U_i+K_i=U_f+K_f\to \ Mgh_i+K_{i,tot}=MgH+K_{f,tot}\] 
Let the lowest point of the hill be the base of the gravitational potential so $U_i=0$. At the final point the rolling object comes to a stop, $K_{f,tot}=0$. Therefore 
\[0+188=\left(10\right)\left(9.8\right)H+0\to H=\frac{188}{98}=1.92\mathrm{\ m}\] 
(d) Use again the conservation of mechanical energy, but in this case, there is no rotational kinetic energy so 
\[E_i=E_f\to mgh_i+\frac{1}{2}mv^2_{cm}=mgH+K_f\] 
\[\Rightarrow 0+\frac{1}{2}mv^2_{cm}=mgH+0\Rightarrow H=\frac{v^2_{cm}}{2g}=\frac{5^2}{2\left(9.8\right)}=1.28\ \mathrm{m}\] 
 

(a) From the geometry first find the moment arm $r$, which is the distance from the pivot point to the point where the force $F$ acts.

\[{\cos 30{}^\circ \ }=\frac{\frac{L}{2}}{r}\Rightarrow r=\frac{L}{2{\cos 30{}^\circ \ }}\] 
Now use the definition of the torque $\vec{\tau }=\vec{r}\times \vec{F}=rF\,\sin\ \theta $ to find its magnitude. $\theta $ is the angle between the radial direction and the direction of the force.
\begin{align*}\left|{\vec {\tau }_1}\right|=rF_1\,{\sin \theta \ } &=\left(\frac{L}{2{\cos 30{}^\circ \ }}\right)\left(\underbrace{F_1{\cos 30{}^\circ \ }}_{F_{\bot }}{\sin 90{}^\circ \ }+\underbrace{F_1{\sin 30{}^\circ \ }}_{F_{\parallel }}{\sin 0{}^\circ \ }\right)\\&=\left(\frac{2}{2{\cos 30{}^\circ \ }}\right)\left(100\times {\cos 30{}^\circ \ }\times 1\right)\\& =100\ \mathrm{N.m}\end{align*} Now put your fingers of the right hand along with the moment arm then curl them in the direction of the applied force. The thumb points in the direction of torque (right-hand rule). In this case, the torque is directed in the plane or is clockwise.
 

(b) Similar as part (a) 
\begin{align*} \left|{\vec \tau }_2 \right|&=r_2F_2{\sin \theta \ }\\\\&=\left(\frac{L}{2{\cos 30{}^\circ \ }}\right)\left(\underbrace{F_2{\sin (30+60){}^\circ \ }}_{F_{\bot }}\right){\sin 90{}^\circ \ }\\\\&=\frac{100}{\frac{\sqrt{3}}{2}}\\\\&=\frac{200}{\sqrt{3}}\quad{\rm N.m}\end{align*} 

Using the right hand rule, we obtain its direction out of the plane or counter clockwise. Let us assigning a minus sign for clockwise and positive for counter clockwise that is 
\[{\vec{\tau }}_1=-100\ \mathrm{N.m\ \ and\ \ } \vec{\tau }_2=+\frac{200}{\sqrt 3}\ \mathrm{N.m}\] 
(c) Therefore, the total torque due to the forces $F_1$ and $F_2$ is 
\begin{align*} {\vec{\tau }}_{tot}&={\vec{\tau }}_1+{\vec{\tau }}_2\\\\&=-100+\frac{200}{\sqrt{3}}\\\\ &\cong +15\quad \mathrm{N.m\ counter\ clockwise}\end{align*}
(d) The total torque acting on a body equals the product of the moment of inertia of its and its angular acceleration. This is Newton's second law for rotation. 
\[{\tau }_{net}=I\alpha \] 
Where $\alpha $ is related to the angular coordinate by $\theta ={\theta }_0+\frac{1}{2}\alpha t^2+{\omega }_0t$, which is analogous to those for straight-line motion with constant linear acceleration. 
In the problem is said that the body rotates one complete cycle so $\theta =2\pi \ \mathrm{rad}$. Now find the angular acceleration of the triangle as 
\begin{align*} \theta &={\theta }_0+\frac{1}{2}\alpha t^2+{\omega }_0t\to 2\pi \\\\&=0+\frac{1}{2}\alpha {\left(4\right)}^2+0\left(4\right)\\\\ \Rightarrow \quad \alpha &=\frac{\pi }{4}\quad {\rm rad/s^2}\end{align*}
Where we have supposed that the initial angular coordinate and velocity be zero.
By substituting $\alpha $ into the equation ${\tau }_{net}$ , we get
\[{\tau }_{net}=I\alpha \Rightarrow I_{cm}=\frac{15}{\frac{\pi }{4}}=\frac{60}{\pi }\ \mathrm{kg.}{\mathrm{m}}^{\mathrm{2}}\] 

In the rotational motions, the kinetic energy has two contributions one is due to the translational motion and the other is due to the pure rotational motion. Thus 
\[K=\underbrace{\frac{1}{2}mV^2}_{trans}+\underbrace{\frac{1}{2}I{\omega }^2}_{rotation}\] 
Where $I$ is the moment of inertial of the object. 
In the special motion of rolling without slipping, the point of the wheel in contact with the ground is momentarily at rest and the wheel rotates about a rotation axis through the contact points. In this motion, there is a relationship that is unique that is $v=r\omega $ where $r$ is the radial distance from the rotation axis to any point of the wheel.

In the case of wheel or cylinder the geometric center of them is coincident with the center of mass. So the nonslip condition is satisfied as $v_{CM}=R\omega$, where $R$ is the radius of them. Therefore, 
\[K=\frac{1}{2}mV^2_{CM}+\frac{1}{2}I{\omega }^2=\frac{1}{2}mV^2_{CM}+\frac{1}{2}\left(\frac{1}{2}mR^2\right){\left(\frac{V_{CM}}{R}\right)}^2=\frac{3}{4}mV^2_{CM}\] 
In above, we have used the fact that the moment of inertia of the cylinder about the central axis is $I_{cyl}=\frac{1}{2}mR^2$.