91Ó°ÊÓ

The given data can be listed below as,

• The length of the stiff rod in the pendulum is,$L$
• The mass hanging from the end of the rod is,$m$
• The arc length of the pendulum is,$s=L\mathrm{θ}$
• The initial speed for hitting the hanging mass is,$v_i$

The principle states energy can only be transformed from one form to another form but it cannot be destroyed or created.

The law of the energy principle gives the gravitational potential energy, tangential force's component and minimum initial speed.

(a)

At the angle $\mathrm{θ}$, according to the given figure, the vertical distance of the mass to the pivotal axle is $L\cos \mathrm{θ}$. But, at the distance $U=0$, the vertical distance is $L$. Hence, the vertical distance of the point from the lowest point that makes an angle $\mathrm{θ}$with the pendulum is $L-L\cos \mathrm{θ}$

According to the energy principle, the equation of the change in the potential energy is expressed as follows,

$U=mg\mathrm{Δ}h$

Here, $m=$The mass of the object, $\mathrm{Δ}h=$The vertical distance of the point from the lowest point, and $\mathrm{g}$is the acceleration due to gravity.

For $\mathrm{Δ}h=L-L\cos \mathrm{θ}$.

$$\begin{gathered} U=mg(L-L\cos \mathrm{θ}) \\ U=mgL(1-\cos \mathrm{θ}) \end{gathered}$$

Thus, the gravitational potential energy as a function of the angle $\mathrm{θ}$, measured from the vertical is $mgL(1-\cos \mathrm{θ})$.

(b)

The diagram of the potential energy as a function of the angle $\mathrm{θ}$, for angles from $-210°$to $-210°$is provided below,

(c)

The free body diagramof the tangential component of the force is shown below_,

Here, from the above diagram, the mass $m$is producing two components of force that are $mg\cos \mathrm{θ}$and $mg\sin \mathrm{θ}$respectively. However, as the force gathered by taking the negative gradient of the energy with respect to the arc length $s$, then the tangential component of the force is $-mg\sin \mathrm{θ}$as this force is acting tangentially with the mass according to the above diagram.

Thus, the tangential component of the force on the mass by taking the (negative) gradient of the energy with respect to $s$is $-mg\sin \mathrm{θ}$.

(d)

From the energy principle, the equation of the minimum initial speed needed for the pendulum is expressed as,

$\frac{1}{2}m{v}_i^2+mgl=\frac{1}{2}m{v}^{\text{'}2}$

Here, $m$is the mass of the body, $v_i$is the initial velocity of the mass (initially the body was at rest), $g$is the acceleration due to gravity, $l$is the length of the rod ($l=2L$that is the total length of the pendulum after rotating $180°$), and $v^{\text{'}}$the minimum initial speed required to go over the top.

For $v_i=0,l=2L$.

$$\begin{gathered} \frac{1}{2}m{\left(0\right)}^2+mg2L=\frac{1}{2}mv{\text{'}}^2 \\ v{\text{'}}^2=4gL \\ v\text{'}=\sqrt{4gL} \end{gathered}$$

Thus, the initial speed needed for the pendulum to go over the top is $\sqrt{4gL}$

The energy levels in the potential energy diagram have been provided below-

Here, $v_C$is the critical initial speed and $v_i$is the initial speed.