91Ó°ÊÓ

1. Mass of the ball m = 0.092 kg
2. Length of the rod L = 0.62 m

If the system is isolated, then only conservative forces will act on it. For such a system, sum of kinetic and potential energy for any state of a system = Sum of kinetic and potential energy for any other state of the system.

Formula:

$$\begin{gathered} K_a+U_a=K_b+U_b \\ K=\frac{1}{2}m{v}^2 \\ U=mgh \end{gathered}$$

In the first situation, the rod is rotated such as it is straight up and then released. The situation is shown in Figure a.

Figure a

This is the isolated system with conservative forces. From Figure a, let A be the lowest point and B be the height point; then

$$\begin{gathered} K_a+U_a=K_b+U_b \\ {U}_a=0\mathrm{as}\mathrm{A}\mathrm{is}\mathrm{the}\mathrm{lowest}\mathrm{point},\mathrm{h}=0 \\ {\mathrm{K}}_{\mathrm{b}}=0\mathrm{as}\mathrm{B}\mathrm{is}\mathrm{the}\mathrm{lowest}\mathrm{point},{\mathrm{V}}_{\mathrm{b}}=0 \\ {\mathrm{U}}_{\mathrm{b}}=\mathrm{mgh} \end{gathered}$$

From the figure, for point B, h =2 L. Hence,

$\frac{1}{2}m{v}_a^2=mg2L$

Rearranging the equation,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{a}}=\sqrt{2\mathrm{g}2\mathrm{L}} \\ ⇒{\mathrm{v}}_{\mathrm{a}}=\sqrt{\left(2×9.8×2×0.62\right)} \\ ⇒{\mathrm{v}}_{\mathrm{a}}\sqrt{24.304} \\ ⇒{\mathrm{v}}_{\mathrm{a}}=4.9\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$ (i)

Applying Newton’s second law at point A, in Figure a,

$$\begin{gathered} ma=T-mg \\ ⇒m\frac{v_a^2}{r}=T-mg \end{gathered}$$

$⇒\mathrm{T}=\mathrm{m}\left[\frac{{\mathrm{v}}_{\mathrm{a}}^2}{\mathrm{r}}+\mathrm{g}\right]$

Using given values and equation (i),

$$\begin{gathered} T=0.092×\left[\frac{24.304}{0.62}+9.8\right] \\ ⇒T=4.56\mathrm{N} \end{gathered}$$

In the first situation, the rod is rotated such as it is horizontal and then released. The situation is shown in Figure b. We need to find the angle at which the tension in the rod is equal to the mass of the ball, i.e.,

T = mg

(ii)

Figure b

Now consider the coordinate system as shown in fig (b), and find the components of weight.

Applying Newton’s second law along x axis, $F_{net}=maanda=\frac{v^2}{r}$

$m\frac{v_a^2}{L}=T-mg\cos \mathrm{θ}$

Using equation (ii),

$m\frac{v_a^2}{L}=mg-mg\cos \mathrm{θ}$

Cancelling m and rearranging the equation,

$v_a^2=gL(1-\cos \mathrm{θ})$

(iii)

Applying law of conservation of energy at point A and B in fig. b,

role="math" localid="1661404579656" $$\begin{gathered} K_a+U_a=K_b+U_b\left(\mathrm{iv}\right) \\ {K}_b=0\mathrm{as}Bisthehighestpoint \\ {\mathrm{v}}_{\mathrm{b}}=0\left(\mathrm{v}\right) \\ {\mathrm{U}}_{\mathrm{b}}=\mathrm{mgh} \\ =\mathrm{mgL} \end{gathered}$$

(vi)

$⇒U_b=mgh$

To find h, in figure b,

$$\begin{gathered} \cos \mathrm{θ}=\frac{d}{L} \\ h=L-d \end{gathered}$$

Using value of d,

$h=L-L\cos \mathrm{θ}=L\left(1-\cos \mathrm{θ}\right)$

Using this in $U_b$,

$$\begin{gathered} U_b=mgh \\ ⇒U_b=mgL\left(1-\cos \mathrm{θ}\right) \end{gathered}$$

vii)

Using equation (v), (vi), (vii) in equation (iv),

$\frac{1}{2}m{v}_a^2+mgL\left(1-\cos \mathrm{θ}\right)=mgL$

Cancelling m and multiplying both the sides by 2 to,

$v_a^2+2gL\left(1-\cos \mathrm{θ}\right)=2gL$

(viii)

Using equation (iii),

$gL\left(1-\cos \mathrm{θ}\right)+2gL\left(1-\cos \mathrm{θ}\right)+2gL$

Cancelling g L from both sides and rearranging the equation,

$$\begin{gathered} \cos \mathrm{θ}=\frac{1}{3} \\ ⇒\mathrm{θ}={\cos }^{-1}\left(\frac{1}{3}\right) \\ ⇒\mathrm{θ}=71° \end{gathered}$$

(d)

As in equation 8, mass m gets cancelled from both sides, it becomes independent of mass m. Hence, if the mass of the ball increased, the value $\mathrm{θ}=71°$will remain constant.