Circular Motion

For the pendulum bob in the turning car,

$\tan \theta =\frac{v^2}{rg}$

Given:

$v=54\times \frac{5}{18}=15m/s$

So,

$\tan \theta =\frac{{15}^2}{20\times 10}=1.125$

$\theta ={\tan }^{-1}(1.125)$

Banking angle formula:

$\tan \theta =\frac{v^2}{rg}$

From geometry (small angles:)

$\tan \theta \approx \sin \theta =\frac{h}{l}$

Equating the two expressions:

$\frac{h}{l}=\frac{v^2}{rg}\Rightarrow h=\frac{v^{2}l}{rg}$

Substituting the values:

$h=\frac{{\left(12\right)}^2\times 1.5}{400\times 10}$

$h=\frac{144\times 1.5}{4000}$

$h=\frac{216}{4000}=0.054\text{ m}$

Convert to cm:

$h=0.054\times 100=5.4\text{ cm}$

$$\omega =2\pi f=2\pi \left(\frac{1}{6}\right)=\frac{\pi }{3}\text{rad/s}$$ $$T-mg=\frac{m{v}^2}{r}=m{\omega }^{2}r$$ $$T=mg+m{\omega }^{2}r$$ $$T=m(g+{\omega }^{2}r)$$ $$T=0.9[9.8+{\left(\frac{\pi }{3}\right)}^2(1\left)\right]$$ $$T=0.9[9.8+\frac{{\pi }^2}{9}]$$ $$T=0.9[9.8+\frac{9.8}{9}]$$ $$T=0.9\times 9.8(1+\frac{1}{9})$$ $$T=0.9\times 9.8\times \left(\frac{10}{9}\right)$$ $$T=1\times 9.8=9.8\text{N}$$

In a vertical circular motion, energy conservation is used to relate the velocities at the topmost and bottommost points.

• Given velocity at the topmost point:

  $$v_{\text{top}} = n \sqrt{g l}, \quad n > 1$$

• Let velocity at the bottommost point be $v_{\text{bottom}}$

• Applying energy conservation between the top and bottom points:

  Total Energy at the Top = Total Energy at the Bottom

• Total energy at the topmost point:

  $$E_{\text{top}} = KE_{\text{top}} + PE_{\text{top}}$$ $$= \frac{1}{2} m v_{\text{top}}^2 + mg(2l)$$

  Since $v_{\text{top}} = n \sqrt{g l}$​we substitute:

  $$KE_{\text{top}} = \frac{1}{2} m (n^2 g l) = \frac{1}{2} m n^2 g l$$
• $$E_{\text{top}} = \frac{1}{2} m n^2 g l + 2 m g l$$

• Total energy at the bottommost point:

  $$E_{\text{bottom}} = KE_{\text{bottom}} + PE_{\text{bottom}}$$​

  Since at the bottommost point, PE = 0, we get:

  $$E_{\text{bottom}} = KE_{\text{bottom}} = \frac{1}{2} m v_{\text{bottom}}^2$$​

• By energy conservation:

  $$E_{\text{bottom}} = E_{\text{top}}$$
• $$\frac{1}{2} m v_{\text{bottom}}^2 = \frac{1}{2} m n^2 g l + 2 m g l$$

  Canceling $m$m and multiplying by 2:

  $$v_{\text{bottom}}^2 = n^2 g l + 4 g l$$
• $$v_{\text{bottom}}^2 = (n^2 + 4) g l$$
• $$v_{\text{bottom}} = \sqrt{(n^2 + 4) g l}$$

$$\frac{KE_{\text{bottom}}}{KE_{\text{top}}} = \frac{\frac{1}{2} m v_{\text{bottom}}^2}{\frac{1}{2} m v_{\text{top}}^2}$$​​

Canceling $\frac{1}{2} m$:

$$\frac{KE_{\text{bottom}}}{KE_{\text{top}}} = \frac{(n^2 + 4) g l}{n^2 g l}$$ $$= \frac{n^2 + 4}{n^2}$$​

Comparing with the options, the correct answer is:

A.$\boxed{\frac{n^2+4}{n^2}}$​​