Question

A machine in an amusement park, consists of a cage at the end of one arm, hinged at $O$. The cage revolves along a vertical circle $\text{(ABCDEFGH)}$ of radius $r$ about its hinge $O$, at constant linear speed $v=\sqrt{gr}$. The cage is so attached that the man of weight $W$ standing on a weighing machine, inside the cage, is always vertical. Then which of the following is/are correct.

• A. the weight reading at $\text{A}$ is greater than the weight reading at $\text{E}$ by $2W$.
• B. the weight reading at $\text{G}$ is $W$.
• C. the ratio of the weight reading at $\text{E}$ to that at $\text{A}$ is $0$.
• D. the ratio of the weight reading at $\text{A}$ to that at $\text{C}$ is $2$.

Answer 1

Answer: (A), (B), (C), (D)

the ratio of the weight reading at $\text{A}$ to that at $\text{C}$ is $2$.

Given: $v=\sqrt{gr}$ and weight $=W$

Let $N$ be the normal reaction which is the reading of the weighing machine.

At every point a centripetal force of magnitude $\frac{m{v}^2}{r}$ experienced by the man toward the centre which is caused by the net external force acting toward the centre.

Now, At point $\text{A}$,

$N_A-mg=\frac{m{v}^2}{r}=\frac{m(\sqrt{gr}{)}^2}{r}=mg$

$\Rightarrow N_A=2mg=2W............\left(1\right)$

Similarly, at point $\text{C}$,

$N_C-mg=0$

$\Rightarrow N_C=mg=W...........\left(2\right)$

Now, At point $\text{E}$,

$mg-N_E=\frac{m{v}^2}{r}=\frac{m(\sqrt{gr}{)}^2}{r}=mg$

$\Rightarrow N_E=0...........\left(3\right)$

At point $\text{G}$,

$N_G-mg=0$

$\Rightarrow N_G=mg=W.............\left(4\right)$

So, from Eqs. (1), (2), (3) and (4)

$\frac{N_E}{N_A}=\frac{0}{2W}=0$

And, $\frac{N_A}{N_C}=\frac{2W}{W}=2$

Hence, all options are correct.