Centre of Mass 2 Question 1
1. A person of mass $M$ is sitting on a swing to length $L$ and swinging with an angular amplitude $\theta _0$. If the person stands up when the swing passes through its lowest point, the work done by him, assuming that his centre of mass moves by a distance $l(l \ll L)$, is close to
(2019 Main, 12 April I)
(a) $\operatorname{Mgl}\left(1-\theta _0^{2}\right)$
(b) $M g l\left(1+\theta _0^{2}\right)$
(c) $M g l$
(d) $M g l \quad 1+\frac{\theta _0^{2}}{2}$
Show Answer
Answer:
Correct Answer: 1. (b)
Solution:
- Initially, centre of mass is at distance $L$ from the top end of the swing. It shifts to $(L-l)$ distance when the person stands up on the swing.
$\therefore$ Using angular momentum conservation law, if $v _0$ and $v _1$ are the velocities before standing and after standing of the person, then
$$ \begin{aligned} \quad M v _0 L & =M v _1(L-l) \\ \Rightarrow \quad v _1 & =\frac{L}{L-l} v _0 \end{aligned} $$
Now, total work done by (person + gravitation) system will be equal to the change in kinetic energy of the person, i.e.
$$ \begin{array}{ll} & W _g+W _p=KE _1-KE _0 \\ \Rightarrow & -M g l+W _p=\frac{1}{2} M v _1^{2}-\frac{1}{2} M v _0^{2} \\ \Rightarrow & W _p=M g l+\frac{1}{2} M\left(v _1^{2}-v _0^{2}\right) \end{array} $$
$$ \begin{aligned} & =M g l+\frac{1}{2} M \frac{L}{L-l} v _0^{2}-v _0^{2} \quad \text { [from Eq. (i)] } \\ & =M g l+\frac{1}{2} M v _0^{2} \frac{L-l^{-2}}{L}-1 \\ & =M g l+\frac{1}{2} M v _0^{2} \quad 1-\frac{l}{L}^{-2}-1 \\ & =M g l+\frac{1}{2} M v _0^{2} \quad 1+\frac{2 l}{L}-1 \end{aligned} $$
[using $(1+x)^{n}=1+n x$ as higher terms can be neglected, if $n<1$ ]
$$ \Rightarrow \quad W _p=M g l+\frac{1}{2} M v _0^{2} \times \frac{2 l}{L} $$
$$ \text { or } \quad W _p=M g l+M v _0^{2} \frac{l}{L} $$
Here,
$$ v _0=W A=\sqrt{\frac{g}{L}}\left(\theta _0 L\right) $$
$\Rightarrow \quad v _0=\theta _0 \sqrt{g L}$
$\therefore$ Using this value of $v _0$ in Eq.(ii), we get
$$ \begin{aligned} W _p & =M g l+M \theta _0^{2} g L \cdot \frac{l}{L} \\ \Rightarrow \quad W _p & =M g l\left[1+\theta _0^{2}\right] \end{aligned} $$
Key Idea Since, the ground is frictionless, so whe the particle will collide and climb over the wedge, then the wedge will also move. Thus, by using conservation laws for momentum and energy, maximum height climbed by the particle can be calculated.
Initial condition can be shown in the figure below
As mass $m$ collides with wedge, let both wedge and mass move with speed $v^{\prime}$. Then,
By applying linear momentum conservation, we have
Initial momentum of the system $=$
Final momentum of the system
$$ \begin{array}{rlrl} & & m v+0 & =(m+4 m) v^{\prime} \\ \Rightarrow & v^{\prime} & =\frac{v}{5} \end{array} $$
Now, if $m$ rises upto height $h$ over wedge, then by applying conservation of mechanical energy, we have
Initial energy of the system =Final energy of the system
$$ \begin{aligned} & \frac{1}{2} m v^{2}+0=\frac{1}{2} m v^{\prime 2}+m g h+\frac{1}{2}(4 m) v^{\prime 2} \\ & m v^{2}=(m+4 m) v^{\prime 2}+2 m g h \\ & \Rightarrow \quad v^{2}=5 v^{\prime 2}+2 g h \\ & \Rightarrow \quad v^{2}=\frac{1}{5} v^{2}+2 g h \\ & \Rightarrow \quad \frac{4}{5} v^{2}=2 g h \Rightarrow h=\frac{2 v^{2}}{5 g} \end{aligned} $$
