SATHEE: Centre of Mass 2 Question 1

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 $θ_{0}$ . 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 $l (l ≪ L)$ , is close to

(2019 Main, 12 April I)

(a) $Mgl (1 - θ_{0}^{2})$

(b) $M g l (1 + θ_{0}^{2})$

(d) $M g l 1 + \frac{θ_{0}^{2}}{2}$

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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.

$∴$ Using angular momentum conservation law, if $v_{0}$ and $v_{1}$ are the velocities before standing and after standing of the person, then

$\begin{aligned} M v_{0} L & = M v_{1} (L - l) \\ \Rightarrow v_{1} & = \frac{L}{L - l} v_{0} \end{aligned}$

Now, total work done by (person + gravitation) system will be equal to the change in kinetic energy of the person, i.e.

$\begin{array}{ll} W_{g} + W_{p} = K E_{1} - K E_{0} \\ \Rightarrow & - M g l + W_{p} = \frac{1}{2} M v_{1}^{2} - \frac{1}{2} M v_{0}^{2} \\ \Rightarrow & W_{p} = M g l + \frac{1}{2} M (v_{1}^{2} - v_{0}^{2}) \end{array}$

$\begin{aligned} = M g l + \frac{1}{2} M \frac{L}{L - l} v_{0}^{2} - v_{0}^{2} [from Eq. (i)] \\ = M g l + \frac{1}{2} M v_{0}^{2} \frac{L - l^{- 2}}{L} - 1 \\ = M g l + \frac{1}{2} M v_{0}^{2} 1 - {\frac{l}{L}}^{- 2} - 1 \\ = M g l + \frac{1}{2} M v_{0}^{2} 1 + \frac{2 l}{L} - 1 \end{aligned}$

[using $(1 + x)^{n} = 1 + n x$ as higher terms can be neglected, if $n < 1$ ]

$\Rightarrow W_{p} = M g l + \frac{1}{2} M v_{0}^{2} \times \frac{2 l}{L}$

$or W_{p} = M g l + M v_{0}^{2} \frac{l}{L}$

Here,

$v_{0} = W A = \sqrt{\frac{g}{L}} (θ_{0} L)$

$\Rightarrow v_{0} = θ_{0} \sqrt{g L}$

$∴$ Using this value of $v_{0}$ in Eq.(ii), we get

$\begin{aligned} W_{p} & = M g l + M θ_{0}^{2} g L \cdot \frac{l}{L} \\ \Rightarrow W_{p} & = M g l [1 + θ_{0}^{2}] \end{aligned}$

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 $v^{'}$ . Then,

By applying linear momentum conservation, we have

Initial momentum of the system $=$

Final momentum of the system

$\begin{aligned} m v + 0 & = (m + 4 m) v^{'} \\ \Rightarrow & v^{'} & = \frac{v}{5} \end{aligned}$

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

$\begin{aligned} \frac{1}{2} m v^{2} + 0 = \frac{1}{2} m v^{' 2} + m g h + \frac{1}{2} (4 m) v^{' 2} \\ m v^{2} = (m + 4 m) v^{' 2} + 2 m g h \\ \Rightarrow v^{2} = 5 v^{' 2} + 2 g h \\ \Rightarrow v^{2} = \frac{1}{5} v^{2} + 2 g h \\ \Rightarrow \frac{4}{5} v^{2} = 2 g h \Rightarrow h = \frac{2 v^{2}}{5 g} \end{aligned}$

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 $θ_{0}$ . 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 $l (l ≪ L)$ , is close to

Answer:

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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 θ0. 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≪L), is close to

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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1. A person of mass $M$ is sitting on a swing to length $L$ and swinging with an angular amplitude $θ_{0}$ . 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 $l (l ≪ L)$ , is close to