Numerical Problems-1 - Chapter 5 Motion of System of Particles and Rigid Bodies 11th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Numerical Problems
Question 1.
A uniform disc of mass $100 \mathrm{~g}$ has a diameter of $10 \mathrm{~cm}$. Calculate the total energy of the disc when rolling along a horizontal table with a velocity of $20 \mathrm{~cm} \mathrm{~s}^{-2}$.
Answer:
Given,
Mass of the disc $=100 \mathrm{~g}=100 \times 10^{-3} \mathrm{~kg}=\frac{1}{10} \mathrm{~kg}$
Velocity of disc $=20 \mathrm{~cm} \mathrm{~s}^{-1}=20 \times 10^{-2} \mathrm{~ms}^{-1}=0.2 \mathrm{~ms}^{-1}$
$
r=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}, \quad \omega=\frac{v}{r}=\frac{20 \times 10^{-2}}{5 \times 10^{-2}}=4
$
Energy $=\frac{1}{2} m V^2+\frac{1}{2} \mathrm{I} \omega^2=\frac{1}{2}\left(m \mathrm{~V}^2+\mathrm{I} \omega^2\right)$, where $\mathrm{I}=\frac{1}{2} m r^2$
$
=\frac{1}{2}\left[\frac{1}{10} \times 0.2 \times 0.2+\frac{1}{2} \times \frac{1}{10} \times 25 \times \frac{1}{10^4} \times 16\right]
$
$
=\frac{1}{2}\left[\frac{4}{1000}+\frac{2}{1000}\right]=\frac{1}{2}\left[\frac{6}{1000}\right]
$
Energy $=3 \times 10^{-3} \mathrm{~J}$
Question 2.
A particle of mass 5 units is moving with a uniform speed of $\mathrm{v}=3 \sqrt{2}$ units in the XOY plane along the line $y=x+4$. Find the magnitude of angular momentum.
Answer:
Given,
Mass $=5$ units
Speed $=\mathrm{v}=3 \sqrt{2}$ units
$\mathrm{Y}=\mathrm{X}+4$
Angular momentum $=\mathrm{L}=\mathrm{m}(\bar{r} \times \bar{v})$
$=\mathrm{m}(\mathrm{x} \hat{i}+\mathrm{y} \hat{j}) \mathrm{x}(\mathrm{v} \hat{i}+\mathrm{v} \hat{j})=\mathrm{m}[\mathrm{xv} \hat{k}-\mathrm{vy} \hat{k}]=\mathrm{m}[\mathrm{xv} \hat{k}-\mathrm{v}(\mathrm{x}+4) \hat{k}]$
$\mathrm{L}=-\mathrm{mv} \hat{k}=-4 \times 5 \times 3 \sqrt{2} \hat{k}=-60 \sqrt{2} \hat{k}$
$\mathrm{L}=60 \sqrt{2}$ units.
Question 3.
A fly wheel rotates with a uniform angular acceleration. If its angular velocity increases from $20 \pi$ $\mathrm{rad} / \mathrm{s}$ to $40 \pi \mathrm{rad} / \mathrm{s}$ in 10 seconds, find the number of rotations in that period.
Answer:
Given,
Initial angular velocity $\omega_0=20 \pi \mathrm{rad} / \mathrm{s}$

Final angular velocity $\omega=40 \pi \mathrm{rad} / \mathrm{s}$
Time $\mathrm{t}=10 \mathrm{~s}$
Solution:
Angular acceleration $\alpha=\frac{\omega-\omega_0}{t}=\frac{40 \pi-20 \pi}{10}$
$\alpha=2 \pi \mathrm{rad} / \mathrm{s}^2$
According to equation of motion for rotational motion
$\theta=\omega_0 t+\frac{1}{2} \alpha t^2=20 \pi \times 10+\frac{1}{2} 2 \pi \times 100=300 \pi \mathrm{rad}$
The number of rotations $=\mathrm{n}=\frac{\theta}{2 \pi}$
$\mathrm{n}=\frac{300 \pi}{2 \pi}=150$ rotations.
Question 4.
A uniform rod of mass $\mathrm{m}$ and length / makes a constant angle 0 with an axis of rotation which passes through one end of the rod. Find the moment of inertia about this gravity is.
Answer:
Moment of inertia of the rod about the axis which is passing through its center of gravity is

$
=\mathrm{I}_o=\frac{\mathrm{M} l^2}{12}=\mathrm{M} l^2 \sin ^2 \theta / 12
$
Moment of inertia of a uniform rod of mass $m$ and length 1 about one axis which passes through one end of the rod
$
\begin{aligned}
& =\mathrm{I}_o+\mathrm{M} \frac{l^2 \sin ^2 \theta}{4} \\
& =\frac{\mathrm{M} l^2 \sin ^2 \theta}{12}+\frac{\mathrm{M} l^2 \sin ^2 \theta}{4} \\
& =\frac{1}{3} \mathrm{M} l^2 \sin ^2 \theta
\end{aligned}
$
Question 5.
Two particles $P$ and $Q$ of mass $1 \mathrm{~kg}$ and $3 \mathrm{~kg}$ respectively start moving towards each other from rest under mutual attraction. What is the velocity of their center of mass?
Answer:
Given,
Mass of particle $P=1 \mathrm{~kg}$ Mass of particle $Q=3 \mathrm{~kg}$
Solution:
Particles $\mathrm{P}$ and $\mathrm{Q}$ forms a system. Here no external force is acting on the system,

We know that $\mathrm{M}=\frac{d}{d t}\left(\mathrm{~V}_{\mathrm{CM}}\right)=\mathrm{f}$
It means that, C.M. of an isolated system remains at rest when no external force is acting and internal forces do not change its center of mass.
Question 6.
Find the moment of inertia of a hydrogen molecule about an axis passing through its center of mass and perpendicular to the inter-atomic axis.
Given: mass of hydrogen atom $1.7 \times 10^{27} \mathrm{~kg}$ and inter atomic distance is equal to $4 \times 10^{-10} \mathrm{~m}$.
Answer:
Given,
Inter-atomic distance : $4 \times 10^{-10} \mathrm{~m}$
Mass of $\mathrm{H}_2$ atom : $1.7 \times 10^{-27} \mathrm{~kg}$
Moment of inertia of $\mathrm{H}_2=$

$
\begin{aligned}
& =2 \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2=\frac{1}{2} \mathrm{MR}^2 \\
& =\frac{1}{2} \times 1.7 \times 10^{-27} \times 16 \times 10^{-20} \\
& =1.86 \times 10^{-46} \mathrm{~kg} \mathrm{~m}^2
\end{aligned}
$
Question 7.
On the edge of a wall, we build a brick tower that only holds because of the bricks' own weight. Our goal is to build a stable tower whose overhang $\mathrm{d}$ is greater than the length 1 of a single brick. What is the minimum number of bricks you need? (Hint: Find the center of mass for each brick and add.)
Answer:
Given:

Length of the brick $=1$
Length of the overhang $=\mathrm{d}$
The mono of bricks can be decided only by using the concept of position of center of gravity. The first brick is in contact with the ground and it will not fall over.
Let one end of brick 2 is coinciding with the center of brick 1 i.e. $\mathrm{x}=0$.
$\therefore$ The position of $n$ brick is
$
\mathrm{x}_{\mathrm{n}}=(\mathrm{n}-1) \frac{L}{4}
$
The center of gravity is in the midway between the center of brick 2 and the center of brick $n$. position of $\mathrm{G}=$
$
\begin{aligned}
& =\frac{x_2+x_n}{2} \\
& =\frac{\frac{4}{4}+(n-1) \frac{L}{4}}{2} \\
& =\frac{(1+n-1) \frac{L}{4}}{2}=n \frac{L}{8}
\end{aligned}
$
brick tower will fall when $\mathrm{G}>\frac{L}{4}$ it shows that $\mathrm{n}>4$.
Question 8.
The 747 boing plane is landing at a speed of $70 \mathrm{~m} \mathrm{~s}^{-1}$. Before touching the ground, the wheels are not rotating. How long a skid mark do the wing wheels leave (assume their mass is $100 \mathrm{~kg}$ which is distributed uniformly, radius is $0.7 \mathrm{~m}$, and the coefficient of friction with the ground is 0.5 )?
Answer:
The types of the plane will leave a skid mark if the speed of the types in contact with ground is lesser than the velocity of the plane. The condition for this is -
$\mathrm{v}>\omega$
(When the type attained an angular velocity of V/R)
The types will stop the skidding and starts the rolling.
The forces acting on the wheel after the plane touches down are,
$\mathrm{N}-\mathrm{P}$ Normal force W - weight
The wheel is not accelerating means
$\mathrm{N}=\omega$
The torque about the center of the wheel is
$\tau=\mathrm{RF}=\mu \omega \mathrm{R}$
The angular acceleration is

$
\alpha=\frac{\tau}{\mathrm{I}}=\frac{\mu \omega \mathrm{R}}{\frac{1}{2} m \mathrm{R}^2}=\frac{2 \mu \omega}{m \mathrm{R}}
$
According to equation of motion, time taken to stop the skidding by the wheel is,
$
\begin{gathered}
\omega=\omega_0+\alpha t=\alpha t=\frac{\mathrm{V}}{\mathrm{R}} \\
\therefore \quad t=\frac{\mathrm{V}}{\alpha \mathrm{R}}=\frac{\mathrm{V}}{\mathrm{R}}\left(\frac{m \mathrm{R}}{2 \mu \omega}\right)=\frac{m \mathrm{~V}}{2 \mu \omega} \\
t=\frac{m \mathrm{~V}}{2 \mu \omega}=\frac{100 \times 70}{2 \times 0.5 \times 232 \times 10^3}=0.03 \mathrm{~s}
\end{gathered}
$
The six mark will have a length of
$
1=\mathrm{vt}=70 \times 0.03=2.1 \mathrm{~m}
$
Note:
The 747 is resting on the runway, supported by 16 wheels under the wing, and 2 under the nose total length is $68.63 \mathrm{~m}$. The normal force experienced by plane through its 16 wheels is $\omega=232$ $\mathrm{KN}$.