Numerical Problems-2 - 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 41.
Three masses $3 \mathrm{~kg}, 4 \mathrm{~kg}$ and $5 \mathrm{~kg}$ are located at the comers of an equilateral triangle of side $1 \mathrm{~m}$.
Locate the center of mass of the system.
Answer:
$
(\mathrm{x}, \mathrm{y})=(0.54 \mathrm{~m}, 0.36 \mathrm{~m})
$

Question 42 .
Two particles mass $100 \mathrm{~g}$ and $300 \mathrm{~g}$ at a given time have velocities $10 \hat{j}-7 \hat{j}-3 \hat{j}$ and $7 \hat{i}-9 \hat{j}+6 \hat{k}$ $\mathrm{ms}^{-1}$ respectively. Determine velocity of center of mass.
Answer:
Velocity of center of mass $=\frac{31 \hat{i}-34 \hat{j}+15 \hat{k}}{2} \mathrm{~ms}^{-1}$
Question 43 .
From a uniform disc of radius $\mathrm{R}$, a circular disc of radius $\mathrm{R} / 2$ is cut out. The center of the hole is at $\mathrm{R} / 2$ from the center of original disc. Locate the center of gravity of the resultant flat body.
Answer:
Center of mass of resulting portion lies at $\mathrm{R} / 6$ from the center of the original disc in a direction opposite to the center of the cut out portion.
Question 44.
The angular speed of a motor wheel is increased from $1200 \mathrm{rpm}$ to $3120 \mathrm{rpm}$ in 16 seconds,
1. What is its angular acceleration (assume the acceleration to be uniform)
2. How many revolutions does the wheel make during this time ?
Answer:
$
\begin{aligned}
& \mathrm{a}=4 \pi \mathrm{rad} \mathrm{s}^{-2} \\
& \mathrm{n}=576
\end{aligned}
$
Question 45 .
A meter stick is balanced on a knife edge at its center. When two coins, each of mass $5 \mathrm{~g}$ are put one on top of the other at the $12.0 \mathrm{~cm}$ mark, the stick is found to be balanced at $45.0 \mathrm{~cm}$, what is the mass of the meter stick?
Answer:
$\mathrm{m}=66.0 \mathrm{gm}$
Question 46.
A solid sphere is rolling op a friction less plane surface about its axis of symmetry. Find ratio of its rotational energy to its total energy.

Answer:
$
\begin{aligned}
& \text { Rot. K.E. }=\frac{1}{2} \mathrm{I} \omega^2=\frac{1}{2} \times \frac{2}{5} m \mathrm{R}^2 \times \frac{v^2}{\mathrm{R}^2} \\
& =\frac{1}{5} m v^2 \\
& \left(\text { as } \omega=\frac{v}{\mathrm{R}}, \mathrm{I}=\frac{2}{5} m \mathrm{R}^2\right) \\
& \text { Total energy }=\text { Translational K.E. }+ \text { Rot. K.E. }=\frac{1}{2} m v^2+\frac{1}{5} m v^2=\frac{7}{10} m v^2 \\
& \frac{\text { Rot. K.E. }}{\text { Total Energy }}=\frac{\frac{1}{5} m v^2}{\frac{7}{10} m v^2}=\frac{2}{7} \\
&
\end{aligned}
$
Question 47.
Calculate the ratio of radii of gyration of a circular ring and a disc of the same radius with respect to the axis passing through their centers and perpendicular to their planes.
Answer:
$2: 1$
Question 48.
Two discs of moments of inertia $I_1$ and $I_2$ about their respective axes (normal to the disc and passing through the center), and rotating with angular speed col and $\omega_2$ are brought into contact face to face with their axes of rotation coincident,
1. What is the angular speed of the two - disc system?
2. Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy ? Take $\omega_1 \neq \omega_2$.
Answer:
1. Let co be the angular speed of the two-disc system. Then by conservation of angular momentum
$
\left(\mathrm{I}_1+\mathrm{I}_2\right) \omega=\mathrm{I}_1 \omega_1+\mathrm{I}_2 \omega_2 \text { or } \quad \omega=\frac{\mathrm{I}_1 \omega_1+\mathrm{I}_2 \omega_2}{\mathrm{I}_1+\mathrm{I}_2}
$
2. Initial K.E. of the two discs.
$
\begin{aligned}
\mathrm{K}_1 & =\frac{1}{2}\left(\mathrm{I}_1+\mathrm{I}_2\right) \omega^2=\frac{1}{2}\left(\mathrm{I}_1+\mathrm{I}_2\right)\left(\frac{\mathrm{I}_1 \omega_1+\mathrm{I}_2 \omega_2}{\mathrm{I}_1+\mathrm{I}_2}\right)^2 \\
\text { Loss in K.E. } & =\mathrm{K}_1-\mathrm{K}_2=\frac{1}{2}\left(\mathrm{I}_1 \omega_1^2+\mathrm{I}_2 \omega_2^2\right)-\frac{1}{2\left(\mathrm{I}_1+\mathrm{I}_2\right)}\left(\mathrm{I}_1 \omega_1+\mathrm{I}_2 \omega_2\right)^2
\end{aligned}
$

$
=\frac{\mathrm{I}_1 \mathrm{I}_2}{2\left(\mathrm{I}_1+\mathrm{I}_2\right)}\left(\omega_1-\omega_2\right)^2=\text { a positive quantity }\left[\because \omega_1 \neq \omega_2\right]
$
Hence there is a loss of rotational K.E. which appears as heat.
When the two discs are brought together, work is done against friction between the two discs.
Question 49.
In the HCL molecule, the separating between the nuclei of the two atoms is about $1.27 Å\left(1 Å=10^{-}\right.$ $10 \mathrm{~m}$ ). Find the approximate location of the $\mathrm{CM}$ of the molecule, given that the chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in all its nucleus.
Answer:
As shown in Fig. suppose the $\mathrm{H}$ nucleus is located at the origin. Then,
$
\mathrm{x}_1=0, \mathrm{x}_2=1.27 Å, \mathrm{m}_1=1, \mathrm{~m}_2=35.5
$
The position of the $\mathrm{CM}$ of $\mathrm{HCl}$ molecule is
$
\begin{aligned}
& x=\frac{m_1 x_1+m_2 x_2}{m_1+m_2} \\
& \frac{1 \times 0+35.5 \times 1.27}{1+35.5}=1.239 Å
\end{aligned}
$

Thus the $\mathrm{CM}$ of $\mathrm{HCl}$ is located on the line joining $\mathrm{H}$ and $\mathrm{Cl}$ nuclei at a distance of $1.235 Å$ from the $\mathrm{H}$ nucleus.
Question 50.
A child stands at the center of turn table with his two arms out stretched. The turn table is set rotating with an angular speed of $40 \mathrm{rpm}$. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to $2 / 3$ times the initial value?
1. Assume that the turn table rotates without friction
2. Show that the child's new kinetic energy of rotation is more than the initial kinetic energy of rotation.
How do you account for this increase in kinetic energy ?
Answer:
Here $\omega=40 \mathrm{rpm}, \mathrm{I}_2=\frac{1}{2} \mathrm{I}_1$
By the principle of conservation of angular momentum,
$\mathrm{I}_1 \omega_1=\mathrm{I}_{2 \omega 2}$ or $\mathrm{I}_1 \times 40=\frac{2}{5} \mathrm{I}_1 \omega_1$ or $\omega_2=100 \mathrm{rpm}$
(ii) Initial kinetic energy of rotation -
$
\frac{2}{5} \mathrm{I}_1 \omega_1^2=\frac{2}{5} \mathrm{I}_1(40)^2=800 \mathrm{I}_1
$

new kinetic energy of rotation -
$
\frac{2}{5} \mathrm{I}_2 \omega_2^2=\frac{1}{2} \times \frac{2}{3} \mathrm{I}_1(100)^2=2000 \mathrm{I}_1
$
$\frac{\text { New K.E. }}{\text { Initial K.E. }}=\frac{2000 \mathrm{I}_1}{800 \mathrm{I}_1}=2.5$
Thus the child's new kinetic energy of rotation is 2.5 times its initial kinetic energy of rotation. This increase in kinetic energy is due to the internal energy of the child which he uses in folding his hands back from the out stretched position.
Question 51.
To maintain a rotor at a uniform angular speed of $200 \mathrm{rad} \mathrm{s}^{-1}$ an engine needs to transmit a torque of $180 \mathrm{~N} \mathrm{~m}$. What is the power required by the engine? Assume that the engine is $100 \%$ efficient.
Here $\omega=200 \mathrm{rad} \mathrm{s}^{-1}, \tau=180 \mathrm{~N} \mathrm{~m}$ Power, $\mathrm{P}=\tau \omega=180 \times 200=36,000 \mathrm{~W}=36 \mathrm{~kW}$.
Question 52.
A car weighs $1800 \mathrm{~kg}$. The distance between its front and back axles is $1.8 \mathrm{~m}$. Its center of gravity is $1.05 \mathrm{~m}$ behind the front axle. Determine the force exerted by the level ground on each front and back wheel.
Answer:

For transnational equilibrium of car
$
\mathrm{N}_{\mathrm{F}}+\mathrm{N}_{\mathrm{B}}=\mathrm{W}=1800 \times 9.8=17640 \mathrm{~N}
$
For rotational equilibrium of car
$
\begin{aligned}
& 1.05 \mathrm{~N}_{\mathrm{F}}=0.75 \mathrm{~N}_{\mathrm{B}} \\
& 1.05 \mathrm{~N}_{\mathrm{F}}=0.75\left(17640-\mathrm{N}_{\mathrm{F}}\right) \\
& 1.8 \mathrm{~N}_{\mathrm{F}}=13230 \\
& \mathrm{~N}_{\mathrm{F}}=13230 / 1.8=7350 \mathrm{~N} \\
& \mathrm{~N}_{\mathrm{B}}=17640-7350=10290 \mathrm{~N}
\end{aligned}
$
Force on each front wheel $=\frac{7350}{2}=3675 \mathrm{~N}$
Force on each back wheel $=\frac{10290}{2}=5145 \mathrm{~N}$
Long Answer Questions (5 Marks)
Question 1.
Derive an expression for center of mass for distributed point masses.
Answer:
A point mass is a hypothetical point particle which has nonzero mass and no size or shape. To find
the center of mass for a collection of $n$ point masses, say, $\mathrm{m}_1, \mathrm{~m}_2, \mathrm{~m}_3 \ldots \ldots \mathrm{m}_{\mathrm{n}}$ we have to first choose an origin and an appropriate coordinate system as shown in Figure. Let, $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3 \ldots \ldots \ldots \mathrm{x}_{\mathrm{n}}$ be the $\mathrm{X}$ - coordinates of the positions of these point masses in the $\mathrm{X}$ direction from the origin. The equation for the $\mathrm{X}$ coordinate of the center of mass is,
$
x_{\mathrm{CM}}=\frac{\sum m_i x}{\sum m_i}
$
where, $\sum m i$ is the total mass $M$ of all the particles. ( $\sum \mathrm{mi}=\mathrm{M}$ ).Hence,
$
x_{\mathrm{CM}}=\frac{\sum m_i x_i}{\sum \mathrm{M}}
$

Similarly, we can also find $y$ and $z$ coordinates of the center of mass for these distributed point masses as indicated in figure.
$
\begin{aligned}
& y_{\mathrm{CM}}=\frac{\sum m_i y_i}{\sum \mathrm{M}} \\
& z_{\mathrm{CM}}=\frac{\sum m_i z_i}{\sum \mathrm{M}}
\end{aligned}
$
Hence, the position of center of mass of these point masses in a Cartesian coordinate system is $\left(\mathrm{x}_{\mathrm{CM}}, \mathrm{y}_{\mathrm{CM}} \mathrm{z}_{\mathrm{CM}}\right)$. in general, the position of center of mass can be written in a vector form as, $\vec{r}_{\mathrm{CM}}=\frac{\sum m_1 \vec{r}_i}{\mathrm{M}}$
where, is the position vector of the center of mass and $\vec{r}_i=\mathrm{x}_{\mathrm{i}} \hat{j}+\mathrm{y}_{\mathrm{i}} \hat{j}+\mathrm{z}_{\mathrm{i}} \hat{k}$ is the position vector of the distributed point mass; where, $\hat{i}, \hat{j}$, and $\hat{j}$ are the unit vectors along $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$-axis respectively.
Question 2.
Discuss the center of mass of two point masses with pictorial representation.
Answer:
With the equations for center of mass, let us find the center of mass of two point masses $m_1$ and
$\mathrm{m}_2$, which are at positions $\mathrm{x}_1$ and $\mathrm{x}_2$ respectively on the $\mathrm{X}$ - axis. For this case, we can express the position of center of mass in the following three ways based on the choice of the coordinate system.
(1) When the masses are on positive $\mathrm{X}$-axis:
The origin is taken arbitrarily so that the masses $\mathrm{m}_1$ and $\mathrm{m}_2$ are at positions $\mathrm{x}_1$ and $\mathrm{x}_2$ on the positive $\mathrm{X}$-axis as shown in figure (a). The center of mass will also be on the positive $\mathrm{X}$-axis at $\mathrm{x}_{\mathrm{CM}}$ as given by the equation,
$
x_{\mathrm{CM}}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}
$

(2) When the origin coincides with any one of the masses:
The calculation could be minimized if the origin of the coordinate system is made to coincide with any one of the masses as shown in figure (b). When the origin coincides with the point mass $m_1$, its position $x_1$ is zero, (i.e. $x_1=0$ ). Then,
$
x_{\mathrm{CM}}=\frac{m_1(0)+m_2 x_2}{m_1+m_2}
$
The equation further simplifies as,
$
\mathrm{x}_{\mathrm{CM}}=\frac{m_2 x_2}{m_1+m_2}
$
(3) When the origin coincides with the center of mass itself: If the origin of the coordinate system is made to coincide with the center of mass, then, $\mathrm{x}_{\mathrm{CM}}=\mathrm{O}$ and the mass $\mathrm{m} 1$ is found to be on the negative $\mathrm{X}$ - axis as shown in figure (c). Hence, its position $\mathrm{x} 1$ is negative, (i.e. $-\mathrm{x}_1$ ).
$
\begin{aligned}
& 0=\frac{m_1\left(-x_1\right)+m_2 x_2}{m_1+m_2} \\
& 0=m_1\left(-x_1\right)+m_2 x_2 \\
& m_1 x_1=m_2 x_2
\end{aligned}
$
The equation given above is known as principle of moments.

Question 3.
Derive an expression for kinetic energy in rotation and establish the relation between rotational kinetic energy and angular momentum.
Answer:
Let us consider a rigid body rotating with angular velocity $\omega$ about an axis as shown in figure. Every particle of the body will have the same angular velocity $\omega$ and different tangential velocities $v$ based on its positions from the axis of rotation. Let us choose a particle of mass $m_i$ situated at distance $r_i$ from the axis of rotation. It has a tangential velocity $v_i$ given by the relation, $v_i=r_i \omega$. The kinetic energy $\mathrm{KE}_{\mathrm{i}}$. of the particle is,
$
\mathrm{KE}_{\mathrm{i}}=\frac{1}{2} m_i v_i^2
$
Writing the expression with the angular velocity,
$
\mathrm{KE}=\frac{1}{2} \mathrm{~m}_{\mathrm{i}}\left(\mathrm{r}_{\mathrm{i}} \omega\right)^2=\frac{1}{2} m_i r_i^2 \omega^2
$
For the kinetic energy of the whole body, which is made up of large number of such particles, the equation is written with summation as,
$
\mathrm{KE}=\frac{1}{2}\left(\sum m_i r_i^2\right) \omega^2
$
where, the term $\sum m_i \mathrm{r}_{\mathrm{r}}^{\mathrm{r}}$ is the moment of inertia $\mathrm{I}$ of the whole body. $\sum \mathrm{m}_{\mathrm{i}} \mathrm{r}_{\mathrm{r}}^{\mathrm{r}}$
Hence, the expression for $\mathrm{KE}$ of the rigid body in rotational motion is -
$
\mathrm{KE}=\frac{1}{2} \mathrm{I} \omega^2
$
This is analogous to the expression for kinetic energy in transnational motion.
$
\mathrm{KE}=\frac{1}{2} \mathrm{Mv}^2
$

Relation between rotational kinetic energy and angular momentum
Let a rigid body of moment of inertia I rotate with angular velocity $\omega$.
The angular momentum of a rigid body is, $\mathrm{L}=\mathrm{I} \omega$
The rotational kinetic energy of the rigid body is, $\mathrm{KE}=\frac{1}{2} I \omega^2$
By multiplying the numerator and denominator of the above equation with I, we get a relation between $\mathrm{L}$ and $\mathrm{KE}$ as,
$
\begin{aligned}
& \mathrm{KE}=\frac{1}{2} \frac{\mathrm{I}^2 \omega^2}{\mathrm{I}}=\frac{1}{2} \frac{(\mathrm{I} \omega)^2}{\mathrm{I}} \\
& \mathrm{KE}=\frac{\mathrm{L}^2}{2 \mathrm{I}}
\end{aligned}
$
Question 4.
Discuss how the rolling is the combination of transnational and rotational and also be possibilities of velocity of different points in pure rolling.
Answer:
The rolling motion is the most commonly observed motion in daily life. The motion of wheel is an example of rolling motion. Round objects like ring, disc, sphere etc. are most suitable for rolling.
Let us study the rolling of a disc on a horizontal surface. Consider a point $\mathrm{P}$ on the edge of the disc. While rolling, the point undergoes transnational motion along with its center of mass and rotational motion with respect to its center of mass.
Combination of Translation and Rotation:
We will now see how these transnational and rotational motions arc related in rolling. If the radius of the rolling object is $\mathrm{R}$, in one full rotation, the center of mass is displaced by $2 \pi \mathrm{R}$ (its circumference). One would agree that not only the center of mass. but all the points Of 1 the disc are displaced by the same $2 \pi \mathrm{R}$ after one full rotation. The only difference is that the center of mass takes a straight path; but, all the other points undergo a path which has a combination of the transnational and rotational motion. Especially the point on the edge undergoes a path of a cyclonic as shown in the figure.

As the center of mass takes only a straight line path. its velocity $\mathrm{v}_{\mathrm{CM}}$ is only transnational velocity $\mathrm{v}_{\text {TRANS }}\left(\mathrm{v}_{\mathrm{CM}}=\mathrm{v}_{\text {TRANS }}\right)$. All the other points have two velocities. One is the transnational velocity $\mathrm{v}_{\text {TRANS }}$ (which is also the velocity of center of mass) and the other is the rotational velocity $v_{R O T}\left(v_{R O T}=r \omega\right)$. Here, $r_i ;$ the distance of the point from the center of mass and $o$ is the angular velocity. The rotational velocity $\mathrm{v}_{\mathrm{ROT}}$ is perpendicular to the instantaneous position vector from the center of mass as shown in figure (a). The resultant of these two velocities is $v$. This resultant velocity y is perpendicular to the position vector from the point of contact of the rolling object with the surface on which it is rolling as shown in figure (b).

We shall now give importance to the point of contact. In pure rolling, the point of the rolling object which comes in contact with the surface is at momentary rest. This is the case with every point that is on the edge of the rolling object. As the rolling proceeds, all'the points on the edge, one by one come in contact with the surface; remain at momentary rest at the time of contact and then take the path of the cycloid as already mentioned.
Hence, we can consider the pure rolling in two different ways.
(i) The combination of transnational motion and rotational motion about the center of mass.
(or)
(ii) The momentary rotational motion about the point of contact.
As the point of contact is at momentary rest in pure rolling, its resultant velocity $v$ is zero $(v=0)$. For example, in figure, at the point of contact, $\mathrm{v}_{\text {TRANS }}$ is forward (to right) and $\mathrm{v}_{\mathrm{ROT}}$ is backwards (to the left).

rolling we have,
$
\mathrm{v}_{\mathrm{CM}}=\mathrm{R} \omega
$
We should remember the special feature of the above equation. In rotational motion, as per the relation $\mathrm{v}=\mathrm{r} \omega$, the center point will not have any velocity as $\mathrm{r}$ is zero. But in rolling motion, it suggests that the center point has a velocity $\mathrm{v}_{\mathrm{CM}}$ given by above equation $\mathrm{v}_{\mathrm{CM}}-R \omega$. For the topmost point, the two velocities $v_{\text {TRANS }}$ and $v_{R O T}$ are equal in magnitude and in the same direction (to the right). Thus, the resultant velocity $\mathrm{v}$ is the sum of these two velocities, $\mathrm{v}=\mathrm{v}_{\text {TRANS }}$ $+\mathrm{v}_{\mathrm{ROT}}$ In other form, $\mathrm{v}=2 \mathrm{v}_{\mathrm{CM}}$ as shown in figure below.
Question 5.
Derive an expression for kinetic energy in pure rolling.
Answer:
As pure is the combination of transnational and rotational motion, we can write the total kinetic energy ( $\mathrm{KE}$ ) as the sum of kinetic energy due to transnational motion ( $\mathrm{KE}_{\mathrm{TRANS}}$ ) and kinetic energy due to rotational motion $\left(\mathrm{KE}_{\mathrm{ROT}}\right)$.
$
\mathrm{KE}=\mathrm{KE}_{\text {TRANS }}+\mathrm{KE}_{\mathrm{ROT}} \ldots \ldots \ldots \text { (i) }
$
If the mass of the rolling object is $\mathrm{M}$, the velocity of center of mass is $\mathrm{v}_{\mathrm{CM}}$, its moment of inertia about center of mass is $\mathrm{I}_{\mathrm{CM}}$ and angular velocity is $\omega$, then

With center of mass as reference:
The moment of inertia $\left(\mathrm{I}_{\mathrm{CM}}\right)$ of a rolling object about the center of mass is, $\mathrm{I}_{\mathrm{CM}}=\mathrm{MK}^2$ and $\mathrm{v}_{\mathrm{CM}}=$
$\mathrm{R} \omega$. Here, $K$ is radius of gyration.
$
\begin{aligned}
\mathrm{KE} & =\frac{1}{2} \mathrm{M}_{\mathrm{CM}}^2+\frac{1}{2}\left(\mathrm{MK}^2\right) \frac{v_{\mathrm{CM}}^2}{\mathrm{R}^2} \\
\mathrm{KE} & =\frac{1}{2} \mathrm{M} v_{\mathrm{CM}}^2+\frac{1}{2} \mathrm{M}_{\mathrm{CM}}^2\left(\frac{\mathrm{K}^2}{\mathrm{R}^2}\right) \\
\mathrm{KE} & =\frac{1}{2} \mathrm{M} v_{\mathrm{CM}}^2\left(1+\frac{\mathrm{K}^2}{\mathrm{R}^2}\right) \quad \ldots(i v)
\end{aligned}
$
With point of contact as reference:
We can also arrive at the same expression by taking the momentary rotation happening with respect to the point of contact (another approach to rolling). If we take the point of contact as $o$, then, $\mathrm{KE}=\frac{1}{2} \mathrm{I}_0 \omega^2$
Here, $\mathrm{I}_0$ is the moment of inertia of the object about the point of contact. By parallel axis theorem, $\mathrm{I}_0=\mathrm{I}_{\mathrm{CM}}+\mathrm{MK}^2$ Further we can write, $\mathrm{I}_0 \mathrm{MK}^2+\mathrm{MR}^2$. With $\mathrm{v}_{\mathrm{CM}}=\mathrm{R} \omega$ or $\omega=\frac{v_{\mathrm{CM}}}{\mathrm{R}}$
$
\begin{aligned}
& \mathrm{KE}=\frac{1}{2}\left(\mathrm{MK}^2+\mathrm{MR}^2\right) \frac{\mathrm{v}_{\mathrm{CM}}^2}{\mathrm{R}^2} \\
& \mathrm{KE}=\frac{1}{2} M v_{\mathrm{CM}}^2\left(1+\frac{\mathrm{K}^2}{\mathrm{R}^2}\right) \ldots
\end{aligned}
$

As the two equations (v) and (vi) are the same, it $j$ s once again confirmed that the pure tolling problems could be solved by considering the motion as any one of the following two cases.
(i) The combination of transnational motion and rotational motion about the center of mass. (or)
(ii) The momentary rotational motion about the point of contact.
Question 6.
1. Can a body in translator y motion have angular momentum? Explain.
2. Why is it more difficult to revolve a stone by tying it to a longer string than by tying it to a shorter string?
Answer:
(1) Yes, a body in translatory motion shall have angular momentum unless fixed point about which angular momentum is taken lies on the line of motion of body
$|\overrightarrow{\mathrm{L}}|=\mathrm{rp} \sin \theta$
$=0$ only when $\theta=\mathrm{O}^{\circ}$ or $180^{\circ}$
(2) MI of stone $\mathrm{I}=\mathrm{ml}^2$ ( 1 - length of string) 1 is large, a is very small
$\tau=\mathrm{I} \alpha$
$\alpha=\frac{\tau}{I}=\frac{\tau}{m l^2}$
if 1 is large a is very small.
$\therefore$ more difficult to revolve.