5 Marks Questions

Question 15 Marks

Show that the angular momentum of a particle is the product of its linear momentum and moment arm. Also, show that the angular momentum is produced only by the angular component of linear momentum. What is the physical meaning of angular momentum?

Answer

Physical meaning of angular momentum. Consider a particle P of mass m whose position vector relative to the origin O is $\vec{r}$. Suppose the momentum vector $\vec{p}$ of the particle makes angle $\theta$ with the position vector $\vec{r}$ as shown in Fig.

Draw ON perpendicular to the line of action of linear momentum $\vec{p}$. From right angled $\Delta ONP$, we get
$
\frac{O N}{O P}=\frac{d}{r}=\sin \theta \text { or } d=r \sin \theta
$
This is the perpendicular distance of the line of action of linear momentum from the point of rotation O and is called moment arm of the momentum.
The magnitude of the angular momentum about the point $O$ is
$
L=r p \sin \theta=p(r \sin \theta)=p d
$
Angular momentum $=$ Linear momentum $\times$ moment arm
This is the physical meaning of angular momentum. According to it, angular momentum is the moment of linear momentum and is a measure of the turning motion of the object. In contrast to it, we know that
$
\text { Torque }=\text { Force } \times \text { moment arm }
$
Thus torque is the moment of force and is a measure of the turning effect of force.
Moreover, as shown in Fig. the momentum vector $\vec{p}$ can be resolved into two rectangular components:
i. Radial component, $p _{ r }$ along the direction of the position vector $\vec{r}$.
ii. Angular or tangential component, $p_\theta$ perpendicular to $\vec{r}$
or Angular momentum = Angular component of linear momentum
Hence only the angular component and not the radial component of the linear momentum contributes towards the angular momentum.

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Question 25 Marks

A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to $10 \pi rad s ^{-1}$. Which of the two will start to roll earlier? The co-efficient of kinetic friction is $\mu_k=$

Answer

Radii of the ring and the disc, $r=10 cm=0.1 m$
Initial angular speed, $\omega_z=10 \pi rad s ^{-1}$
Coefficient of kinetic friction, $\mu_k=0.2$
Initial velocity of both the objects, $u =0$
Motion of the two objects is caused by frictional force. As per Newton's second law of motion, we have frictional force, $f=m a$
$
\mu_k mg=ma
$
Where,
$a=$ Acceleration produced in the objects
$m =$ Mass
$
\therefore a=\mu_k g \ldots(i)
$
As per the first equation of motion, the final velocity of the objects can be obtained as:
$
\begin{aligned}
& v=u+a t \\
& =0+\mu_{k} gt \\
& =\mu_{k} gt \ldots(ii)
\end{aligned}
$
The torque applied by the frictional force will act in a perpendicularly outward direction and cause a reduction in the initial angular speed.
Torque, $T =- I \alpha$
$\alpha=$ Angular acceleration
$
\begin{aligned}
& u_{Z} mgr=-I \alpha \\
& \therefore a=\frac{-\mu_k m g r}{I}\ldots(iii)
\end{aligned}
$
Using the first equation of rotational motion to obtain the final angular speed:
$
\begin{aligned}
& \omega=\omega_e+a t \\
& =\omega_x+\frac{-\mu_k m g r}{I} t\ldots(iv)
\end{aligned}
$
Rolling starts when linear velocity, $v=r u$
$
\therefore v=r\left(\omega_0-\frac{\mu_k g m r t}{I}\right)\ldots(v)
$
Equating equations (ii) and (v), we get:
$
\begin{aligned}
& \mu_k g t=r\left(\omega_0-\frac{\mu_k g m r t}{I}\right) \\
& =r \omega_0-\frac{\mu_i g m r^2 t}{I} \ldots \ldots .(vi)
\end{aligned}
$
For the ring $I=m r^2$
$
\begin{aligned}
& \therefore \mu_k g t=r \omega_0-\frac{\mu_k g m r^2 t}{m r^2} \\
& =r \omega_0=u_k-\frac{u_k g m r^2 t}{m r^2}
\end{aligned}
$
$
\begin{aligned}
& 2 \mu_k g t=r \omega_0 \\
& \therefore t_r=\frac{r \omega_0}{2 \mu_k g} \\
& =\frac{0.1 \times 10 \times 3.14}{2 \times 0.2 \times 9.8}=0.80 s\ldots(vii)
\end{aligned}
$
For the ring $I=\frac{1}{2} m r^2$
$
\begin{aligned}
& \therefore \mu_k g t_d=r \omega_0-\frac{\mu_k g m r^2 t}{\frac{1}{2} m r^2} \\
& =r \omega_0-2 \mu_k g t \\
& 3 \mu_k g t_d=r \omega_0 \\
& \therefore t_d=\frac{r \omega_0}{3 \mu_k g} \\
& =\frac{0.1 \times 10 \times 3.14}{3 \times 0.2 \times 9.8}=0.53 s \ldots(viii)
\end{aligned}
$
Sincet $t_d>t_r$, the disc will start rolling before the ring.

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Question 35 Marks

i. Pick out only the vector quantities from the following: Temperature, pressure, impulse, time, power, charge.
ii. Show by drawing a neat diagram that the flight of a bird is an example of composition of vectors.
iii. A man is travelling at $10.8 km h ^{-1}$ in a topless car on a rainy day. He holds his umbrella at an angle $37^{\circ}$ to the vertical to protect himself from the rain which is falling vertically downwards. What is the velocity of the rain? $\left[\right.$ Given $\cos 37^{\circ}=\frac{4}{5}$ ]

Answer

i. Impulse
ii. Flight of a bird. When a bird flies, it pushes the air with forces $F_1$ and $F_2$ in the downward direction with its wings $W_1$ and $W _2$. The lines of action of these two forces meet at point O . In accordance with Newton's third law of motion, the air exerts equal and opposite reactions $R_1$ and $R_2$. According to the parallelogram law, the resultant $R$ of the reactions $R_1$ and $R_2$ acts on the bird in the upward direction and helps the bird to fly upward.

iii. $v _{ R }=10.8 km h ^{-1}=3 ms^{-1}$
Given: $\cos 37^{\circ}=\frac{4}{5} \therefore \tan 37^{\circ}=\frac{3}{4}$
But $\tan 37^{\circ}=\frac{v_R}{v_M}$ or $\frac{3}{4}=\frac{v_R}{3 m s^{-1}}$
or $v_R=\frac{9}{4}=2.25 ms^{-1}$

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Question 45 Marks

i. Analytically, find the resultant $\vec{R}$ of two vectors $\vec{A}$ and $\vec{B}$ inclined at an angle $\theta$.
ii. Find the angle between two vectors $\vec{P}$ and $\vec{Q}$ if resultant of the vectors is given by $R ^2= P ^2+ Q ^2$.

Answer

Let two vectors A and B are inclined at an angle $\theta$ as shown in figure, Extend the vector A and drop perpendicular from S on vector A,As diagonal represents the resultant vector so OS is the magnitude of resultant vector.

$\begin{aligned} & \text { i. In } \triangle PSN \\ & \frac{S N}{S P}=\sin \theta \\ & \text { so } SN = SP \sin \theta= B \sin \theta \\ & \frac{P N}{S P}=\cos \theta\end{aligned}$
$
\begin{aligned}
& PN=SP \cos \theta=B \cos \theta \\
& \text { In } \triangle OSN, OS^2=ON^2+NS^2 \\
& OS^2=(OP+PN)^2+SN^2 \\
& R^2=(A+B \cos \theta)^2+(B \sin \theta)^2 R^2=A^2+B^2+2 AB \cos \theta \\
& R=\sqrt{A^2+B^2+2 A B \cos \theta}
\end{aligned}
$
The direction of result tan vector can be found by $\tan \alpha=\frac{S N}{O N}=\frac{B \sin \theta}{A+B \cos \theta}$ we have nothing to do with the magnitude R.
ii. Let $\theta$ be the angle between the vectors $\vec{P}$ and $\vec{Q}$, Then Result $\tan t$ vector magnitude is given as
$
\begin{aligned}
& R=\sqrt{P^2+Q^2+2 P Q \cos \theta} \\
& R^2=P^2+Q^2+2 PQ \cos \theta \ldots \text { (i) }
\end{aligned}
$
but it is given,
$
R^2=P^2+Q^2 \ldots(ii)
$
from (i) and (ii)
$
2 PQ \cos \theta=0
$
therefore the angle between two vector $\theta=90^{\circ}$

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Question 55 Marks

Let us take the position of mass when the spring is unstretched as $x=0$ and the direction from left to right as the positive direction of x -axis. Given x as a function of time t for the oscillating mass, if at the moment we start the stopwatch $(t=0)$, the mass is
i. at the mean position,
ii. at the maximum stretched position and
iii. at the maximum compressed position.
In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

Answer

Assuming the standard equation
$
x(t)=A \sin (\omega t+\phi)
$
i. When $t =0, x =0$ [mean position]
$
\begin{aligned}
& \Rightarrow 0=A \sin (\omega \times 0+\phi) \\
& \text { Asin } \phi=0[\text { as } A \neq 0] \\
& \text { or } \sin \phi=0 \therefore \phi=0
\end{aligned}
$
$\therefore$ Required function is
$
x(t)=A \sin (\omega t+0) \text { or } x(t)=A \sin \omega t
$
where, $\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{1200}{3.0}}=20 rad / s$
$
\therefore x(t)=A \sin 20 t \text { or } x(t)=2 \sin 20 t
$
ii. When $t =0, x =+ A$ (maximum stretched position)
$
\begin{aligned}
& x(t)=A \sin (\omega t+\phi) \text { at } t=0 \text { and } x=+A \\
& +A=A \sin (\omega \times 0+\phi) \text { or } 1=\sin \phi \Rightarrow \phi=\frac{\pi}{2} \\
& \therefore x(t)=A \sin \left(\omega t+\frac{\pi}{2}\right) \\
& =A \cos \omega t=A \cos 20 t=2 \cos 20 t
\end{aligned}
$
iii. When the spring is at maximum compressed position.
$
\begin{aligned}
& \text { At } t=0, x(t)=-A \\
& \Rightarrow-A=A \sin (\omega \times 0+\phi) \text { or }-1=\sin \phi \text { or } \phi=\frac{3 \pi}{2}
\end{aligned}
$
So, the equations only differ in initial phase and in no other factors.

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Question 65 Marks

A cylindrical piece of cork of density of base area $A$ and height $h$ floats in a liquid of density $\rho_l$. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period $T=2 \pi \sqrt{\frac{h \rho}{\rho, g}}$ Where $\rho$ is the density of cork. (Ignore damping due to viscosity of the liquid).

Answer

This numerical can be solved using concept of Simple Harmonic Motion of floating object in which an object is dipped into the liquid and released by pushing it down,due to increased buoyant force it will move upward due to which excess force will push it downward.This repeated up and down movement of the object is governed by the laws of Simple Harmonic Motion assuming viscous forces are absent.
so area of the cork = A
Height of the cork $= h$
Density of the liquid $=\rho_l$
Density of the cork $=\omega$
In equilibrium:
Weight of the cork = Weight of the liquid displaced by the floating cork
Let the cork be depressed slightly by x . As a result, some extra water of a certain volume is displaced. Hence, an extra up-thrust acts upward and provides the restoring force to the cork.
Up-thrust $=$ Restoring force, $F =$ Weight of the extra water displaced
$
F=-(\text { Volume } \times \text { Density } \times g)
$
Volume $=$ Area $\times$ Distance through which the cork is depressed
Volume $= Ax$
$
\therefore F=-A \times \rho_l g
$
According to the force law:
$
\begin{aligned}
& F=kx \\
& k=\frac{F}{x}
\end{aligned}
$
Where, k is a constant
$
k=\frac{F}{x}=-A \rho_l g\ldots(i)
$
The time period of the oscillations of the cork:
$
T=2 \pi \sqrt{\frac{m}{k}}\ldots(ii)
$
Where,
$m =$ Mass of the cork
$=$ Volume of the cork $\times$ Density
$=$ Base area of the cork $\times$ Height of the cork $\times$ Density of the cork
$
=\text { Ah } \rho
$
Hence, the expression for the time period will be -
$
T=2 \pi \sqrt{\frac{A h \rho}{A \rho_l g}}=2 \pi \sqrt{\frac{h \rho}{\rho_l g}}
$
From the above expression it is proved that time period of the fork does not depend on the mass of the object rather depends on specific gravity of the cork and height of the cork and acceleration due to gravity.

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