Question 15 Marks
State the law of conservation of angular momentum and illustrate it with the example of planetary motion.
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Question 25 Marks
A particle of mass $m$ is released from point P at $x = x _0$ on the X -axis from origin O and falls vertically along the Y-axis, as shown in Fig.
i. Find the torque $t$ acting on the particle at a time $t$ when it is at point Q with respect to O .
ii. Find the angular momentum L of the particle about O at this time t .
iii. Show that $\tau=\frac{d L}{d t}$ in this example.
i. Find the torque $t$ acting on the particle at a time $t$ when it is at point Q with respect to O .
ii. Find the angular momentum L of the particle about O at this time t .
iii. Show that $\tau=\frac{d L}{d t}$ in this example.
Answer
View full question & answer→i. The force of gravity, $F = mg$ produces the torque $\tau$. Let $\vec{r}$ be the position vector of Q . Then the magnitude of the torque is given by
$
\begin{aligned}
& \tau=r F \sin \theta \\
& =r \times m g \times \frac{x_0}{r}=m g x_0 \quad\left[\because \sin \theta=\frac{x_0}{r}\right]
\end{aligned}
$
The direction of the torque is directed into the plane of paper and perpendicular to it, as shown by $\otimes$.
ii. The magnitude of the angular momentum is $L=r p \sin \theta=r m v \sin \theta$
But the velocity v at point Q is given by $v = u + at =0+ gt = gt$
$
\therefore \quad L=r m g t . \frac{x_0}{r}=m g x_0 t
$
The direction of angular momentum is the same as that of torque.
iii. Now $L = mgx _0 t$
Differentiating both sides with respect to $f$, we get
$
\frac{d L}{d t}=\frac{d}{d t}\left(m g x_0 t\right)=m g x_0=\tau
$
Hence the relation $\tau=\frac{d L}{d}$ holds in this example.
$
\begin{aligned}
& \tau=r F \sin \theta \\
& =r \times m g \times \frac{x_0}{r}=m g x_0 \quad\left[\because \sin \theta=\frac{x_0}{r}\right]
\end{aligned}
$
The direction of the torque is directed into the plane of paper and perpendicular to it, as shown by $\otimes$.
ii. The magnitude of the angular momentum is $L=r p \sin \theta=r m v \sin \theta$
But the velocity v at point Q is given by $v = u + at =0+ gt = gt$
$
\therefore \quad L=r m g t . \frac{x_0}{r}=m g x_0 t
$
The direction of angular momentum is the same as that of torque.
iii. Now $L = mgx _0 t$
Differentiating both sides with respect to $f$, we get
$
\frac{d L}{d t}=\frac{d}{d t}\left(m g x_0 t\right)=m g x_0=\tau
$
Hence the relation $\tau=\frac{d L}{d}$ holds in this example.
Question 35 Marks
On an open ground, a motorist follows a track that turns to his left by an angle of $60^{\circ}$ after every 500 m . Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
Answer
The path followed by the motorist is a regular hexagon with side 500 m , as shown in the given figure.
Let the motorist start from point P . The motorist takes the third turn at S .
Magnitude of displacement $= PS = PV + VS =500+500=1000 m(\because P V=Q R, V S=S R)$
Total path length, $d _1=P Q+Q R+R S=500+500+500=1500 m$
The motorist take the sixth turn at point P , which is the starting point
$\therefore$ Magnitude of displacement $=0$
Total path length, $d _2= PQ + QR + RS + ST + TU + UP$
$
d_2=500+500+500+500+500=3000 m
$
The motorist takes the eight turn at point R
$\therefore$ Magnitude of displacement $= PR$
$
\begin{aligned}
& P R=\sqrt{P Q^2+Q R^2+2(P Q) \cdot(Q R) \cos 60^{\circ}} \\
& P R=\sqrt{500^2+500^2+\left(2 \times 500 \times 500 \times \cos 60^{\circ}\right)} \\
& P R=\sqrt{250000+250000+\left(500000 \times \frac{1}{2}\right)} \\
& P R=866.03 m \\
& \beta=\tan ^{-1}\left(\frac{500 \sin 60^{\circ}}{500+500 \cos 60^{\circ}}\right)=30^{\circ}
\end{aligned}
$
Therefore, the magnitude of displacement is 866.03 m at an angle of $30^{\circ}$ with PR.
Total path length $=$ Circumference of the hexagon $+ PQ + QR$
Total path length $=6 \times 500+500+500=4000 m$
The magnitude of displacement and the total path length corresponding to the required turns is shown in the given table
View full question & answer→The path followed by the motorist is a regular hexagon with side 500 m , as shown in the given figure.
Let the motorist start from point P . The motorist takes the third turn at S .
Magnitude of displacement $= PS = PV + VS =500+500=1000 m(\because P V=Q R, V S=S R)$
Total path length, $d _1=P Q+Q R+R S=500+500+500=1500 m$
The motorist take the sixth turn at point P , which is the starting point
$\therefore$ Magnitude of displacement $=0$
Total path length, $d _2= PQ + QR + RS + ST + TU + UP$
$
d_2=500+500+500+500+500=3000 m
$
The motorist takes the eight turn at point R
$\therefore$ Magnitude of displacement $= PR$
$
\begin{aligned}
& P R=\sqrt{P Q^2+Q R^2+2(P Q) \cdot(Q R) \cos 60^{\circ}} \\
& P R=\sqrt{500^2+500^2+\left(2 \times 500 \times 500 \times \cos 60^{\circ}\right)} \\
& P R=\sqrt{250000+250000+\left(500000 \times \frac{1}{2}\right)} \\
& P R=866.03 m \\
& \beta=\tan ^{-1}\left(\frac{500 \sin 60^{\circ}}{500+500 \cos 60^{\circ}}\right)=30^{\circ}
\end{aligned}
$
Therefore, the magnitude of displacement is 866.03 m at an angle of $30^{\circ}$ with PR.
Total path length $=$ Circumference of the hexagon $+ PQ + QR$
Total path length $=6 \times 500+500+500=4000 m$
The magnitude of displacement and the total path length corresponding to the required turns is shown in the given table
| Turn | Magnitude of displacement (m) | Total path length (m) |
| Third | 1000 | 1500 |
| Sixth | 0 | 3000 |
| Eighth | $866.03 ; 30^{\circ}$ | 4000 |
Question 45 Marks
A marble rolls along a table at a constant speed of $1.00 m / s$ and then falls off the edge of the table to the floor 1.00 m below,
i. How long does the marble take to reach the floor?
ii. At what horizontal distance from the edge of the table does the marble land?
iii. What is its velocity as it strikes the floor?
i. How long does the marble take to reach the floor?
ii. At what horizontal distance from the edge of the table does the marble land?
iii. What is its velocity as it strikes the floor?
Answer
View full question & answer→As the marble was rolling on the table, therefore it has horizontal velocity and it will act as a projectile as soon as it leaves the edge of the table and fall freely under the effect of gravity.
Since, the marble is initially moving horizontally, $v _{ y 0}=0$ and $v _{ x 0}=1.00 m / s$. We must consider the origin to be at the edge of the table, so that $x _0= y _0=0$
i. $t=?$ and $y=-1.00 m$
$
\begin{aligned}
& \therefore y=\frac{-1}{2} gt^2 \\
& \Rightarrow t=\sqrt{\frac{-2 y}{g}}=\sqrt{\frac{(-2)(-1.00)}{9.8}}=0.452 s
\end{aligned}
$
ii. $x =$ ?, when $t =0.452 s$
$
\therefore x=v_{x 0} t=1.00 \times 0.452 s=0.452 m
$
iii. $v =$ ?, $\theta=$ ? at $t =0.452 s$
The $x$-component of velocity is constant throughout the motion,
$
v_{x}=v_{x 0}=1.00 m / s
$
The $y$-component of velocity is given by
$
\begin{aligned}
& v_{y}=v_{y 0}-gt=0-9.8 \times 0.452=-4.43 m / s \\
& \therefore v=\sqrt{v_x^2+v_y^2}=\sqrt{(1.00)^2+(-4.43)^2}=4.54 m / s \text {, the magnitude of the resultant velocity of the motion. } \\
& \theta=\tan ^{-1}\left|\frac{v_y}{v_x}\right|=\frac{4.43}{1.00}=77.3^{\circ}
\end{aligned}
$
As the marble hits the floor, its velocity is $4.54 m / s$ directed $77.3^{\circ}$ downward with respect to the horizontal.
Since, the marble is initially moving horizontally, $v _{ y 0}=0$ and $v _{ x 0}=1.00 m / s$. We must consider the origin to be at the edge of the table, so that $x _0= y _0=0$
i. $t=?$ and $y=-1.00 m$
$
\begin{aligned}
& \therefore y=\frac{-1}{2} gt^2 \\
& \Rightarrow t=\sqrt{\frac{-2 y}{g}}=\sqrt{\frac{(-2)(-1.00)}{9.8}}=0.452 s
\end{aligned}
$
ii. $x =$ ?, when $t =0.452 s$
$
\therefore x=v_{x 0} t=1.00 \times 0.452 s=0.452 m
$
iii. $v =$ ?, $\theta=$ ? at $t =0.452 s$
The $x$-component of velocity is constant throughout the motion,
$
v_{x}=v_{x 0}=1.00 m / s
$
The $y$-component of velocity is given by
$
\begin{aligned}
& v_{y}=v_{y 0}-gt=0-9.8 \times 0.452=-4.43 m / s \\
& \therefore v=\sqrt{v_x^2+v_y^2}=\sqrt{(1.00)^2+(-4.43)^2}=4.54 m / s \text {, the magnitude of the resultant velocity of the motion. } \\
& \theta=\tan ^{-1}\left|\frac{v_y}{v_x}\right|=\frac{4.43}{1.00}=77.3^{\circ}
\end{aligned}
$
As the marble hits the floor, its velocity is $4.54 m / s$ directed $77.3^{\circ}$ downward with respect to the horizontal.
Question 55 Marks
Find the total energy of the particle executing S.H.M. and show graphically the variation of P.E. and K.E. with time in S.H.M. What is the frequency of these energies with respect to the frequency of the particle executing S.H.M?
Answer
View full question & answer→The total energy of the system of a block and a spring is equal to the sum of the potential energy stored in the spring plus the kinetic energy of the block and is proportional to the square of the amplitude.
$
\begin{aligned}
& \frac{1}{2} m \omega^2\left(A^2-x^2\right)+\frac{1}{2} m \omega^2 x^2 \\
& E=\frac{1}{2} m \omega^2 A^2
\end{aligned}
$
Hence, the total energy of the particle in SHM is constant and it is independent of the instantaneous displacement. Relationship between potential energy, kinetic energy, and time in Simple Harmonic Motion at $t =0$, when $x = \pm A$.
The frequency of kinetic energy is twice the frequency of SHM.
$
\begin{aligned}
& \frac{1}{2} m \omega^2\left(A^2-x^2\right)+\frac{1}{2} m \omega^2 x^2 \\
& E=\frac{1}{2} m \omega^2 A^2
\end{aligned}
$
Hence, the total energy of the particle in SHM is constant and it is independent of the instantaneous displacement. Relationship between potential energy, kinetic energy, and time in Simple Harmonic Motion at $t =0$, when $x = \pm A$.
The frequency of kinetic energy is twice the frequency of SHM.
Question 65 Marks
Show that for a particle in linear S.H.M., the average kinetic energy over a period of oscillation is equal to the average potential energy over the same period. At what distance from the mean position is the kinetic energy in simple harmonic oscillator equal to potential energy?
Answer
View full question & answer→Let $x = A \sin \omega t$
Then, $v =\frac{d x}{d t}= A \omega \cos (\omega t )$
K. E. $=\frac{1}{2} mv ^2$
Average kinetic energy is
$
\frac{1}{T} \int_0^{T} \frac{1}{2} mv^2 dt=\frac{1}{4} mA^2 \omega^2
$
Average potential energy is
$
\frac{1}{T} \int_0^{T} \frac{1}{2} kx^2 dt=\frac{1}{4} mA^2 \omega^2
$
Let x be the distance where KE is equal to PE . Kinetic energy of particles executing SHM is $KE =\frac{1}{2} m \omega^2\left( a ^2- x ^2\right)$
Potential energy of particles is given by $PE =\frac{1}{2} m \omega^2 x ^2$ Let x be the distance where $KE = PE$
$
\begin{aligned}
& \frac{1}{2} m \omega^2\left(a^2-x^2\right)=\frac{1}{2} m \omega^2 x^2 \\
& \left(a^2-x^2\right)=x^2 \\
& x=\frac{a}{\sqrt{2}}
\end{aligned}
$
Kinetic energy equal to its potential energy at $x =\frac{a}{\sqrt{2}}$
Then, $v =\frac{d x}{d t}= A \omega \cos (\omega t )$
K. E. $=\frac{1}{2} mv ^2$
Average kinetic energy is
$
\frac{1}{T} \int_0^{T} \frac{1}{2} mv^2 dt=\frac{1}{4} mA^2 \omega^2
$
Average potential energy is
$
\frac{1}{T} \int_0^{T} \frac{1}{2} kx^2 dt=\frac{1}{4} mA^2 \omega^2
$
Let x be the distance where KE is equal to PE . Kinetic energy of particles executing SHM is $KE =\frac{1}{2} m \omega^2\left( a ^2- x ^2\right)$
Potential energy of particles is given by $PE =\frac{1}{2} m \omega^2 x ^2$ Let x be the distance where $KE = PE$
$
\begin{aligned}
& \frac{1}{2} m \omega^2\left(a^2-x^2\right)=\frac{1}{2} m \omega^2 x^2 \\
& \left(a^2-x^2\right)=x^2 \\
& x=\frac{a}{\sqrt{2}}
\end{aligned}
$
Kinetic energy equal to its potential energy at $x =\frac{a}{\sqrt{2}}$