Question 14 Marks
Express the kinetic energy of a rotating body in terms of its angular momentum.
AnswerThe kinetic energy of a body of moment of inertia I and rotating with a constant angular velocity $\omega$ is
$
\mathrm{E}=\frac{1}{2} I \omega^2
$
The angular momentum of the body, $L=\mid \omega$.
$
\therefore \mathrm{E}=\frac{1}{2}(I \omega) \omega=\frac{1}{2} L \omega
$
This is the required relation.
View full question & answer→Question 24 Marks
Define the angular momentum of a particle.
Question 34 Marks
A thin rod of uniform cross section is made up of two sections made of wood and steel. The wooden section has length $50 \mathrm{~cm}$ and mass $0.6 \mathrm{~kg}$. The steel section has length $30 \mathrm{~cm}$ and mass $3 \mathrm{~kg}$. Find the moment of inertia of the rod about a transverse axis passing through the junction of the two sections.
View full question & answer→Question 44 Marks
The radius of gyration of a disc about its transverse symmetry axis is $2 \mathrm{~cm}$. Determine its radius of gyration about a diameter.
AnswerData : $\mathrm{k} \mathrm{CM}=2 \mathrm{~cm}$
Let $\mathrm{M}$ and $\mathrm{R}$ be the mass and radius of the disc. Let $\mathrm{I}_{\mathrm{CM}}$ and $\mathrm{K}_{\mathrm{CM}}$ be the $\mathrm{MI}$ and radius of gyration of the disc about its transverse symmetry axis. Let I and $\mathrm{k}$ be the $\mathrm{MI}$ and radius of gyration of the disc about its diameter. Then
$
\begin{aligned}
& I_{\mathrm{CM}}=\frac{M R^2}{2}=M k_{\mathrm{CM}}^2 \text { and } I=\frac{M R^2}{4}=M k^2 \\
& \therefore k_{\mathrm{CM}}^2=\frac{R^2}{2} \text { and } k^2=\frac{R^2}{4} \\
& \therefore \frac{k^2}{k_{\mathrm{CM}}^2}=\frac{R^2}{4} \times \frac{2}{R^2}=\frac{1}{2} \\
& \therefore k^2=\frac{k_{\mathrm{CM}}^2}{2} \quad \therefore k=\frac{k_{\mathrm{CM}}}{\sqrt{2}} \\
& \therefore k=\frac{2}{\sqrt{2}}=\sqrt{2}=1.414 \mathrm{~cm} \\
&
\end{aligned}
$
View full question & answer→Question 54 Marks
The radius of gyration of a body about an axis at $6 \mathrm{~cm}$ from its centre of mass is $10 \mathrm{~cm}$. Find its radius of gyration about a parallel axis through its centre of mass.
AnswerLet $O$ be a point at $6 \mathrm{~cm}$ from the centre of mass of the body.
Let $\mathrm{I}=\mathrm{Ml}$ about an axis through $\mathrm{O}$,
$\mathrm{k}=$ radius of gyration about the axis through $\mathrm{O}$,
$\mathrm{I}_{\mathrm{CM}}=\mathrm{Ml}$ about a parallel axis through the centre of mass of the body,
$\mathrm{k}_{\mathrm{CM}}=$ radius of gyration about a parallel axis through the centre of mass,
$M=$ mass of the body,
$\mathrm{h}=$ distance between the two axes.
Data : $\mathrm{h}=6 \mathrm{~cm}, \mathrm{k}=10 \mathrm{~cm}$
By the theorem of parallel axis,
$
\mathrm{I}=\mathrm{I}_{\mathrm{CM}}+\mathrm{Mh}^2
$
Also, $\mathrm{I}=\mathrm{Mk}^2$ and $\mathrm{I}_{\mathrm{CM}}=M k_{\mathrm{CM}}^2$
$
\begin{aligned}
& \therefore M k^2=M k_{\mathrm{CM}}^2+M h^2 \\
& \therefore k^2=k_{\mathrm{CM}}^2+h^2 \\
& \therefore 100=k_{\mathrm{CM}}^2+36 \quad \therefore k_{\mathrm{CM}}=8 \mathrm{~cm}
\end{aligned}
$
The radius of gyration about a parallel axis through its centre of mass is $8 \mathrm{~cm}$.
View full question & answer→Question 64 Marks
A compound object is formed of a thin rod and a disc attached at the end of the rod. The rod is $0.5 \mathrm{~m}$ long and has mass $2 \mathrm{~kg}$. The disc has mass of $1 \mathrm{~kg}$ and its radius is $20 \mathrm{~cm}$. Find the moment of inertia of the compound object about an axis passing through the free end of the rod and perpendicular to its length.
AnswerData : $L=0.5 \mathrm{~m}, R=0.2 \mathrm{~m}, M_{\text {rod }}=2 \mathrm{~kg}, M_{\text {disk }}=1 \mathrm{~kg}$ About a transverse axis through $C M$,
$
I_{\mathrm{CM}, \text { rod }}=\frac{1}{2} M_{\text {rod }} L^2 \quad \text { and } \quad I_{\mathrm{CM}, \text { disk }}=\frac{1}{2} M_{\text {disk }} R^2
$
The MI of the compound object about the given axis,
$
\begin{aligned}
& I_{\text {total }}=I_{\text {rod }}+I_{\text {disk }} \\
& =\left[I_{\mathrm{CM}, \mathrm{rod}}+M_{\mathrm{rod}}\left(\frac{L}{2}\right)^2\right] \\
& +\left[I_{\mathrm{CM}, \text { disc }}+M_{\text {disk }}(L+R)^2\right] \\
& =\left(\frac{1}{12} M_{\text {rod }} L^2+\frac{1}{4} M_{\text {rod }} L^2\right) \\
& +\left[\frac{1}{2} M_{\text {disc }} R^2+M_{\text {disc }}(L+R)^2\right] \\
& =\frac{1}{3} M_{\text {rod }} L^2+M_{\text {disc }}\left[\frac{1}{2} R^2+M_{\text {disk }}(L+R)^2\right] \\
& =\frac{1}{3}(2)(0.5)^2+(1)\left[\frac{1}{2}(0.2)^2+(0.5+0.2)^2\right] \\
& =\frac{1}{6}+\frac{0.04}{2}+0.49=0.167+0.02+0.49 \\
& =0.677 \mathrm{~kg} \cdot \mathrm{m}^2 \\
&
\end{aligned}
$
View full question & answer→Question 74 Marks
A thin uniform rod $1 \mathrm{~m}$ long has mass $1 \mathrm{~kg}$. Find its moment of inertia and radius of gyration for rotation about a transverse axis through a point midway between its centre and one end.
AnswerData : $M=1 \mathrm{~kg}, \mathrm{~L}=1 \mathrm{~m}$
Let $\mathrm{I}_{\mathrm{CM}}$ and $\mathrm{I}$ be the moments of inertia of the rod about a transverse axis through its centre, and a parallel axis midway between its centre and one end.
Then, $I_{\mathrm{CM}}=\frac{M L^2}{12}$ and $h=\frac{L}{4}$
By the theorem of parallel axis,
$
\begin{aligned}
I & =I_{\mathrm{CM}}+M h^2 \\
& =\frac{M L^2}{12}+\frac{M L^2}{16}=\frac{4 M L^2+3 M L^2}{48} \\
& =\frac{7}{48} M L^2=\frac{7}{48}(1)(1)^2=0.1458 \mathrm{~kg} \cdot \mathrm{m}^2
\end{aligned}
$
If $k$ is the radius of gyration about the parallel axis,
$
\begin{gathered}
I=M k^2 \quad \therefore M k^2=\frac{7}{48} M L^2 \\
\therefore k=\sqrt{\frac{7}{48}} L=\sqrt{\frac{7}{48}} \times 1=0.3818 \mathrm{~m}
\end{gathered}
$
View full question & answer→Question 84 Marks
A solid sphere of radius $R$, rotating with an angular velocity $\omega$ about its diameter, suddenly stops rotating and $75 \%$ of its $\mathrm{KE}$ is converted into heat. If $\mathrm{c}$ is the specific heat capacity of the material in SI units, show that the temperature of $3 \mathrm{R}^2 \mathrm{CO}^2$ the sphere rises by $\frac{3 R^2 \omega^2}{20 c}$.
AnswerThe $\mathrm{Ml}$ of a solid sphere about its diameters, $\mathrm{I}=\frac{2}{5} \mathrm{MR}^2$ where $M$ is its mass.
The rotational KE of the sphere,
$
\begin{aligned}
E=\frac{1}{2} I \omega^2 & =\frac{1}{2}\left(\frac{2}{5} M R^2\right) \omega^2 \\
& =\frac{1}{5} M R^2 \omega^2
\end{aligned}
$
If $\Delta \theta$ is the rise in temperature,
$
\begin{aligned}
& \quad M c \Delta \theta=\frac{3}{4} E=\frac{3}{4}\left(\frac{1}{5} M R^2 \omega^2\right) \\
& \therefore \Delta \theta=\frac{3 R^2 \omega^2}{20 c}
\end{aligned}
$
View full question & answer→Question 94 Marks
The mass and the radius of the Moon are, respectively, about $\frac{1}{81}$ time and about $\frac{1}{3.7}$ time those of the Earth. Given that the rotational period of the Moon is 27.3 days, compare the rotational kinetic energy of the Earth with that of the Moon.
Answer$
\text { Data : } M_M=\frac{1}{81} M_E, R_M=\frac{1}{3.7} R_E, T_M=27.3 \text { days, } T_E=1 \text { day }
$
Let $I_E$ and $I_M$ be the moments of inertia of the Earth and the Moon about their respective axes of rotation, and $\omega_E$ and $\omega_M$ be their respective rotational angular speeds. Assuming the Earth and the Moon to be solid spheres of uniform densities,
$
I_{\mathrm{E}}=\frac{2}{5} M_{\mathrm{E}} R_{\mathrm{E}}^2 \text { and } I_{\mathrm{M}}=\frac{2}{5} M_{\mathrm{M}} R_{\mathrm{M}}^2
$
Kinetic energy of rotation, $E_{\mathrm{rot}}=\frac{1}{2} \mathrm{I} \omega^2$
$
=\frac{1}{2} I\left(\frac{2 \pi}{T}\right)^2=2 \pi^2 \frac{I}{T^2}
$
Therefore, the ratio of the rotational KE of the Earth to that of the Moon is MaharashtraBoardSolutions.Guru
$
\begin{aligned}
\frac{E_{\text {rot (Earth) }}}{E_{\text {rot (Moon) }}} & =\frac{I_{\mathrm{E}}}{I_{\mathrm{M}}} \cdot\left(\frac{T_{\mathrm{M}}}{T_{\mathrm{E}}}\right)^2=\frac{M_{\mathrm{E}}}{M_{\mathrm{M}}} \cdot\left(\frac{R_{\mathrm{E}}}{R_{\mathrm{M}}}\right)^2 \cdot\left(\frac{T_{\mathrm{M}}}{T_{\mathrm{E}}}\right)^2 \\
& =81 \times(3.7)^2 \times(27.3)^2=8.264 \times 10^5
\end{aligned}
$
View full question & answer→Question 104 Marks
Calculate the moment of inertia of a ring of mass $500 \mathrm{~g}$ and radius $0.5 \mathrm{~m}$ about an axis of rotation passing through
(i) its diameter
(ii) a tangent perpendicular to its plane.
AnswerData : $M=500 \mathrm{~g} 0.5 \mathrm{~kg}, \mathrm{R}=0.5 \mathrm{~m}$
(i) The moment of inertia of the ring about its
$
\begin{aligned}
& \text { diameter }=\frac{M R^2}{2}=\frac{0.5 \times(0.5)^2}{2} \\
& =0.0625 \mathrm{~kg} \mathrm{~m}{ }^2=6.25 \times 10^{-2} \mathrm{~kg} \cdot \mathrm{m}^2
\end{aligned}
$
(ii) The moment of inertia of the ring about a tangent perpendicular to its plane $=2 \mathrm{MR}^2=2 \times 0.5 \times(0.5)^2=0.25 \mathrm{~kg} \cdot \mathrm{m}^2$
View full question & answer→Question 114 Marks
State the MI of a thin rectangular plate-of mass M, length l and breadth b about its transverse axis passing through its centre. Hence find its MI about a parallel axis through the midpoint of edge of length b.
Question 124 Marks
State the MI of a thin rectangular plate-of mass M, length l and breadth b- about an axis passing through its centre and parallel to its length. Hence find its MI about a parallel axis along one edge.
Question 134 Marks
Find the ratio of the radius of gyration of a solid sphere about its diameter to the radius of gyration of a hollow sphere about its tangent, given that both the spheres have the same radius.
AnswerThe radius of gyration of a body about a given axis, $\mathrm{k}=\sqrt{I / M}$, where $\mathrm{M}$ and $\mathrm{I}$ are respectively the mass of the body and its moment of inertia (MI) about the axis.
For a solid sphere rotating about its diameter,
$
k_{\mathrm{SS}}=\sqrt{\frac{2}{5}} R
$
For a hollow sphere rotating about its diameter,
$
I_{\mathrm{HS}}=\frac{2}{3} M R^2
$
For a hollow sphere rotating about its tangent,
$
I_{\mathrm{HS}}=\frac{2}{3} M R^2+M R^2=\frac{5}{3} M R^2
$
so that, its radius of gyration for rotation about a tangent is
$
k_{\mathrm{HS}}^{\prime}=\sqrt{\frac{5}{3}} R
$
The required ratio, $\frac{k_{\mathrm{SS}}}{k_{\mathrm{HS}}^{\prime}}=\sqrt{\frac{2}{5}} \times \sqrt{\frac{3}{5}}=\frac{\sqrt{6}}{5}$
View full question & answer→Question 144 Marks
State the expression for the Ml of a thin spherical shell (i.e., a thin-walled hollow sphere) about its diameter. Hence obtain the expression for its $\mathrm{Ml}$ about a tangent.
AnswerConsider a uniform, thin-walled hollow sphere radius $R$ and mass M. An axis along its diameter is an axis of spherical symmetry through its centre of mass. The $\mathrm{Ml}$ of the thin spherical shell about its diameter is
$
\mathrm{I}_{\mathrm{CM}}=\frac{2}{3} M R^2
$
Let I be its $\mathrm{Ml}$ about a tangent parallel to the diameter. Here, $\mathrm{h}=\mathrm{R}=$ distance between the two axes. Then, according to the theorem of parallel axis,
$
\begin{aligned}
I & =I_{\mathrm{CM}}+M h^2 \\
& =\frac{2}{3} M R^2+M R^2=\frac{5}{3} M R^2
\end{aligned}
$
[Note : The corresponding radii of gyration are
$
k_{\mathrm{CM}}=\sqrt{\frac{I_{\mathrm{CM}}}{M}}=\sqrt{\frac{2}{3}} R \simeq 0.8165 R \text { and }
$
$
\left.k=\sqrt{\frac{1}{M}}=\sqrt{\frac{5}{3}} R \simeq 1.291 R\right]
$
View full question & answer→Question 154 Marks
The radius of gyration of a uniform solid sphere of radius $R$ is $\sqrt{\frac{2}{5}} R$ for rotation about its diameter. Show that its radius of gyration for rotation about a tangential axis of rotation is $\sqrt{\frac{7}{5}} R$
AnswerLet the mass of the uniform solid sphere of radius R be $\mathrm{M}$. Let $\mathrm{I}_{\mathrm{CM}}$ and $\mathrm{k}_{\mathrm{d}}$ be its $\mathrm{Ml}$ about any diameter and the corresponding radius of gyration, respectively. Then,
$
I_{\mathrm{CM}}=M k_{\mathrm{d}}^2=\frac{2}{5} M R^2 \quad\left(\because k_{\mathrm{d}}=\sqrt{\frac{2}{5}} R, \text { given }\right)
$
Let $\mathrm{I}$ and $\mathrm{k}_{\mathrm{t}}$ be its $\mathrm{Ml}$ about a parallel tangential axis and the corresponding radius of gyration, respectively. Here, $h=R=$ distance between the two axis.
$
\therefore \mathrm{I}=M k_{\mathrm{t}}^2
$
By the theorem of parallel axis,
$
\begin{aligned}
& I=I_{C M}+M h^2 \\
& \therefore M k_{\mathrm{t}}^2=\frac{2}{5} M R^2+M R^2 \quad \therefore k_{\mathrm{t}}^2=\frac{2}{5} R^2+R^2=\frac{7}{5} R^2 \\
& \therefore k_{\mathrm{t}}=\sqrt{\frac{7}{5}} R
\end{aligned}
$
View full question & answer→Question 164 Marks
Assuming the expression for the MI of a uniform solid sphere about its diameter, obtain the expression for its moment of inertia about a tangent.
Question 174 Marks
State an expression for the moment of inertia of a solid sphere about its diameter. Write the expression for the corresponding radius of gyration.
Question 184 Marks
Assuming the expression for the moment of inertia of a thin uniform disc about its diameter, show that the moment of inertia of the disc about a tangent in its plane is $\mathrm{MR}^2$. Write the expression for the corresponding radius of gyration.
View full question & answer→Question 194 Marks
Given the moment of inertia of a thin uniform disc about its diameter to be $\frac{1}{4} M R^2$, where $M$ and $R$ are respectively the mass and radius of the disc, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.
AnswerConsider a thin uniform disc of mass $\mathrm{M}$ and radius $\mathrm{R}$ in the $\mathrm{xy}$ plane. Let $\mathrm{I} \mathrm{x}$, ly and $\mathrm{Iz}$ be the moments of inertia of the disc about the $x, y$ and $z$ axes respectively.
Now, $I_x=I_y$
since each represents the moment of inertia (MI) of the disc about its diameter and, by symmetry, the $\mathrm{Ml}$ of the disc about any diameter is the same.
$\therefore \mathrm{I}_{\mathrm{x}}=\mathrm{I}_{\mathrm{y}}=\frac{1}{4} \mathrm{MR}^2$ (Given)
According to the theorem of perpendicular axes,
$\mathrm{I}_z=\mathrm{I}_{\mathrm{x}}+\mathrm{I}_{\mathrm{y}}=2\left(\frac{1}{4} M R^2\right)=\frac{1}{2} \mathrm{MR}^2$
Let I be the $\mathrm{Ml}$ of the disc about a tangent normal to the disc and passing through a point on its edge (i.e., a tangent perpendicular to its plane). According to the theorem of parallel axis,
$\mathrm{I}=\mathrm{I}_{\mathrm{CM}}+\mathrm{Mh}^2$
Here, $I_{C M}=I_z=\frac{1}{2} M R^2$ and $h=R$.
$\therefore \mathrm{I}=\frac{1}{2} \mathrm{MR}^2+\mathrm{MR}^2=\frac{3}{2} \mathrm{MR}^2$
which is the required expression.
View full question & answer→Question 204 Marks
State the expression for the Ml of a thin spherical shell (i.e., a thin-walled hollow sphere) about its diameter. Hence obtain the expression for its $\mathrm{Ml}$ about a tangent.
AnswerConsider a uniform, thin-walled hollow sphere radius $\mathrm{R}$ and mass $\mathrm{M}$. An axis along its diameter is an axis of spherical symmetry through its centre of mass. The Ml of the thin spherical shell about its diameter is
$
\mathrm{I}_{\mathrm{CM}}=\frac{2}{3} \mathrm{MR}^2
$
Let $\mathrm{I}$ be its $\mathrm{MI}$ about a tangent parallel to the diameter. Here, $\mathrm{h}=\mathrm{R}=$ distance between the two axes. Then, according to the theorem of parallel axis,
$
\begin{aligned}
& k_{\mathrm{CM}}=\sqrt{\frac{I_{\mathrm{CM}}}{M}}=\sqrt{\frac{2}{3}} R \simeq 0.8165 R \text { and } \\
& \left.k=\sqrt{\frac{I}{M}}=\sqrt{\frac{5}{3}} R \simeq 1.291 R\right]
\end{aligned}
$
View full question & answer→Question 214 Marks
About which axis of rotation is the radius of gyration of a body the least ?
AnswerThe radius of gyration of a body is the least about an axis through the centre of mass (CM) of the body.
From the parallel axis theorem, we know that a given body has the smallest possible moment of inertia about an axis through its $\mathrm{CM}$. The radius of gyration of a body about a given axis is directly proportional to the square root of its moment of inertia about that axis. Hence, the conclusion.
$
\left\{O R I=I_{C M}+M h^2 . \therefore M k^2=\backslash\left([/ \text { latexM k_\{ } \operatorname{mathrm}\{C M\}\}^{\wedge}\{2\}\right]+M^2\right. \text {. }
$
$\therefore \mathrm{k}^2=\left[\right.$ latex] $\mathrm{k}_{-}\{\mathrm{mathrm}\{\mathrm{CM}\}\}^{\wedge}\{2\} \mathrm{N}+\mathrm{h}^2$, which shows that $\mathrm{k}$ is minimum, equal to $\mathrm{k}_{\mathrm{CM}}$ when $\mathrm{h}=0$.
View full question & answer→Question 224 Marks
State an expression for the radius of gyration of
(i) a thin ring
(ii) a thin disc, about respective transverse symmetry axis.
OR
Show that for rotation about respective transverse symmetry axis, the radius of gyration of a thin disc is less than that of a thin ring.
Answer(i) The $\mathrm{Ml}$ of the ring about the transverse symmetry axis is
$\mathrm{I}_{\mathrm{CM}}=\mathrm{MR}^2 \ldots(1)$
Radius of gyration : The radius of gyration of the ring about the transverse symmetry axis is $\mathrm{K}=\sqrt{I_{\mathrm{CM}} / M}=\sqrt{R^2}=\mathrm{R}$
(ii) The Ml of the disc about the transverse symmetry axis is
$\mathrm{I}_{\mathrm{CM}}=\frac{1}{2} \mathrm{MR}<^2$..
Radius of gyration : The radius of gyration of the disc for the given rotation axis is
$
k=\sqrt{\frac{I}{M}}=\sqrt{\frac{R^2}{2}}=\frac{R}{\sqrt{2}}
$
Therefore, $k_{\text {disc }}<k_{\text {ring }}$.
View full question & answer→Question 234 Marks
Explain the linear velocity of a particle performing circular motion.
OR
Derive the relation between the linear velocity and the angular velocity of a particle performing circular motion.
Question 244 Marks
State an expression for the moment of inertia of a thin ring about its transverse symmetry axis.
Question 254 Marks
Four particles of masses 0.2 kg, 0.3 kg, 0.4 kg and 0.5 kg respectively are kept at comers A, B, C and D of a square ABCD of side 1 m. Find the moment of inertia of the system about an axis passing through point A and perpendicular to the plane of the square.
Question 264 Marks
Find the moment of inertia of a hydrogen molecule about its centre of mass if the mass of each hydrogen atom is $m$ and the distance between them is $R$.
AnswerWe assume the rotation axis to be a transverse axis through the centre of mass of the linear molecule $\mathrm{H}_2$. Then, each of the hydrogen atom is a distance $\frac{1}{2} \mathrm{R}$ from the $\mathrm{CM}$. Therefore, the $\mathrm{Ml}$ of the molecule about this axis,
$
I=2 m\left(\frac{R}{2}\right)^2=\frac{1}{2} m R^2
$
Notes :
1. For a $\mathrm{H}_2$ molecule, $\mathrm{mH}=1.674 \times 10^{-27} \mathrm{~kg}$ and bond length $=7.774 \times 10^{-11} \mathrm{~m}$, so that $\mathrm{I}=5.065 \times 10^{-48} \mathrm{~kg} \cdot \mathrm{m}^2$.
2. As atoms are treated as particles, we do not consider rotation about the line passing through the atoms.
View full question & answer→Question 274 Marks
Three point masses $M_1, M_2$ and $M_3$ are located at the vertices of an equilateral triangle of side $a$. What is the moment of inertia of the system about an axis along the altitude of the triangle passing through $M _1$ ?
View full question & answer→Question 284 Marks
A stone of mass $100 \mathrm{~g}$ attached to a string of length $50 \mathrm{~cm}$ is whirled in a vertical circle by giving it a velocity of $7 \mathrm{~m} / \mathrm{s}$ at the lowest point. Find the velocity at the highest point.
AnswerData : $m=0.1 \mathrm{~kg}, \mathrm{r}=\mathrm{l}=0.5 \mathrm{~m}, \mathrm{v}_2=7 \mathrm{~m} / \mathrm{s}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$
The total energy at the bottom, $E_{\text {bot }}$
$
=\mathrm{KE}+\mathrm{PE}=\frac{1}{2} m v_2^2+0=\frac{1}{2}(0.1)(7)^2=2.45 \mathrm{~J}
$
The total energy at the top, $\mathrm{E}_{\text {top }}=\mathrm{KE}+\mathrm{PE}=\frac{1}{2} m v_1^2+\mathrm{mg}(2 \mathrm{r})$
$
\begin{aligned}
& =\frac{1}{2}(0.1) v_1^2+(0.1)(10)(2 \times 0.5) \\
& =0.05 v_1^2+1
\end{aligned}
$
By the principle of conservation of energy,
$
\begin{aligned}
& \quad E_{\text {top }}=E_{\text {bot }} \\
& \therefore 0.05 v_1^2+1=2.45 \\
& \therefore v_1^2=\frac{2.45-1}{0.05}=\frac{145}{5}=29 \\
& \therefore v_1=\sqrt{29}=5.385 \mathrm{~m} / \mathrm{s}
\end{aligned}
$
View full question & answer→Question 294 Marks
A small body of mass $0.3 \mathrm{~kg}$ oscillates in a vertical plane with the help of a string $0.5 \mathrm{~m}$ long with a constant speed of $2 \mathrm{~m} / \mathrm{s}$. It makes an angle of $60^{\circ}$ with the vertical. Calculate the tension in the string.
Answer$
\begin{aligned}
\text { Data }: \mathrm{m} & =0.3 \mathrm{~kg}, \mathrm{r}=0.5 \mathrm{~m}, \mathrm{v}=2 \mathrm{~m} / \mathrm{s}, \theta=60^{\circ}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 \\
\frac{m v^2}{r} & =T-m g \cos \theta
\end{aligned}
$
Tension in the string, $T=\frac{m v^2}{r}+m g \cos \theta$
$
\begin{aligned}
& =\frac{(0.3)(2)^2}{0.5}+\begin{array}{c} \\
(0.3)(10) \cos 60^{\circ}
\end{array} \\
& =2.4+0.3 \times 10 \times \frac{1}{2}=2.4+0.3 \times 5=3.9 \mathrm{~N}
\end{aligned}
$
View full question & answer→Question 304 Marks
A bucket of water is tied to one end of a rope $8 \mathrm{~m}$ long and rotated about the other end in a vertical circle. Find the number of revolutions per minute such that water does not spill.
Answer[Important note: The circular motion of the bucket in a vertical plane under gravity is not a uniform circular motion. Assuming the critical case of the motion such that the bucket has the minimum speed at the highest point required for the water to stay put in the bucket, we can find the minimum frequency of revolution. ]
Data $: r=8 \mathrm{~m}, \mathrm{~g}=9.8 \mathrm{~m} / \mathrm{s}^2, \pi=3.142$
Assuming the bucket has a minimum speed $v=\sqrt{r g}$ at the highest point, the corresponding angular speed is
$
\omega=2 \pi f=\frac{v}{r}=\frac{\sqrt{r g}}{r}=\sqrt{\frac{g}{r}}
$
The minimum frequency of revolution,
$
\begin{aligned}
f & =\frac{1}{2 \pi} \sqrt{\frac{g}{r}} \\
& =\frac{1}{2 \times 3.142} \sqrt{\frac{9.8}{8}} \\
& =\frac{1}{6.284} \sqrt{1.225}=0.1761 \mathrm{rps} \\
& =0.1761 \times 60 \mathrm{rpm}=10.566 \mathrm{rpm}
\end{aligned}
$
View full question & answer→Question 314 Marks
A small body of mass $m=0.1 \mathrm{~kg}$ at the end of a cord $1 \mathrm{~m}$ long swings in a vertical circle. Its speed is $2 \mathrm{mls}$ when the cord makes an angle $\theta=30^{\circ}$ with the vertical. Find the tension in the cord.
AnswerData: $\mathrm{m}=0.1 \mathrm{~kg}, \mathrm{r}=1 \mathrm{~m}, \mathrm{y}=2 \mathrm{~m} / \mathrm{s}, \theta=30^{\circ}$,
$\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2$
When the cord makes an angle $\theta$ with the vertical, the centripetal force on the body is
$
\frac{m v^2}{r}=T-m g \cos \theta
$
The tension in the cord,
$
\begin{aligned}
T & =\frac{m v^2}{r}+m g \cos \theta \\
& =0.1\left(\frac{2^2}{1}+9.8 \times \cos 30^{\circ}\right) \\
& =0.1\left(4+9.8 \times \frac{\sqrt{3}}{2}\right)=0.1(4+4.9 \times 1.732) \\
& =0.1(4+8.486)=1.2486 \mathrm{~N}
\end{aligned}
$
View full question & answer→Question 324 Marks
A loop-the-loop cart runs down an incline into a vertical circular track of radius 3 m and then describes a complete circle. Find the minimum height above the top of the circular track from which the cart must be released.
Question 334 Marks
Explain angular velocity. State the right hand thumb rule for the direction of angular velocity.
Question 344 Marks
What is vertical circular motion? Comment on its two types.
AnswerA body revolving in a vertical circle in the gravitational field of the Earth is said to perform vertical circular motion.
A vertical circular motion controlled only by gravity is a nonuniform circular motion because the linear speed of the body does not remain constant although the motion can be periodic.
In a controlled vertical circular motion, such as that a body attached to a rod, the linear speed of the body can be constant (including zero) so that such a motion can be uniform and periodic.
View full question & answer→Question 354 Marks
A stone of mass $1 \mathrm{~kg}$, attached at the end of a $1 \mathrm{~m}$ long string, is whirled in a horizontal circle. If the string makes an angle of $30^{\circ}$ with the vertical, calculate the centripetal force acting on the stone.
Answer$
\begin{aligned}
& \text { Data : } \mathrm{m}=1 \mathrm{~kg}, \mathrm{~L}=1 \mathrm{~m}, \theta=30^{\circ}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 \\
& \therefore \theta=\sin ^{-1}\left(\frac{1}{6}\right)=\sin ^{-1} 0.1667=9^{\circ} 36^{\prime} \\
& \therefore \cos \theta=\cos 9^{\circ} 36^{\prime}=0.9860 \\
\end{aligned}
$
The tension in the string,
$
\begin{aligned}
& F=\frac{m g}{\cos \theta}=\frac{0.15 \times 9.8}{0.9860}=1.491 \mathrm{~N} \\
& v=\sqrt{r g \tan \theta}
\end{aligned}
$
The centripetal force $=\frac{m v^2}{r}=\frac{m(r g \tan \theta)}{r}$
$
\begin{aligned}
& =m g \tan \theta \quad \text { MaharashtraBoardSolutions.Guru } \\
& =(1)(10)\left(\tan 30^{\circ}\right) \\
& =10 \times \frac{1}{\sqrt{3}}=\frac{10}{1.732}=\mathbf{5 . 7 7 4} \mathrm{N}
\end{aligned}
$
View full question & answer→Question 364 Marks
A string of length $0.5 \mathrm{~m}$ carries a bob of mass $0.1 \mathrm{~kg}$ at its end. If this is to be used as a conical pendulum of period $0.4 \pi \mathrm{s}$, calculate the angle of inclination of the string with the vertical and the tension in the string.
AnswerData : $L=0.5 \mathrm{~m}, \mathrm{~m}=0.1 \mathrm{~kg}, \mathrm{~T}=0.4 \pi \mathrm{s}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$
(i) Period, $T=2 \pi \sqrt{\frac{L \cos \theta}{g}}$
$
\begin{aligned}
\therefore \cos \theta & =\frac{g T^2}{4 \pi^2 L} \\
& =\frac{10(0.4 \pi)^2}{4 \pi^2 \times 0.5} \\
& =\frac{10 \times 0.16 \pi^2}{2 \pi^2}=10 \times 0.08=0.8
\end{aligned}
$
The inclination of the string with the vertical,
$
\theta=36^{\circ} 5^{\prime}
$
(ii) The tension in the string,
$
F=\frac{m g}{\cos \theta}=\frac{0.1 \times 10}{0.8}=1.25 \mathrm{~N}
$
View full question & answer→Question 374 Marks
A stone of mass $2 \mathrm{~kg}$ is whirled in a horizontal circle attached at the end of a $1.5 \mathrm{~m}$ long string. If the string makes an angle of $30^{\circ}$ with the vertical, compute its period.
AnswerData : $L=1.5 \mathrm{~m}, \theta=30^{\circ}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$
The period of the conical pendulum,
$
\begin{aligned}
T & =2 \pi \sqrt{\frac{L \cos \theta}{g}}=2 \times 3.142 \times \sqrt{\frac{1.5 \cos 30^{\circ}}{10}} \\
& =6.284 \times \sqrt{\frac{1.5 \times 0.866}{10}}=6.284 \sqrt{\frac{1.299}{10}} \\
& =2.265 \mathrm{~s} \quad
\end{aligned}
$
View full question & answer→Question 384 Marks
Explain angular displacement in circular motion.
Question 394 Marks
Derive an expression for the angular speed of the bob of a conical pendulum.
OR
Derive an expression for the frequency of revolution of the bob of a corical pendulum.
Question 404 Marks
A motor van weighing $4400 \mathrm{~kg}$ (i.e., a motor van of mass $4400 \mathrm{~kg}$ ) rounds a level curve of radius $200 \mathrm{~m}$ on an unbanked road at $60 \mathrm{~km} / \mathrm{h}$. What should be the minimum value of the coefficient of friction to prevent skidding? At what angle should the road be banked for this velocity?
AnswerData : $\mathrm{m}=4400 \mathrm{~kg}, \mathrm{r}=200 \mathrm{~m}$,
$v=60 \mathrm{~km} / \mathrm{h}=60 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}=\frac{50}{3} \mathrm{~m} / \mathrm{s}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$
(i) $\frac{m v^2}{r}=\mu_s m g$
The minimum value of the coefficient of friction is
$
\mu_s=\frac{v^2}{r g}=\frac{(50 / 3)^2}{200 \times 10}=\frac{25}{18 \times 10}=\mathbf{0 . 1 3 8 9}
$
(ii) $\tan \theta=\frac{v^2}{r g}=0.1389$
The angle of banking, $\theta=\tan ^{-1} 0.1389$
$
=7^{\circ} 5^{\prime}
$
[Note: In part (ii), v is to be taken as the optimum speed.]
View full question & answer→Question 414 Marks
A coin kept at a distance of $5 \mathrm{~cm}$ from the centre of a turntable of radius $1.5 \mathrm{~m}$ just begins to slip when the turntable rotates at a speed of $90 \mathrm{rpm}$. Calculate the coefficient of static friction between the coin and the turntable. $\left[\mathrm{g}=\pi^2 \mathrm{~m} / \mathrm{s}^2\right.$ ]
AnswerData: $\mathrm{r}=5 \mathrm{~cm}=0.05 \mathrm{~m}, \mathrm{f}=90 \mathrm{rpm}=\frac{90}{60} \mathrm{rps}=1.5 \mathrm{rps}, \mathrm{g}=\pi^2 \mathrm{~m} / \mathrm{s}^2$
The centripetal force for the circular motion of the coin is provided by the friction between the coin and the turntable. The coin is just about to slip off the turntable when the limiting force of friction is equal to the centripetal force.
$
\therefore \mu_s m g=\frac{m v^2}{r}
$
The coefficient of static friction,
$
\begin{aligned}
\mu_s & =\frac{v^2}{r g}=\frac{(r \omega)^2}{r g} \\
& =\frac{\omega^2 r}{g}=\frac{(2 \pi f)^2 \times r}{g} \quad(\because \omega=2 \pi f) \\
& =\frac{4 \pi^2 f^2 r}{g}=\frac{4 \pi^2 \times(1.5)^2 \times 0.05}{\pi^2} \\
& =0.2 \times 2.25=0.45
\end{aligned}
$
View full question & answer→Question 424 Marks
On a dry day, the maximum safe speed at which a car can be driven on a curved horizontal road without skidding is $7 \mathrm{~m} / \mathrm{s}$. When the road is wet, the frictional force between the tyres and road reduces by $25 \%$. How fast can the car safely take the turn on the wet road?
AnswerLet subscripts 1 and 2 denote the values of a quantity under dry and wet conditions, respectively.
Data : $v_1=7 \mathrm{~m} / \mathrm{s}_2 \mathrm{f}_2=\mathrm{f}_1,-0.25 \mathrm{f}_1=0.75 \mathrm{f}_1$
On a dry horizontal curved road, the frictional force between the tyres and road is $f_1=\mu_1 m g$, where $\mathrm{m}$ is the mass of the car and $\mathrm{g}$ is the gravitational acceleration.
The maximum safe speed for taking a turn of radius $r$ on a dry horizontal curved road is
$
v_1=\sqrt{\mu_1 r g}=\sqrt{\frac{r}{m}} \sqrt{f_1}
$
If the road is wet, the corresponding quantities
are
$
f_2=\mu_2 m g \text { and } v_2=\sqrt{\frac{r}{m}} \sqrt{f_2}
$
Thus, for $m$ and $r$ remaining the same,
$
\begin{aligned}
\frac{v_1}{v_2} & =\sqrt{\frac{f_1}{f_2}} \\
\therefore v_2 & =\sqrt{\frac{f_2}{f_1}} \cdot v_1=\sqrt{\frac{0.75 f_1}{f_1}} \cdot(7) \\
& =7 \sqrt{0.75}=7 \times 0.866=6.062 \mathrm{~m} / \mathrm{s}
\end{aligned}
$
View full question & answer→Question 434 Marks
An amusement park ride (known variously as the Rotor, the Turkish Twist and the Gravitron) consists of a large vertical cylinder that is spun about it axis fast enough such that the riders remain pinned against its inner wall. The floor drops away once the cylinder has attained its full rotational speed. The radius of the cylinder is $R$ and the coefficient of static friction between a rider and the wall is $\mu_{\mathrm{s}}$.
(i) Show that the minimum angular speed necessary to keep a rider from falling is given by $\omega$ $=\sqrt{g / \mu_B R}$.
(ii) Obtain a numerical value for the frequency of rotation of the cylinder in rotations per minute if $R=4 \mathrm{~m}$ and $\mu_5=0.4$
View full question & answer→Question 444 Marks
A banked circular road is designed for traffic moving at an optimum or most safe speed vo. Obtain an expression for
(a) the minimum safe speed
(b) the maximum safe speed with which a vehicle can negotiate the curve without skidding.
Question 454 Marks
A road at a bend should be banked for an optimum or most safe speed $v_0$. Derive an expression for the required angle of banking.
OR
Obtain an expression for the optimum or most safe speed with which a vehicle can be driven along a curved banked road. Hence show that the angle of banking is independent of the mass of a vehicle.
View full question & answer→Question 464 Marks
A carnival event known as a "well of death" consists of a large vertical cylinder inside which usually a stunt motorcyclist rides in horizontal circles. Show that the minimum speed necessary to keep the rider from falling is given by $\mathrm{v}=\sqrt{\mathrm{rg} / \mu_s}$, in usual notations.
View full question & answer→Question 474 Marks
Derive an expression for the maximum safe speed for a vehicle on a circular horizontal road without toppling/overturning/rollover.
Question 484 Marks
Derive an expression for the maximum safe speed for a vehicle on a horizontal circular road without skidding off. State its significance.
Question 494 Marks
A disc of radius $15 \mathrm{~cm}$ rotates with a speed of $33 \frac{1}{3} \mathrm{rpm}$. Two coins are placed on it at $4 \mathrm{~cm}$ and $14 \mathrm{~cm}$ from its centre. If the coefficient of friction between the coins and the disc is 0.15 , which of the two coins will revolve with the disc?
AnswerData : $\mathrm{r}=15 \mathrm{~cm}=0.15 \mathrm{~m}$,
$
\mathrm{f}=33 \frac{1}{3} \mathrm{rpm}=\frac{100}{3 \times 60} \mathrm{rev} / \mathrm{s}=\frac{5}{9} \mathrm{~Hz}, \mu_{\mathrm{s}}=0.15, \mathrm{r}_1=4 \mathrm{~cm}=0.04 \mathrm{~m}, \mathrm{r}_2=14 \mathrm{~cm}=0.14 \mathrm{~m}
$
Angular speed, $\omega=2 \pi f=2 \times 3.142 \times \frac{5}{9}=\frac{31.42}{9}$
$
=3.491 \mathrm{rad} / \mathrm{s}
$
To revolve with the disc without slipping, the necessary centripetal force must be less than or equal to the limiting force of static friction.
Limiting force of static friction, $\mathrm{f}_{\mathrm{s}}=\mu_{\mathrm{s}} \mathrm{N}=\mu_{\mathrm{s}}(\mathrm{mg})$ where $\mathrm{m}$ is the mass of the coin and $\mathrm{N}=$ $\mathrm{mg}$ is the normal force on the coin.
$
\begin{aligned}
& \therefore \mathrm{m} \omega^2 r \leq \mu_s(\mathrm{mg}) \text { or } \omega^2 r \leq \mu_{\mathrm{s}} g \\
& \mu_{\mathrm{s}} \mathrm{g}=0.15 \times 9.8=1.47 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$
For the first coin, $r_1=0.04 \mathrm{~m}$.
$
\therefore \omega^2 r_1=(3.491)^2 \times 0.04=12.19 \times 0.04=0.4876 \mathrm{~m} / \mathrm{s}^2
$
Since, $\omega^2 r_1<\mu_s g$, this coin will revolve with the disc. For the second coin, $r_2=0.14 \mathrm{~m}$.
$
\therefore \omega^2 r_2=(3.491)^2 \times 0.14=12.19 \times 0.14=1.707 \mathrm{~m} / \mathrm{s}^2
$
Since, $\omega^2 r_2>\mu_5 g$, this coin will not revolve with the disc.
Thus, only the coin placed at $4 \mathrm{~cm}$ from the centre will revolve with the disc.
View full question & answer→Question 504 Marks
Distinguish between centripetal force and centrifugal force. State any two points of distinction.