Question 13 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 23 Marks
Define the angular momentum of a particle.
Question 33 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 43 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 53 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 63 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 73 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 83 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 93 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 103 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 113 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 123 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 133 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 143 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 153 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 163 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 173 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 183 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 193 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 203 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 213 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 223 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 233 Marks
State an expression for the moment of inertia of a thin ring about its transverse symmetry axis.
Question 243 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 253 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 263 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 273 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 283 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 293 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 303 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 313 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 323 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 333 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 343 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 353 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 363 Marks
A motorcyclist is describing a circle of radius $25 \mathrm{~m}$ at a speed of $5 \mathrm{~m} / \mathrm{s}$. Find his inclination with the vertical. What is the value of the coefficient of friction between the tyres and ground ?
AnswerData : $v=5 \mathrm{~m} / \mathrm{s}, \mathrm{r}=25 \mathrm{~m}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^2$
(i)
$
\begin{aligned}
& \tan \theta=\frac{v^2}{r g}=\frac{(5)^2}{25 \times 10}=0.10 \\
& \therefore \theta=\tan ^{-1} 0.10=5^{\circ} \mathbf{4}^{\prime}
\end{aligned}
$
(inclination with the vertical)
(ii) $\frac{m v^2}{r}=\mu_s m g$
where $\mu_{\mathrm{s}}$ is the coefficient of friction.
$
\therefore \mu_{\mathrm{s}}=\frac{v^2}{r g}=0.10
$
View full question & answer→Question 373 Marks
A thin cylindrical shell of inner radius 1.5 m rotates horizontally, about a vertical axis, at an angular speed ω. A wooden block rests against the inner surface and rotates with it. If the coefficient of static friction between block and surface is 0.3, how fast must the shell be rotating if the block is not to slip and fall ?
AnswerData : $\mathrm{r}=1.5 \mathrm{~m}, \mu_{\mathrm{s}}=0.3$
The normal force $\vec{N}$ of the shell on the block is the centripetal force which holds the block in place. $\vec{N}$ determines the friction on the block, which in turn keeps it from sliding downward.
If the block is not to slip, the friction force $\overrightarrow{f_{\mathrm{s}}}$ must balance the weight $m \vec{g}$ of the block.
$
\begin{aligned}
& \therefore N=m \omega^2 r \text { and } f_s=\mu_s N=m g \\
& \therefore \mu_s\left(\omega^2 r\right)=m g \\
& \therefore \omega=\sqrt{\frac{g}{\mu_s r}}=\sqrt{\frac{10}{(0.3)(1.5)}}=\sqrt{\frac{10}{0.45}} \\
& \quad=\sqrt{22.22}=4.714 \mathrm{rad} / \mathrm{s}=\frac{4.714}{2 \pi}=0.75 \mathrm{rev} / \mathrm{s}
\end{aligned}
$
This gives the required angular speed.
View full question & answer→Question 383 Marks
A circular race course track has a radius of $500 \mathrm{~m}$ and is banked at $10^{\circ}$. The coefficient of static friction between the tyres of a vehicle and the road surface is 0.25 . Compute
(i) the maximum speed to avoid slipping
(ii) the optimum speed to avoid wear and tear of the tyres.
Answer$
\text { Data : } r=500 \mathrm{~m}, \theta=10^{\circ}, \mu_{\mathrm{s}}=0.25, \mathrm{~g}=9.8 \mathrm{~m} / \mathrm{s}^2, \tan 10^{\circ}=0.1763
$
(i) On the banked track, the maximum speed of the vehicle without slipping (skidding) is
$
\begin{aligned}
& v_{\max }=\sqrt{\frac{r g\left(\mu_{\mathrm{s}}+\tan \theta\right)}{1-\mu_{\mathrm{s}} \tan \theta}} \\
& =\sqrt{\frac{500 \times 10(0.25+0.1763)}{1-(0.25 \times 0.1763)}} \\
& \text { MaharashtraBoardSolutions. Guru } \\
& =\sqrt{\frac{500 \times 10 \times 0.4263}{0.9559}}=\sqrt{2230} \\
& =47.22 \mathrm{~m} / \mathrm{s} \\
&
\end{aligned}
$
(ii) The optimum speed of the vehicle on the track is
$
\begin{aligned}
v_{\text {opt }} & =\sqrt{r g \tan \theta} \\
& =\sqrt{500 \times 10 \times 0.1763} \\
& =\sqrt{881.5}=29.69 \mathrm{~m} / \mathrm{s}
\end{aligned}
$
View full question & answer→Question 393 Marks
A certain string $500 \mathrm{~cm}$ long breaks under a tension of $45 \mathrm{~kg}$ wt. An object of mass $100 \mathrm{~g}$ is attached to this string and whirled in a horizontal circle. Find the maximum number of revolutions that the object can make per second without breaking the string, $\left[\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\right.$ ]
AnswerData : $\mathrm{m}=100 \mathrm{~g}=0.1 \mathrm{~kg}, \mathrm{r}=500 \mathrm{~cm}=5 \mathrm{~m}, \mathrm{~g}=9.8 \mathrm{~m} / \mathrm{s}^2, \mathrm{~F}=45 \mathrm{~kg} w \mathrm{t}=45 \times 9.8 \mathrm{~N}$ The breaking tension is equal to the maximum centripetal force that can be applied. $\therefore \mathrm{F}=\mathrm{m} \omega^2 \mathrm{r}$,
But $\omega=2 \pi \mathrm{f}$, where/is the corresponding frequency of revolution.
$
\begin{aligned}
& \therefore F=m(2 \pi f)^2 r=4 \pi^2 m f^2 r \\
& \therefore f=\sqrt{\frac{F}{4 \pi^2 m r}}=\sqrt{\frac{45 \times 9.8}{4 \times(3.142)^2 \times 0.1 \times 5}}
\end{aligned}
$
The maximum number of revolutions per second, $f=4.726 \mathrm{~Hz}$
View full question & answer→Question 403 Marks
A coin is placed on a stationary disc at a distance of $1 \mathrm{~m}$ from the disc's centre. At time $t=0$ $\mathrm{s}$, the disc begins to rotate with a constant angular acceleration of $2 \mathrm{rad} / \mathrm{s}^2$ around a fixed vertical axis through its centre and perpendicular to its plane.
Find the magnitude of the linear acceleration of the coin at $t=1.5 \mathrm{~s}$. Assume the coin does not slip.
AnswerData : $r=1 \mathrm{~m}, \alpha=2 \mathrm{rad} / \mathrm{s}^2, \omega_0=0, t=1.5 \mathrm{~s}$
$
a_{\mathrm{t}}=\alpha r=(2)(1)=2 \mathrm{~m} / \mathrm{s}^2
$
Angular speed at $t=1.5 \mathrm{~s}$,
$
\begin{aligned}
\omega & =\omega_{\mathrm{o}}+\alpha t=0+(2)(1.5)=3 \mathrm{rad} / \mathrm{s} \\
\therefore a_{\mathrm{r}} & =\omega^2 r=(3)^2(1)=9 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$
The required linear acceleration is,
$
\begin{aligned}
a & =\sqrt{a_{\mathrm{r}}^2+a_{\mathrm{t}}^2}=\sqrt{9^2+2^2}=\sqrt{85} \\
& =9.22 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$
[OR $v=u+a_1 t=0+(2)(1.5)=3 \mathrm{~m} / \mathrm{s}$
$
\left.\therefore a_{\mathrm{r}}=\frac{v^2}{r}=\frac{3^2}{1}=9 \mathrm{~m} / \mathrm{s}^2\right]
$
View full question & answer→Question 413 Marks
A flywheel slows down uniformly from $1200 \mathrm{rpm}$ to $600 \mathrm{rpm}$ in $5 \mathrm{~s}$. Find the number of revolutions made by the wheel in $5 \mathrm{~s}$.
AnswerData : $\omega_0=1200 \mathrm{rpm}, \omega=600 \mathrm{rpm}, \mathrm{f}=5 \mathrm{~s}$
Since the flywheel slows down uniformly, its angular acceleration is constant. Then, its average angular speed,
$
\begin{aligned}
& \omega_{\mathrm{av}}=\frac{\omega_{\mathrm{o}}+\omega}{2}=\frac{1200+600}{2} \\
& =900 \mathrm{rpm}=\frac{900 \mathrm{rev}}{60 \mathrm{~s}}=15 \mathrm{rps}
\end{aligned}
$
Its angular displacement in time $t$,
$\theta=\omega_{\text {av }} \cdot \mathrm{t}=15 \times 5=75$ revolutions
View full question & answer→Question 423 Marks
The frequency of revolution of a particle performing circular motion changes from $60 \mathrm{rpm}$ to $180 \mathrm{rpm}$ in 20 seconds. Calculate the angular acceleration of the particle.
AnswerData $: \mathrm{f}_1=60 \mathrm{rpm}=\frac{60}{60} \mathrm{rev} / \mathrm{s}=1 \mathrm{rev} / \mathrm{s}, \mathrm{f}_2=180 \mathrm{rpm}=\frac{180}{60} \mathrm{rev} / \mathrm{s}=3 \mathrm{rev} / \mathrm{s}, \mathrm{t}=20 \mathrm{~s}$ The angular acceleration in SI units,
$
\begin{aligned}
\alpha & =\frac{\omega_2-\omega_1}{t}=\frac{2 \pi f_2-2 \pi f_1}{t}=\frac{2 \pi(3)-2 \pi(1)}{20} \\
& =\frac{4 \pi}{20}=\frac{\pi}{5}=\frac{3.14}{5}=0.628 \mathrm{rad} / \mathrm{s}^2
\end{aligned}
$
OR
Using non SI units, the angular frequencies are $\omega_1=60 \mathrm{rpm}=1 \mathrm{rps}$ and $\omega_2=180 \mathrm{rpm}=3$ rps.
$
\therefore \alpha=\frac{\omega_2-\omega_1}{t}=\frac{3-1}{20}=\frac{1}{10}=0.1 \mathrm{rev} / \mathrm{s}^2 .
$
View full question & answer→Question 433 Marks
A body of mass 100 grams is tied to one end of a string and revolved along a circular path in the horizontal plane. The radius of the circle is $50 \mathrm{~cm}$. If the body revolves with a constant angular speed of $20 \mathrm{rad} / \mathrm{s}$, find the
1. period of revolution
2. linear speed
3. centripetal acceleration of the body.
AnswerData : $m=100 \mathrm{~g}=0.1 \mathrm{~kg}, \mathrm{r}=50 \mathrm{~cm}=0.5 \mathrm{~m}, \omega=20 \mathrm{rad} / \mathrm{s}$
1. The period of revolution of the body,
$
T=\frac{2 \pi}{\omega}=\frac{2 \times 3.142}{20}=0.3142 \mathrm{~s}
$
2. Linear speed, $v=\omega r=20 \times 0.5=10 \mathrm{~m} / \mathrm{s}$
3. Centripetal acceleration,
$
a_c=w^2 r=(20)^2 \times 0.5=200 \mathrm{~m} / \mathrm{s}^2
$
View full question & answer→Question 443 Marks
Write the kinematical equations for circular motion in analogy with linear motion.
AnswerFor circular motion of a particle with constant angular acceleration $\alpha$, average angular speed, $\omega_{\mathrm{av}}=\frac{\omega_0+\omega}{2}$
$
\alpha=\frac{\omega-\omega_0}{t} \text { and } \theta-\theta_0=\omega_{\text {av }} \cdot t
$
where $\omega_0$ and $\omega$ are the initial and final angular speeds, $t$ is the time, $\omega_{a v}$ the average angular speed and $\theta_0$ and $\theta$ the initial and final angular positions of the particle.
Then, the angular kinematical equations for the circular motion are (in analogy with linear kinematical equations for constant linear acceleration)
$
\begin{aligned}
& \omega=\omega_0+\alpha t \\
& \theta-\theta_0=\omega_0 t+\frac{1}{2} \alpha t^2 \\
& \omega^2=\omega_0^2+2 \alpha\left(\theta-\theta_0\right)
\end{aligned}
$
View full question & answer→Question 453 Marks
Draw a diagram showing the linear velocity, angular velocity and radial acceleration of a particle performing circular motion with radius r.
Question 463 Marks
A solid sphere of mass $1 \mathrm{~kg}$ rolls on a table with linear speed $2 \mathrm{~m} / \mathrm{s}$, find its total kinetic energy.
AnswerData : $M=1 \mathrm{~kg}, \mathrm{v}=2 \mathrm{~m} / \mathrm{s}$
The total kinetic energy of a rolling body,
$
E=\frac{1}{2} M v^2\left(1+\frac{k^2}{R^2}\right)
$
For a solid sphere, $k^2=\frac{2}{5} R^2$
$
\begin{aligned}
\therefore E & =\frac{1}{2} M v^2\left(1+\frac{2}{5}\right)=\frac{1}{2} \times \frac{7}{5} M v^2 \\
& =\frac{7}{10} M v^2=\frac{7}{10} \times 1 \times 2^2=\frac{7 \times 4}{10}=2.8 \mathrm{~J}
\end{aligned}
$
View full question & answer→Question 473 Marks
A lawn roller of mass $80 \mathrm{~kg}$, radius $0.3 \mathrm{~m}$ and moment of inertia $3.6 \mathrm{~kg} \cdot \mathrm{m}^2$, is drawn along a level surface at a constant speed of $1.8 \mathrm{~m} / \mathrm{s}$. Find
(i) the translational kinetic energy
(ii) the rotational kinetic energy
(iii) the total kinetic energy of the roller.
AnswerData : $M=80 \mathrm{~kg}, \mathrm{R}=0.3 \mathrm{~m}, \mathrm{I}=3.6 \mathrm{~kg} \cdot \mathrm{m}^2, \mathrm{v}=1.8 \mathrm{~m} / \mathrm{s}$
(i) The translational kinetic energy of the centre of mass of the roller,
$
E_{\text {tran }}=\frac{1}{2} M v^2=\frac{1}{2} \times 80 \times(1.8)^2=40 \times 3.24=129.6 \mathrm{~J}
$
(ii) The rotational kinetic energy about the roller's axle,
$
\begin{aligned}
E_{\text {rot }}=\frac{1}{2} I \omega^2=\frac{1}{2} I\left(\frac{v}{R}\right)^2 & =\frac{1}{2} \times 3.6 \times\left(\frac{1.8}{0.3}\right)^2 \\ \\
& =1.8 \times 36=64.8 \mathrm{~J}
\end{aligned}
$
(iii) The total kinetic energy of the roller,
$
E=E_{\operatorname{tran}}+E_{r o t}=129.6+64.8=194.4 \mathrm{~J}
$
View full question & answer→Question 483 Marks
State with reason if the statement is true or false : A wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.
AnswerThe statement is true.
Explanation : Rolling on a surface (horizontal or inclined) without slipping may be viewed as pure rotation about an horizontal axis through the point of contact, when viewed in the inertial frame of reference in which the surface is at rest. The point of contact of the wheel with the surface will be instantaneously at rest, resulting in a rolling motion, provided the wheel is able to 'grip' the surface, i.e., friction is necessary. With little or no friction, the wheel will slip at the point of contact. On an inclined plane, this will result in pure translation along the plane. On a horizontal surface, the wheel will simply rotate about its axis through the centre without translation.
View full question & answer→Question 493 Marks
Define the period and frequency of revolution of a particle performing uniform circular motion (UCM) and state expressions for them. Also state their SI units.
Answer(1) Period of revolution: The time taken by a particle performing UCM to complete one revolution is called the period of revolution or the period $(T)$ of UCM.
$
\mathrm{T}=\frac{2 \pi r}{v}=\frac{2 \pi}{\omega}
$
where $v$ and $w$ are the linear and angular speeds, respectively.
SI unit: the second (s)
Dimensions : $\left[\mathrm{M}^{\circ} \mathrm{L}^{\circ} \mathrm{T}^1\right]$.
(2) Frequency of revolution: The number of revolutions per unit time made by a particle in UCM is called the frequency of revolution ( $f$ ).
The particle completes 1 revolution in periodic time T. Therefore, it completes $1 / T$ revolutions per unit time.
$\therefore$ Frequency $\mathrm{f}=\frac{1}{T}=\frac{v}{2 \pi r}=\frac{\omega}{2 \pi}$
SI unit : the hertz $(\mathrm{Hz}), 1 \mathrm{~Hz}=1 \mathrm{~s}^{-1}$
Dimensions: $\left[\mathrm{M}^{\circ} \mathrm{L}^{\circ} \mathrm{T}^{-1}\right]$
View full question & answer→Question 503 Marks
A torque of $100 \mathrm{~N} . \mathrm{m}$ is applied to a body capable of rotating about a given axis. If the body starts from rest and acquires kinetic energy of $10000 \mathrm{~J}$ in 10 seconds, find
(i) its moment of inertia about the given axis
(ii) its angular momentum at the end of 10 seconds.
AnswerData : $\tau=100 \mathrm{~N} \cdot \mathrm{m}, \omega_i=0, \mathrm{E}_{\mathrm{i}}=0, \mathrm{E}_{\mathrm{f}}=10^4 \mathrm{~J}, \mathrm{t}=10 \mathrm{~s}$
$
\tau=\frac{\Delta L}{\Delta t}=\frac{L_{\mathrm{f}}-L_{\mathrm{i}}}{\Delta t}
$
Since the body starts from rest, its initial angular momentum, $L_i=0$.
The final angular momentum,
$
L_f=\tau \Delta t=(100)(10)=10^3 \mathrm{~kg} \cdot \mathrm{m}^2 / \mathrm{s}
$
The final rotational kinetic energy, $E_f=\frac{1}{2} L_{\mathrm{f}} \omega_{\mathrm{f}}$
$
\begin{aligned}
\therefore \omega_{\mathrm{f}} & =\frac{2 E_{\mathrm{f}}}{L_{\mathrm{f}}} \\
& =\frac{2 \times 10^4}{10^3}=20 \mathrm{rad} / \mathrm{s}
\end{aligned}
$
The moment of inertia of the body,
$
\begin{aligned}
I & =\frac{L_f}{\omega_f} \\
& =\frac{10^3}{20}=50 \mathrm{~kg} \cdot \mathrm{m}^2
\end{aligned}
$
View full question & answer→