150151
The moment of inertia of a rectangular plate of mass \(M\), length \(L\) and breadth \(B\), about an axis passing through its centre and perpendicular to its plane is
C Rectangular EFGH of length (L), breadth (B) and mass (M), Now, differentiate the area, \(\mathrm{dA}=\mathrm{B} \cdot \mathrm{dx} \quad\left(\because \sigma=\frac{\mathrm{M}}{\mathrm{A}}\right)\) \(\sigma=\frac{\mathrm{M}}{\mathrm{L} \times \mathrm{B}} \quad \text { Where, } \sigma=\text { Mass per unit area }\) So, \(\quad \mathrm{dM}=\sigma \mathrm{dA}=\frac{\text { M.B.dx }}{\text { L.B }}\) Now, integrating the values, \(=\frac{M \cdot d x}{L}\) \(\int d I_{x}=\int_{-L / 2}^{L / 2} d M \times x^{2}\) \(I_{x}=\frac{M}{L} \int_{-L / 2}^{L / 2} x^{2} d x\) \(I_{x}=\frac{M}{L}\left[\frac{x^{3}}{3}\right]_{-L / 2}^{L / 2}\) \(I_{x}=\frac{M}{L}\left[\frac{L^{3}}{24}+\frac{L^{3}}{24}\right]=\frac{M}{L} \times \frac{L^{3}}{12}\) \(I_{x}=\frac{M L^{2}}{12}\) Same as for, \(I_{y}=\frac{M^{2}}{12}\) Using perpendicular axis theorem, \(I_{z}=I_{x}+I_{y}\) \(I_{z}=\frac{M L^{2}}{12}+\frac{M B^{2}}{12}=\frac{M\left(L^{2}+B^{2}\right)}{12}\) For a rectangular rod, \(\mathrm{I}=\Sigma \mathrm{M}\left(\frac{\mathrm{L}^{2}+\mathrm{B}^{2}}{12}\right)\) \(\therefore \quad \mathrm{I} =\mathrm{M}\left(\frac{\mathrm{L}^{2}+\mathrm{B}^{2}}{12}\right)\)
AP EAMCET (21.09.2020) Shift-I
Rotational Motion
150152
Three identical uniform solid spheres each of mass \(m\) and radius \(r\) are joined as shown in the figure, with centres lying in the same plane. The moment of inertia of the system about an axis lying in that plane and passing through the centre of sphere \(C\) is
1 \(\frac{16}{5} m r^{2}\)
2 \(\frac{12}{5} m r^{2}\)
3 \(4 m r^{2}\)
4 \(\frac{3}{5} m r^{2}\)
Explanation:
A Given, Mass of solid sphere \(=\mathrm{m}\) Radius of solid sphere \(=r\) We know that, \(\mathrm{I}_{\text {sphere }}=\frac{2}{5} \mathrm{mr}^{2}\) Moment of inertia of solid spheres passing through center- \(\mathrm{I}_{\mathrm{C}}=\frac{2}{5} \mathrm{mr}^{2}\) Moment of inertia of a solid spheres through its tangent is given by \(\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{B}}=\frac{2}{5} \mathrm{mr}^{2}+\mathrm{mr}^{2}\) \(\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{B}}=\frac{7}{5} \mathrm{mr}^{2}\) Net moment of inertia of the system- \(\mathrm{I}_{\mathrm{PQ}} =\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}\) \(\mathrm{I}_{\mathrm{PQ}} =\frac{7}{5} \mathrm{mr}^{2}+\frac{7}{5} \mathrm{mr}^{2}+\frac{2}{5} \mathrm{mr}^{2}\) \(\mathrm{I}_{\mathrm{PQ}} =\frac{16}{5} \mathrm{mr}^{2}\)
AP EAMCET (17.09.2020) Shift-I
Rotational Motion
150153
Four point masses, each of mass \(M\) are placed at the corners of a square of side \(L\). The moment of inertia of the system about one of its diagonals is
1 \(2 \mathrm{ML}\)
2 \(\mathrm{ML}^{2}\)
3 \(4 \mathrm{ML}^{2}\)
4 \(6 \mathrm{ML}^{2}\)
Explanation:
B Given, mass of each point \(=M\) Side of square \(=\mathrm{L}\) \(\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}=\mathrm{L}\) Length of diagonal, \(\mathrm{BD}=\sqrt{\mathrm{DA}^{2}+\mathrm{AB}^{2}}=\sqrt{\mathrm{L}^{2}+\mathrm{L}^{2}}\) \(\mathrm{BD}=\sqrt{2 \mathrm{~L}^{2}}\) \(\mathrm{BD}=\mathrm{L} \sqrt{2}\) \(\therefore \mathrm{OD}=\mathrm{OB}=\frac{\mathrm{BD}}{2}=\frac{\mathrm{L} \sqrt{2}}{2}=\frac{\mathrm{L}}{\sqrt{2}}\) Moment of inertia of given system about the diagonal (AC), \(\mathrm{I}_{\mathrm{AC}}=\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}+\mathrm{I}_{\mathrm{D}}\) \(\mathrm{I}_{\mathrm{AC}}=0+\mathrm{M}\left(\frac{\mathrm{L}}{\sqrt{2}}\right)^{2}+0+\mathrm{M}\left(\frac{\mathrm{L}}{\sqrt{2}}\right)^{2}\) \(\mathrm{I}_{\mathrm{AC}}=\frac{\mathrm{ML}^{2}}{2}+\frac{\mathrm{ML}^{2}}{2}\) \(\mathrm{I}_{\mathrm{AC}}=\mathrm{ML}^{2}\)
AP EAMCET (17.09.2020) Shift-I
Rotational Motion
150154
A solid cylinder of mass \(2 \mathrm{~kg}\) and radius \(4 \mathrm{~cm}\) is rotating about its axis at the rate of \(3 \mathrm{rpm}\). The torque required to stop after \(2 \pi\) revolutions is
1 \(2 \times 10^{-3} \mathrm{~N}-\mathrm{m}\)
2 \(12 \times 10^{-4} \mathrm{~N}-\mathrm{m}\)
3 \(2 \times 10^{6} \mathrm{~N}-\mathrm{m}\)
4 \(2 \times 10^{-6} \mathrm{~N}-\mathrm{m}\)
Explanation:
D Given that, Mass \(\mathrm({m})=2 \mathrm{~kg}\) Radius \(\mathrm({r})=4 \mathrm{~cm} \Rightarrow 0.04 \mathrm{~m}\) Rate of rotation \(\omega)=3 \mathrm{rpm}=3 \times \frac{2 \pi}{60} \Rightarrow \frac{\pi}{10}\) Moment of inertia \(=\frac{1}{2} \mathrm{mr}^{2}\) \(=\frac{1}{2} \times 2 \times(0.04)^{2}\) \(=16 \times 10^{-4} \mathrm{kgm}^{2}\) \(=1.6 \times 10^{-3} \mathrm{kgm}^{2}\) Angular displacement \(=2 \pi \times\) revolution \(=2 \pi \times 2 \pi\) \(=4 \pi^{2}\) From conservation of energy theorem, Change in rotational kinetic energy \(=\) torque \(\times\) angular displacement So, \(\quad \frac{1}{2} \mathrm{I} \omega^{2}=\) torque \(\times 4 \pi^{2}\) \(\frac{1}{2} \times 1.6 \times 10^{-3} \times\left(\frac{\pi}{10}\right)^{2}=\tau \times 4 \pi^{2}\) \(\tau=\frac{\frac{1}{2} \times 1.6 \times 10^{-3} \times\left(\frac{\pi}{10}\right)^{2}}{4 \pi^{2}}\) \(\text { torque }(\tau)=2 \times 10^{-6} \mathrm{~N}-\mathrm{m}\)
NEET (National) - 2019
Rotational Motion
150155
When a 12000 joule of work is done on a flywheel, its frequency of rotation increases from \(10 \mathrm{~Hz}\) to \(20 \mathrm{~Hz}\). The moment of inertia of flywheel about its axis of rotation is \(\left(\boldsymbol{\pi}^{2}=10\right)\)
1 \(1.688 \mathrm{kgm}^{2}\)
2 \(2 \mathrm{kgm}_{2}^{2}\)
3 \(1.5 \mathrm{kgm}\)
4 \(1 \mathrm{kgm}^2\)
Explanation:
B Given, work done on a flywheel (W) \(=12000\) Joule, initial frequency \(\left(f_{1}\right)=10 \mathrm{~Hz}\), final frequency \(\left(f_{2}\right)\) \(=20 \mathrm{~Hz}\) \(\therefore\) Angular velocity for rotational motion- \(\omega_{1}=2 \pi f_{1}=2 \pi \times 10=20 \pi\) And \(\omega_{2}=2 \pi \times 20=40 \pi\) Acording to work energy theorem- Work done on a flywheel \(=\) Change in rotational kinetic energy \(\mathrm{W}=\frac{1}{2} \mathrm{I}\left(\omega_{2}^{2}-\omega_{1}^{2}\right)\) Where, \(\mathrm{I}=\) Moment of inertia \(12000=\frac{1}{2} \mathrm{I}\left[(40 \pi)^{2}-(20 \pi)^{2}\right]\) \(\therefore \quad 12000=\frac{1}{2} \mathrm{I}(1600-400) \pi^{2}\) \(\mathrm{I}=\frac{12000\times 2}{12000} \quad\left[\because \pi^{2}=10\right]\) \(\mathrm{I}=2 \mathrm{kgm}^{2}\)
150151
The moment of inertia of a rectangular plate of mass \(M\), length \(L\) and breadth \(B\), about an axis passing through its centre and perpendicular to its plane is
C Rectangular EFGH of length (L), breadth (B) and mass (M), Now, differentiate the area, \(\mathrm{dA}=\mathrm{B} \cdot \mathrm{dx} \quad\left(\because \sigma=\frac{\mathrm{M}}{\mathrm{A}}\right)\) \(\sigma=\frac{\mathrm{M}}{\mathrm{L} \times \mathrm{B}} \quad \text { Where, } \sigma=\text { Mass per unit area }\) So, \(\quad \mathrm{dM}=\sigma \mathrm{dA}=\frac{\text { M.B.dx }}{\text { L.B }}\) Now, integrating the values, \(=\frac{M \cdot d x}{L}\) \(\int d I_{x}=\int_{-L / 2}^{L / 2} d M \times x^{2}\) \(I_{x}=\frac{M}{L} \int_{-L / 2}^{L / 2} x^{2} d x\) \(I_{x}=\frac{M}{L}\left[\frac{x^{3}}{3}\right]_{-L / 2}^{L / 2}\) \(I_{x}=\frac{M}{L}\left[\frac{L^{3}}{24}+\frac{L^{3}}{24}\right]=\frac{M}{L} \times \frac{L^{3}}{12}\) \(I_{x}=\frac{M L^{2}}{12}\) Same as for, \(I_{y}=\frac{M^{2}}{12}\) Using perpendicular axis theorem, \(I_{z}=I_{x}+I_{y}\) \(I_{z}=\frac{M L^{2}}{12}+\frac{M B^{2}}{12}=\frac{M\left(L^{2}+B^{2}\right)}{12}\) For a rectangular rod, \(\mathrm{I}=\Sigma \mathrm{M}\left(\frac{\mathrm{L}^{2}+\mathrm{B}^{2}}{12}\right)\) \(\therefore \quad \mathrm{I} =\mathrm{M}\left(\frac{\mathrm{L}^{2}+\mathrm{B}^{2}}{12}\right)\)
AP EAMCET (21.09.2020) Shift-I
Rotational Motion
150152
Three identical uniform solid spheres each of mass \(m\) and radius \(r\) are joined as shown in the figure, with centres lying in the same plane. The moment of inertia of the system about an axis lying in that plane and passing through the centre of sphere \(C\) is
1 \(\frac{16}{5} m r^{2}\)
2 \(\frac{12}{5} m r^{2}\)
3 \(4 m r^{2}\)
4 \(\frac{3}{5} m r^{2}\)
Explanation:
A Given, Mass of solid sphere \(=\mathrm{m}\) Radius of solid sphere \(=r\) We know that, \(\mathrm{I}_{\text {sphere }}=\frac{2}{5} \mathrm{mr}^{2}\) Moment of inertia of solid spheres passing through center- \(\mathrm{I}_{\mathrm{C}}=\frac{2}{5} \mathrm{mr}^{2}\) Moment of inertia of a solid spheres through its tangent is given by \(\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{B}}=\frac{2}{5} \mathrm{mr}^{2}+\mathrm{mr}^{2}\) \(\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{B}}=\frac{7}{5} \mathrm{mr}^{2}\) Net moment of inertia of the system- \(\mathrm{I}_{\mathrm{PQ}} =\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}\) \(\mathrm{I}_{\mathrm{PQ}} =\frac{7}{5} \mathrm{mr}^{2}+\frac{7}{5} \mathrm{mr}^{2}+\frac{2}{5} \mathrm{mr}^{2}\) \(\mathrm{I}_{\mathrm{PQ}} =\frac{16}{5} \mathrm{mr}^{2}\)
AP EAMCET (17.09.2020) Shift-I
Rotational Motion
150153
Four point masses, each of mass \(M\) are placed at the corners of a square of side \(L\). The moment of inertia of the system about one of its diagonals is
1 \(2 \mathrm{ML}\)
2 \(\mathrm{ML}^{2}\)
3 \(4 \mathrm{ML}^{2}\)
4 \(6 \mathrm{ML}^{2}\)
Explanation:
B Given, mass of each point \(=M\) Side of square \(=\mathrm{L}\) \(\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}=\mathrm{L}\) Length of diagonal, \(\mathrm{BD}=\sqrt{\mathrm{DA}^{2}+\mathrm{AB}^{2}}=\sqrt{\mathrm{L}^{2}+\mathrm{L}^{2}}\) \(\mathrm{BD}=\sqrt{2 \mathrm{~L}^{2}}\) \(\mathrm{BD}=\mathrm{L} \sqrt{2}\) \(\therefore \mathrm{OD}=\mathrm{OB}=\frac{\mathrm{BD}}{2}=\frac{\mathrm{L} \sqrt{2}}{2}=\frac{\mathrm{L}}{\sqrt{2}}\) Moment of inertia of given system about the diagonal (AC), \(\mathrm{I}_{\mathrm{AC}}=\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}+\mathrm{I}_{\mathrm{D}}\) \(\mathrm{I}_{\mathrm{AC}}=0+\mathrm{M}\left(\frac{\mathrm{L}}{\sqrt{2}}\right)^{2}+0+\mathrm{M}\left(\frac{\mathrm{L}}{\sqrt{2}}\right)^{2}\) \(\mathrm{I}_{\mathrm{AC}}=\frac{\mathrm{ML}^{2}}{2}+\frac{\mathrm{ML}^{2}}{2}\) \(\mathrm{I}_{\mathrm{AC}}=\mathrm{ML}^{2}\)
AP EAMCET (17.09.2020) Shift-I
Rotational Motion
150154
A solid cylinder of mass \(2 \mathrm{~kg}\) and radius \(4 \mathrm{~cm}\) is rotating about its axis at the rate of \(3 \mathrm{rpm}\). The torque required to stop after \(2 \pi\) revolutions is
1 \(2 \times 10^{-3} \mathrm{~N}-\mathrm{m}\)
2 \(12 \times 10^{-4} \mathrm{~N}-\mathrm{m}\)
3 \(2 \times 10^{6} \mathrm{~N}-\mathrm{m}\)
4 \(2 \times 10^{-6} \mathrm{~N}-\mathrm{m}\)
Explanation:
D Given that, Mass \(\mathrm({m})=2 \mathrm{~kg}\) Radius \(\mathrm({r})=4 \mathrm{~cm} \Rightarrow 0.04 \mathrm{~m}\) Rate of rotation \(\omega)=3 \mathrm{rpm}=3 \times \frac{2 \pi}{60} \Rightarrow \frac{\pi}{10}\) Moment of inertia \(=\frac{1}{2} \mathrm{mr}^{2}\) \(=\frac{1}{2} \times 2 \times(0.04)^{2}\) \(=16 \times 10^{-4} \mathrm{kgm}^{2}\) \(=1.6 \times 10^{-3} \mathrm{kgm}^{2}\) Angular displacement \(=2 \pi \times\) revolution \(=2 \pi \times 2 \pi\) \(=4 \pi^{2}\) From conservation of energy theorem, Change in rotational kinetic energy \(=\) torque \(\times\) angular displacement So, \(\quad \frac{1}{2} \mathrm{I} \omega^{2}=\) torque \(\times 4 \pi^{2}\) \(\frac{1}{2} \times 1.6 \times 10^{-3} \times\left(\frac{\pi}{10}\right)^{2}=\tau \times 4 \pi^{2}\) \(\tau=\frac{\frac{1}{2} \times 1.6 \times 10^{-3} \times\left(\frac{\pi}{10}\right)^{2}}{4 \pi^{2}}\) \(\text { torque }(\tau)=2 \times 10^{-6} \mathrm{~N}-\mathrm{m}\)
NEET (National) - 2019
Rotational Motion
150155
When a 12000 joule of work is done on a flywheel, its frequency of rotation increases from \(10 \mathrm{~Hz}\) to \(20 \mathrm{~Hz}\). The moment of inertia of flywheel about its axis of rotation is \(\left(\boldsymbol{\pi}^{2}=10\right)\)
1 \(1.688 \mathrm{kgm}^{2}\)
2 \(2 \mathrm{kgm}_{2}^{2}\)
3 \(1.5 \mathrm{kgm}\)
4 \(1 \mathrm{kgm}^2\)
Explanation:
B Given, work done on a flywheel (W) \(=12000\) Joule, initial frequency \(\left(f_{1}\right)=10 \mathrm{~Hz}\), final frequency \(\left(f_{2}\right)\) \(=20 \mathrm{~Hz}\) \(\therefore\) Angular velocity for rotational motion- \(\omega_{1}=2 \pi f_{1}=2 \pi \times 10=20 \pi\) And \(\omega_{2}=2 \pi \times 20=40 \pi\) Acording to work energy theorem- Work done on a flywheel \(=\) Change in rotational kinetic energy \(\mathrm{W}=\frac{1}{2} \mathrm{I}\left(\omega_{2}^{2}-\omega_{1}^{2}\right)\) Where, \(\mathrm{I}=\) Moment of inertia \(12000=\frac{1}{2} \mathrm{I}\left[(40 \pi)^{2}-(20 \pi)^{2}\right]\) \(\therefore \quad 12000=\frac{1}{2} \mathrm{I}(1600-400) \pi^{2}\) \(\mathrm{I}=\frac{12000\times 2}{12000} \quad\left[\because \pi^{2}=10\right]\) \(\mathrm{I}=2 \mathrm{kgm}^{2}\)
150151
The moment of inertia of a rectangular plate of mass \(M\), length \(L\) and breadth \(B\), about an axis passing through its centre and perpendicular to its plane is
C Rectangular EFGH of length (L), breadth (B) and mass (M), Now, differentiate the area, \(\mathrm{dA}=\mathrm{B} \cdot \mathrm{dx} \quad\left(\because \sigma=\frac{\mathrm{M}}{\mathrm{A}}\right)\) \(\sigma=\frac{\mathrm{M}}{\mathrm{L} \times \mathrm{B}} \quad \text { Where, } \sigma=\text { Mass per unit area }\) So, \(\quad \mathrm{dM}=\sigma \mathrm{dA}=\frac{\text { M.B.dx }}{\text { L.B }}\) Now, integrating the values, \(=\frac{M \cdot d x}{L}\) \(\int d I_{x}=\int_{-L / 2}^{L / 2} d M \times x^{2}\) \(I_{x}=\frac{M}{L} \int_{-L / 2}^{L / 2} x^{2} d x\) \(I_{x}=\frac{M}{L}\left[\frac{x^{3}}{3}\right]_{-L / 2}^{L / 2}\) \(I_{x}=\frac{M}{L}\left[\frac{L^{3}}{24}+\frac{L^{3}}{24}\right]=\frac{M}{L} \times \frac{L^{3}}{12}\) \(I_{x}=\frac{M L^{2}}{12}\) Same as for, \(I_{y}=\frac{M^{2}}{12}\) Using perpendicular axis theorem, \(I_{z}=I_{x}+I_{y}\) \(I_{z}=\frac{M L^{2}}{12}+\frac{M B^{2}}{12}=\frac{M\left(L^{2}+B^{2}\right)}{12}\) For a rectangular rod, \(\mathrm{I}=\Sigma \mathrm{M}\left(\frac{\mathrm{L}^{2}+\mathrm{B}^{2}}{12}\right)\) \(\therefore \quad \mathrm{I} =\mathrm{M}\left(\frac{\mathrm{L}^{2}+\mathrm{B}^{2}}{12}\right)\)
AP EAMCET (21.09.2020) Shift-I
Rotational Motion
150152
Three identical uniform solid spheres each of mass \(m\) and radius \(r\) are joined as shown in the figure, with centres lying in the same plane. The moment of inertia of the system about an axis lying in that plane and passing through the centre of sphere \(C\) is
1 \(\frac{16}{5} m r^{2}\)
2 \(\frac{12}{5} m r^{2}\)
3 \(4 m r^{2}\)
4 \(\frac{3}{5} m r^{2}\)
Explanation:
A Given, Mass of solid sphere \(=\mathrm{m}\) Radius of solid sphere \(=r\) We know that, \(\mathrm{I}_{\text {sphere }}=\frac{2}{5} \mathrm{mr}^{2}\) Moment of inertia of solid spheres passing through center- \(\mathrm{I}_{\mathrm{C}}=\frac{2}{5} \mathrm{mr}^{2}\) Moment of inertia of a solid spheres through its tangent is given by \(\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{B}}=\frac{2}{5} \mathrm{mr}^{2}+\mathrm{mr}^{2}\) \(\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{B}}=\frac{7}{5} \mathrm{mr}^{2}\) Net moment of inertia of the system- \(\mathrm{I}_{\mathrm{PQ}} =\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}\) \(\mathrm{I}_{\mathrm{PQ}} =\frac{7}{5} \mathrm{mr}^{2}+\frac{7}{5} \mathrm{mr}^{2}+\frac{2}{5} \mathrm{mr}^{2}\) \(\mathrm{I}_{\mathrm{PQ}} =\frac{16}{5} \mathrm{mr}^{2}\)
AP EAMCET (17.09.2020) Shift-I
Rotational Motion
150153
Four point masses, each of mass \(M\) are placed at the corners of a square of side \(L\). The moment of inertia of the system about one of its diagonals is
1 \(2 \mathrm{ML}\)
2 \(\mathrm{ML}^{2}\)
3 \(4 \mathrm{ML}^{2}\)
4 \(6 \mathrm{ML}^{2}\)
Explanation:
B Given, mass of each point \(=M\) Side of square \(=\mathrm{L}\) \(\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}=\mathrm{L}\) Length of diagonal, \(\mathrm{BD}=\sqrt{\mathrm{DA}^{2}+\mathrm{AB}^{2}}=\sqrt{\mathrm{L}^{2}+\mathrm{L}^{2}}\) \(\mathrm{BD}=\sqrt{2 \mathrm{~L}^{2}}\) \(\mathrm{BD}=\mathrm{L} \sqrt{2}\) \(\therefore \mathrm{OD}=\mathrm{OB}=\frac{\mathrm{BD}}{2}=\frac{\mathrm{L} \sqrt{2}}{2}=\frac{\mathrm{L}}{\sqrt{2}}\) Moment of inertia of given system about the diagonal (AC), \(\mathrm{I}_{\mathrm{AC}}=\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}+\mathrm{I}_{\mathrm{D}}\) \(\mathrm{I}_{\mathrm{AC}}=0+\mathrm{M}\left(\frac{\mathrm{L}}{\sqrt{2}}\right)^{2}+0+\mathrm{M}\left(\frac{\mathrm{L}}{\sqrt{2}}\right)^{2}\) \(\mathrm{I}_{\mathrm{AC}}=\frac{\mathrm{ML}^{2}}{2}+\frac{\mathrm{ML}^{2}}{2}\) \(\mathrm{I}_{\mathrm{AC}}=\mathrm{ML}^{2}\)
AP EAMCET (17.09.2020) Shift-I
Rotational Motion
150154
A solid cylinder of mass \(2 \mathrm{~kg}\) and radius \(4 \mathrm{~cm}\) is rotating about its axis at the rate of \(3 \mathrm{rpm}\). The torque required to stop after \(2 \pi\) revolutions is
1 \(2 \times 10^{-3} \mathrm{~N}-\mathrm{m}\)
2 \(12 \times 10^{-4} \mathrm{~N}-\mathrm{m}\)
3 \(2 \times 10^{6} \mathrm{~N}-\mathrm{m}\)
4 \(2 \times 10^{-6} \mathrm{~N}-\mathrm{m}\)
Explanation:
D Given that, Mass \(\mathrm({m})=2 \mathrm{~kg}\) Radius \(\mathrm({r})=4 \mathrm{~cm} \Rightarrow 0.04 \mathrm{~m}\) Rate of rotation \(\omega)=3 \mathrm{rpm}=3 \times \frac{2 \pi}{60} \Rightarrow \frac{\pi}{10}\) Moment of inertia \(=\frac{1}{2} \mathrm{mr}^{2}\) \(=\frac{1}{2} \times 2 \times(0.04)^{2}\) \(=16 \times 10^{-4} \mathrm{kgm}^{2}\) \(=1.6 \times 10^{-3} \mathrm{kgm}^{2}\) Angular displacement \(=2 \pi \times\) revolution \(=2 \pi \times 2 \pi\) \(=4 \pi^{2}\) From conservation of energy theorem, Change in rotational kinetic energy \(=\) torque \(\times\) angular displacement So, \(\quad \frac{1}{2} \mathrm{I} \omega^{2}=\) torque \(\times 4 \pi^{2}\) \(\frac{1}{2} \times 1.6 \times 10^{-3} \times\left(\frac{\pi}{10}\right)^{2}=\tau \times 4 \pi^{2}\) \(\tau=\frac{\frac{1}{2} \times 1.6 \times 10^{-3} \times\left(\frac{\pi}{10}\right)^{2}}{4 \pi^{2}}\) \(\text { torque }(\tau)=2 \times 10^{-6} \mathrm{~N}-\mathrm{m}\)
NEET (National) - 2019
Rotational Motion
150155
When a 12000 joule of work is done on a flywheel, its frequency of rotation increases from \(10 \mathrm{~Hz}\) to \(20 \mathrm{~Hz}\). The moment of inertia of flywheel about its axis of rotation is \(\left(\boldsymbol{\pi}^{2}=10\right)\)
1 \(1.688 \mathrm{kgm}^{2}\)
2 \(2 \mathrm{kgm}_{2}^{2}\)
3 \(1.5 \mathrm{kgm}\)
4 \(1 \mathrm{kgm}^2\)
Explanation:
B Given, work done on a flywheel (W) \(=12000\) Joule, initial frequency \(\left(f_{1}\right)=10 \mathrm{~Hz}\), final frequency \(\left(f_{2}\right)\) \(=20 \mathrm{~Hz}\) \(\therefore\) Angular velocity for rotational motion- \(\omega_{1}=2 \pi f_{1}=2 \pi \times 10=20 \pi\) And \(\omega_{2}=2 \pi \times 20=40 \pi\) Acording to work energy theorem- Work done on a flywheel \(=\) Change in rotational kinetic energy \(\mathrm{W}=\frac{1}{2} \mathrm{I}\left(\omega_{2}^{2}-\omega_{1}^{2}\right)\) Where, \(\mathrm{I}=\) Moment of inertia \(12000=\frac{1}{2} \mathrm{I}\left[(40 \pi)^{2}-(20 \pi)^{2}\right]\) \(\therefore \quad 12000=\frac{1}{2} \mathrm{I}(1600-400) \pi^{2}\) \(\mathrm{I}=\frac{12000\times 2}{12000} \quad\left[\because \pi^{2}=10\right]\) \(\mathrm{I}=2 \mathrm{kgm}^{2}\)
150151
The moment of inertia of a rectangular plate of mass \(M\), length \(L\) and breadth \(B\), about an axis passing through its centre and perpendicular to its plane is
C Rectangular EFGH of length (L), breadth (B) and mass (M), Now, differentiate the area, \(\mathrm{dA}=\mathrm{B} \cdot \mathrm{dx} \quad\left(\because \sigma=\frac{\mathrm{M}}{\mathrm{A}}\right)\) \(\sigma=\frac{\mathrm{M}}{\mathrm{L} \times \mathrm{B}} \quad \text { Where, } \sigma=\text { Mass per unit area }\) So, \(\quad \mathrm{dM}=\sigma \mathrm{dA}=\frac{\text { M.B.dx }}{\text { L.B }}\) Now, integrating the values, \(=\frac{M \cdot d x}{L}\) \(\int d I_{x}=\int_{-L / 2}^{L / 2} d M \times x^{2}\) \(I_{x}=\frac{M}{L} \int_{-L / 2}^{L / 2} x^{2} d x\) \(I_{x}=\frac{M}{L}\left[\frac{x^{3}}{3}\right]_{-L / 2}^{L / 2}\) \(I_{x}=\frac{M}{L}\left[\frac{L^{3}}{24}+\frac{L^{3}}{24}\right]=\frac{M}{L} \times \frac{L^{3}}{12}\) \(I_{x}=\frac{M L^{2}}{12}\) Same as for, \(I_{y}=\frac{M^{2}}{12}\) Using perpendicular axis theorem, \(I_{z}=I_{x}+I_{y}\) \(I_{z}=\frac{M L^{2}}{12}+\frac{M B^{2}}{12}=\frac{M\left(L^{2}+B^{2}\right)}{12}\) For a rectangular rod, \(\mathrm{I}=\Sigma \mathrm{M}\left(\frac{\mathrm{L}^{2}+\mathrm{B}^{2}}{12}\right)\) \(\therefore \quad \mathrm{I} =\mathrm{M}\left(\frac{\mathrm{L}^{2}+\mathrm{B}^{2}}{12}\right)\)
AP EAMCET (21.09.2020) Shift-I
Rotational Motion
150152
Three identical uniform solid spheres each of mass \(m\) and radius \(r\) are joined as shown in the figure, with centres lying in the same plane. The moment of inertia of the system about an axis lying in that plane and passing through the centre of sphere \(C\) is
1 \(\frac{16}{5} m r^{2}\)
2 \(\frac{12}{5} m r^{2}\)
3 \(4 m r^{2}\)
4 \(\frac{3}{5} m r^{2}\)
Explanation:
A Given, Mass of solid sphere \(=\mathrm{m}\) Radius of solid sphere \(=r\) We know that, \(\mathrm{I}_{\text {sphere }}=\frac{2}{5} \mathrm{mr}^{2}\) Moment of inertia of solid spheres passing through center- \(\mathrm{I}_{\mathrm{C}}=\frac{2}{5} \mathrm{mr}^{2}\) Moment of inertia of a solid spheres through its tangent is given by \(\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{B}}=\frac{2}{5} \mathrm{mr}^{2}+\mathrm{mr}^{2}\) \(\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{B}}=\frac{7}{5} \mathrm{mr}^{2}\) Net moment of inertia of the system- \(\mathrm{I}_{\mathrm{PQ}} =\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}\) \(\mathrm{I}_{\mathrm{PQ}} =\frac{7}{5} \mathrm{mr}^{2}+\frac{7}{5} \mathrm{mr}^{2}+\frac{2}{5} \mathrm{mr}^{2}\) \(\mathrm{I}_{\mathrm{PQ}} =\frac{16}{5} \mathrm{mr}^{2}\)
AP EAMCET (17.09.2020) Shift-I
Rotational Motion
150153
Four point masses, each of mass \(M\) are placed at the corners of a square of side \(L\). The moment of inertia of the system about one of its diagonals is
1 \(2 \mathrm{ML}\)
2 \(\mathrm{ML}^{2}\)
3 \(4 \mathrm{ML}^{2}\)
4 \(6 \mathrm{ML}^{2}\)
Explanation:
B Given, mass of each point \(=M\) Side of square \(=\mathrm{L}\) \(\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}=\mathrm{L}\) Length of diagonal, \(\mathrm{BD}=\sqrt{\mathrm{DA}^{2}+\mathrm{AB}^{2}}=\sqrt{\mathrm{L}^{2}+\mathrm{L}^{2}}\) \(\mathrm{BD}=\sqrt{2 \mathrm{~L}^{2}}\) \(\mathrm{BD}=\mathrm{L} \sqrt{2}\) \(\therefore \mathrm{OD}=\mathrm{OB}=\frac{\mathrm{BD}}{2}=\frac{\mathrm{L} \sqrt{2}}{2}=\frac{\mathrm{L}}{\sqrt{2}}\) Moment of inertia of given system about the diagonal (AC), \(\mathrm{I}_{\mathrm{AC}}=\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}+\mathrm{I}_{\mathrm{D}}\) \(\mathrm{I}_{\mathrm{AC}}=0+\mathrm{M}\left(\frac{\mathrm{L}}{\sqrt{2}}\right)^{2}+0+\mathrm{M}\left(\frac{\mathrm{L}}{\sqrt{2}}\right)^{2}\) \(\mathrm{I}_{\mathrm{AC}}=\frac{\mathrm{ML}^{2}}{2}+\frac{\mathrm{ML}^{2}}{2}\) \(\mathrm{I}_{\mathrm{AC}}=\mathrm{ML}^{2}\)
AP EAMCET (17.09.2020) Shift-I
Rotational Motion
150154
A solid cylinder of mass \(2 \mathrm{~kg}\) and radius \(4 \mathrm{~cm}\) is rotating about its axis at the rate of \(3 \mathrm{rpm}\). The torque required to stop after \(2 \pi\) revolutions is
1 \(2 \times 10^{-3} \mathrm{~N}-\mathrm{m}\)
2 \(12 \times 10^{-4} \mathrm{~N}-\mathrm{m}\)
3 \(2 \times 10^{6} \mathrm{~N}-\mathrm{m}\)
4 \(2 \times 10^{-6} \mathrm{~N}-\mathrm{m}\)
Explanation:
D Given that, Mass \(\mathrm({m})=2 \mathrm{~kg}\) Radius \(\mathrm({r})=4 \mathrm{~cm} \Rightarrow 0.04 \mathrm{~m}\) Rate of rotation \(\omega)=3 \mathrm{rpm}=3 \times \frac{2 \pi}{60} \Rightarrow \frac{\pi}{10}\) Moment of inertia \(=\frac{1}{2} \mathrm{mr}^{2}\) \(=\frac{1}{2} \times 2 \times(0.04)^{2}\) \(=16 \times 10^{-4} \mathrm{kgm}^{2}\) \(=1.6 \times 10^{-3} \mathrm{kgm}^{2}\) Angular displacement \(=2 \pi \times\) revolution \(=2 \pi \times 2 \pi\) \(=4 \pi^{2}\) From conservation of energy theorem, Change in rotational kinetic energy \(=\) torque \(\times\) angular displacement So, \(\quad \frac{1}{2} \mathrm{I} \omega^{2}=\) torque \(\times 4 \pi^{2}\) \(\frac{1}{2} \times 1.6 \times 10^{-3} \times\left(\frac{\pi}{10}\right)^{2}=\tau \times 4 \pi^{2}\) \(\tau=\frac{\frac{1}{2} \times 1.6 \times 10^{-3} \times\left(\frac{\pi}{10}\right)^{2}}{4 \pi^{2}}\) \(\text { torque }(\tau)=2 \times 10^{-6} \mathrm{~N}-\mathrm{m}\)
NEET (National) - 2019
Rotational Motion
150155
When a 12000 joule of work is done on a flywheel, its frequency of rotation increases from \(10 \mathrm{~Hz}\) to \(20 \mathrm{~Hz}\). The moment of inertia of flywheel about its axis of rotation is \(\left(\boldsymbol{\pi}^{2}=10\right)\)
1 \(1.688 \mathrm{kgm}^{2}\)
2 \(2 \mathrm{kgm}_{2}^{2}\)
3 \(1.5 \mathrm{kgm}\)
4 \(1 \mathrm{kgm}^2\)
Explanation:
B Given, work done on a flywheel (W) \(=12000\) Joule, initial frequency \(\left(f_{1}\right)=10 \mathrm{~Hz}\), final frequency \(\left(f_{2}\right)\) \(=20 \mathrm{~Hz}\) \(\therefore\) Angular velocity for rotational motion- \(\omega_{1}=2 \pi f_{1}=2 \pi \times 10=20 \pi\) And \(\omega_{2}=2 \pi \times 20=40 \pi\) Acording to work energy theorem- Work done on a flywheel \(=\) Change in rotational kinetic energy \(\mathrm{W}=\frac{1}{2} \mathrm{I}\left(\omega_{2}^{2}-\omega_{1}^{2}\right)\) Where, \(\mathrm{I}=\) Moment of inertia \(12000=\frac{1}{2} \mathrm{I}\left[(40 \pi)^{2}-(20 \pi)^{2}\right]\) \(\therefore \quad 12000=\frac{1}{2} \mathrm{I}(1600-400) \pi^{2}\) \(\mathrm{I}=\frac{12000\times 2}{12000} \quad\left[\because \pi^{2}=10\right]\) \(\mathrm{I}=2 \mathrm{kgm}^{2}\)
150151
The moment of inertia of a rectangular plate of mass \(M\), length \(L\) and breadth \(B\), about an axis passing through its centre and perpendicular to its plane is
C Rectangular EFGH of length (L), breadth (B) and mass (M), Now, differentiate the area, \(\mathrm{dA}=\mathrm{B} \cdot \mathrm{dx} \quad\left(\because \sigma=\frac{\mathrm{M}}{\mathrm{A}}\right)\) \(\sigma=\frac{\mathrm{M}}{\mathrm{L} \times \mathrm{B}} \quad \text { Where, } \sigma=\text { Mass per unit area }\) So, \(\quad \mathrm{dM}=\sigma \mathrm{dA}=\frac{\text { M.B.dx }}{\text { L.B }}\) Now, integrating the values, \(=\frac{M \cdot d x}{L}\) \(\int d I_{x}=\int_{-L / 2}^{L / 2} d M \times x^{2}\) \(I_{x}=\frac{M}{L} \int_{-L / 2}^{L / 2} x^{2} d x\) \(I_{x}=\frac{M}{L}\left[\frac{x^{3}}{3}\right]_{-L / 2}^{L / 2}\) \(I_{x}=\frac{M}{L}\left[\frac{L^{3}}{24}+\frac{L^{3}}{24}\right]=\frac{M}{L} \times \frac{L^{3}}{12}\) \(I_{x}=\frac{M L^{2}}{12}\) Same as for, \(I_{y}=\frac{M^{2}}{12}\) Using perpendicular axis theorem, \(I_{z}=I_{x}+I_{y}\) \(I_{z}=\frac{M L^{2}}{12}+\frac{M B^{2}}{12}=\frac{M\left(L^{2}+B^{2}\right)}{12}\) For a rectangular rod, \(\mathrm{I}=\Sigma \mathrm{M}\left(\frac{\mathrm{L}^{2}+\mathrm{B}^{2}}{12}\right)\) \(\therefore \quad \mathrm{I} =\mathrm{M}\left(\frac{\mathrm{L}^{2}+\mathrm{B}^{2}}{12}\right)\)
AP EAMCET (21.09.2020) Shift-I
Rotational Motion
150152
Three identical uniform solid spheres each of mass \(m\) and radius \(r\) are joined as shown in the figure, with centres lying in the same plane. The moment of inertia of the system about an axis lying in that plane and passing through the centre of sphere \(C\) is
1 \(\frac{16}{5} m r^{2}\)
2 \(\frac{12}{5} m r^{2}\)
3 \(4 m r^{2}\)
4 \(\frac{3}{5} m r^{2}\)
Explanation:
A Given, Mass of solid sphere \(=\mathrm{m}\) Radius of solid sphere \(=r\) We know that, \(\mathrm{I}_{\text {sphere }}=\frac{2}{5} \mathrm{mr}^{2}\) Moment of inertia of solid spheres passing through center- \(\mathrm{I}_{\mathrm{C}}=\frac{2}{5} \mathrm{mr}^{2}\) Moment of inertia of a solid spheres through its tangent is given by \(\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{B}}=\frac{2}{5} \mathrm{mr}^{2}+\mathrm{mr}^{2}\) \(\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{B}}=\frac{7}{5} \mathrm{mr}^{2}\) Net moment of inertia of the system- \(\mathrm{I}_{\mathrm{PQ}} =\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}\) \(\mathrm{I}_{\mathrm{PQ}} =\frac{7}{5} \mathrm{mr}^{2}+\frac{7}{5} \mathrm{mr}^{2}+\frac{2}{5} \mathrm{mr}^{2}\) \(\mathrm{I}_{\mathrm{PQ}} =\frac{16}{5} \mathrm{mr}^{2}\)
AP EAMCET (17.09.2020) Shift-I
Rotational Motion
150153
Four point masses, each of mass \(M\) are placed at the corners of a square of side \(L\). The moment of inertia of the system about one of its diagonals is
1 \(2 \mathrm{ML}\)
2 \(\mathrm{ML}^{2}\)
3 \(4 \mathrm{ML}^{2}\)
4 \(6 \mathrm{ML}^{2}\)
Explanation:
B Given, mass of each point \(=M\) Side of square \(=\mathrm{L}\) \(\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}=\mathrm{L}\) Length of diagonal, \(\mathrm{BD}=\sqrt{\mathrm{DA}^{2}+\mathrm{AB}^{2}}=\sqrt{\mathrm{L}^{2}+\mathrm{L}^{2}}\) \(\mathrm{BD}=\sqrt{2 \mathrm{~L}^{2}}\) \(\mathrm{BD}=\mathrm{L} \sqrt{2}\) \(\therefore \mathrm{OD}=\mathrm{OB}=\frac{\mathrm{BD}}{2}=\frac{\mathrm{L} \sqrt{2}}{2}=\frac{\mathrm{L}}{\sqrt{2}}\) Moment of inertia of given system about the diagonal (AC), \(\mathrm{I}_{\mathrm{AC}}=\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}}+\mathrm{I}_{\mathrm{C}}+\mathrm{I}_{\mathrm{D}}\) \(\mathrm{I}_{\mathrm{AC}}=0+\mathrm{M}\left(\frac{\mathrm{L}}{\sqrt{2}}\right)^{2}+0+\mathrm{M}\left(\frac{\mathrm{L}}{\sqrt{2}}\right)^{2}\) \(\mathrm{I}_{\mathrm{AC}}=\frac{\mathrm{ML}^{2}}{2}+\frac{\mathrm{ML}^{2}}{2}\) \(\mathrm{I}_{\mathrm{AC}}=\mathrm{ML}^{2}\)
AP EAMCET (17.09.2020) Shift-I
Rotational Motion
150154
A solid cylinder of mass \(2 \mathrm{~kg}\) and radius \(4 \mathrm{~cm}\) is rotating about its axis at the rate of \(3 \mathrm{rpm}\). The torque required to stop after \(2 \pi\) revolutions is
1 \(2 \times 10^{-3} \mathrm{~N}-\mathrm{m}\)
2 \(12 \times 10^{-4} \mathrm{~N}-\mathrm{m}\)
3 \(2 \times 10^{6} \mathrm{~N}-\mathrm{m}\)
4 \(2 \times 10^{-6} \mathrm{~N}-\mathrm{m}\)
Explanation:
D Given that, Mass \(\mathrm({m})=2 \mathrm{~kg}\) Radius \(\mathrm({r})=4 \mathrm{~cm} \Rightarrow 0.04 \mathrm{~m}\) Rate of rotation \(\omega)=3 \mathrm{rpm}=3 \times \frac{2 \pi}{60} \Rightarrow \frac{\pi}{10}\) Moment of inertia \(=\frac{1}{2} \mathrm{mr}^{2}\) \(=\frac{1}{2} \times 2 \times(0.04)^{2}\) \(=16 \times 10^{-4} \mathrm{kgm}^{2}\) \(=1.6 \times 10^{-3} \mathrm{kgm}^{2}\) Angular displacement \(=2 \pi \times\) revolution \(=2 \pi \times 2 \pi\) \(=4 \pi^{2}\) From conservation of energy theorem, Change in rotational kinetic energy \(=\) torque \(\times\) angular displacement So, \(\quad \frac{1}{2} \mathrm{I} \omega^{2}=\) torque \(\times 4 \pi^{2}\) \(\frac{1}{2} \times 1.6 \times 10^{-3} \times\left(\frac{\pi}{10}\right)^{2}=\tau \times 4 \pi^{2}\) \(\tau=\frac{\frac{1}{2} \times 1.6 \times 10^{-3} \times\left(\frac{\pi}{10}\right)^{2}}{4 \pi^{2}}\) \(\text { torque }(\tau)=2 \times 10^{-6} \mathrm{~N}-\mathrm{m}\)
NEET (National) - 2019
Rotational Motion
150155
When a 12000 joule of work is done on a flywheel, its frequency of rotation increases from \(10 \mathrm{~Hz}\) to \(20 \mathrm{~Hz}\). The moment of inertia of flywheel about its axis of rotation is \(\left(\boldsymbol{\pi}^{2}=10\right)\)
1 \(1.688 \mathrm{kgm}^{2}\)
2 \(2 \mathrm{kgm}_{2}^{2}\)
3 \(1.5 \mathrm{kgm}\)
4 \(1 \mathrm{kgm}^2\)
Explanation:
B Given, work done on a flywheel (W) \(=12000\) Joule, initial frequency \(\left(f_{1}\right)=10 \mathrm{~Hz}\), final frequency \(\left(f_{2}\right)\) \(=20 \mathrm{~Hz}\) \(\therefore\) Angular velocity for rotational motion- \(\omega_{1}=2 \pi f_{1}=2 \pi \times 10=20 \pi\) And \(\omega_{2}=2 \pi \times 20=40 \pi\) Acording to work energy theorem- Work done on a flywheel \(=\) Change in rotational kinetic energy \(\mathrm{W}=\frac{1}{2} \mathrm{I}\left(\omega_{2}^{2}-\omega_{1}^{2}\right)\) Where, \(\mathrm{I}=\) Moment of inertia \(12000=\frac{1}{2} \mathrm{I}\left[(40 \pi)^{2}-(20 \pi)^{2}\right]\) \(\therefore \quad 12000=\frac{1}{2} \mathrm{I}(1600-400) \pi^{2}\) \(\mathrm{I}=\frac{12000\times 2}{12000} \quad\left[\because \pi^{2}=10\right]\) \(\mathrm{I}=2 \mathrm{kgm}^{2}\)
