150391
A small object of uniform density rolls up a curved surface with an initial velocity \(v\). It reaches up to a maximum height of \(\frac{3 v^{2}}{4 g}\) with respect to the initial position. The object is
1 ring
2 solid sphere
3 hollow sphere
4 disc
Explanation:
D Given that, Initial velocity \(=\mathrm{v}\) Height \(_{(\max )}=\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\) If any object rolling on the curved surface then the kinetic energy of object is changes into potential energy. So, by the law of conservation of mechanical energy, Kinetic energy + Rotational kinetic energy \(=\) Potential energy \(\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}=m g h\) \(\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I}\left(\frac{\mathrm{v}}{\mathrm{r}}\right)^2=\mathrm{mg}\left(\frac{3 \mathrm{v}^2}{4 \mathrm{~g}}\right) \quad\left(\begin{array}{ll}\because & \omega=\mathrm{v} / \mathrm{r} \\ \mathrm{h}=\frac{3 \mathrm{v}^2}{4 \mathrm{~g}}\end{array}\right)\) \(\frac{1}{2} I\left(\frac{v^{2}}{r^{2}}\right)=\frac{3 m v^{2}}{4}-\frac{1}{2} m v^{2}\) \(\frac{1}{2} I\left(\frac{v^{2}}{r^{2}}\right) =\frac{3 m^{2}-2 m v^{2}}{4}\) \(\frac{1}{2} I \frac{v^{2}}{r^{2}} =\frac{1}{4} m v^{2}\) \(I =\frac{1}{2} m^{2}\) Hence, object is a disc.
MHT-CET 2013
Rotational Motion
150392
The moment of inertia of two freely rotating bodies \(A\) and \(B\) are \(I_{A}\) and \(I_{B}\), respectively. \(I_{A}>I_{B}\) and their angular moment are equal. If \(K_{A}\) and \(K_{B}\) are their kinetic energies, then
C We know that, The relation between kinetic energy, angular momentum, and moment of intia is \(\mathrm{K}=\frac{\mathrm{L}^{2}}{2 \mathrm{I}}\) The angular momentum is the same for both the bodies, we can write Then, \(\quad \mathrm{K} \propto \frac{1}{\mathrm{I}}\) Kinetic Energy is inversely proportional to the moment of inertia. If \(I_{A}>I_{B}\) Then \(\mathrm{K}_{\mathrm{A}} \lt \mathrm{K}_{\mathrm{B}}\)
MHT-CET 2009
Rotational Motion
150393
A disc of moment of inertia \(\frac{9.8}{\pi^{2}} \mathrm{~kg}-\mathrm{m}^{2}\) is rotating at \(600 \mathrm{rpm}\). If the frequency of rotation changes from \(600 \mathrm{rpm}\) to \(300 \mathrm{rpm}\), then what is the work done ?
1 \(1467 \mathrm{~J}\)
2 \(1452 \mathrm{~J}\)
3 \(1567 \mathrm{~J}\)
4 \(1632 \mathrm{~J}\)
Explanation:
A Using work energy theorem, Work done \(=\) change in rotational kinetic energy Work done \(=\Delta \mathrm{KE}_{\text {rotational }}=\frac{1}{2} \mathrm{I} \omega_{\mathrm{i}}^{2}-\frac{1}{2} \mathrm{I} \omega_{\mathrm{f}}^{2}\) \(\because \quad \omega_{\mathrm{i}}=600 \times \frac{2 \pi}{60}=20 \pi \mathrm{rad} / \mathrm{s}\) Similarly \(\omega_{\mathrm{f}}=10 \pi \mathrm{rad} / \mathrm{s}\) \(I=\frac{9.8}{\pi^{2}} \mathrm{~kg}-\mathrm{m}^{2}\) \(\therefore \quad\) Work Done \(=\frac{1}{2} \mathrm{I}\left[(20 \pi)^{2}-(10 \pi)^{2}\right]\) \(=\frac{1}{2} \times \frac{9.8}{\pi^{2}} \times\left(300 \pi^{2}\right)=1470 \mathrm{~J} \simeq 1467 \mathrm{~J}\) \(=1467 \mathrm{~J}\)
MHT-CET 2004
Rotational Motion
150394
A solid sphere of mass \(M\) and radius \(2 R\) rolls down an inclined plane of height \(h\) without slipping. The speed of its centre of mass when it reaches the bottom is
1 \(\sqrt{\frac{6}{7} \mathrm{gh}}\)
2 \(\sqrt{3 \mathrm{gh}}\)
3 \(\sqrt{\frac{10}{7} \mathrm{gh}}\)
4 \(\sqrt{\frac{4}{3} \mathrm{gh}}\)
Explanation:
C Potential energy of solid Cylinder of mass M at height \(\mathrm{h})=\mathrm{Mgh}\) When the solid sphere rolls down on an inclined plane then it has both rotational and translational kinetic energy. \(\mathrm{K}=\mathrm{K}_{\text {rot }}+\mathrm{K}_{\text {trans }}\) \(\mathrm{K}=\frac{1}{2} \mathrm{I} \omega^{2}+\frac{1}{2} \mathrm{MV}^{2} \quad(\mathrm{~V}=\mathrm{R} \omega\) \(=\frac{1}{2} \times \frac{2}{5} \mathrm{M}(2 \mathrm{R})^{2}\left(\frac{\mathrm{V}}{2 \mathrm{R}}\right)^{2}+\frac{1}{2} \mathrm{MV}^{2} \quad\left(\begin{array}{l}\mathrm{I}=\frac{1}{2} \mathrm{MR}^{2} \\ \text { for solid sphere }\end{array}\right)\) \(=\frac{4}{5} \times \mathrm{MR}^{2} \frac{\mathrm{V}^{2}}{4 \mathrm{R}^{2}}+\frac{1}{2} \mathrm{MV}^{2}\) \(=\frac{1}{5} \mathrm{MV}^{2}+\frac{1}{2} \mathrm{MV}^{2}=\frac{7}{10} \mathrm{MV}^{2}\) Now gain of kinetic energy \(=\) loss in potential energy \(\frac{7}{10} \mathrm{MV}^{2}=\mathrm{Mgh}\) \(\mathrm{V}=\sqrt{\frac{10}{7} \mathrm{gh}}\) (g) Rolling Motion}
JIPMER-2016
Rotational Motion
150395
A small object of uniform density rolls up a curved surface with an initial velocity \(v\). It reaches up to a maximum height of \(\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\) with respect to the initial position. The object is:
1 ring
2 solid sphere
3 hollow sphere
4 disc
Explanation:
D From conservation of energy, \(\text { K.E. })_{\mathrm{t}}+(\text { K.E. })_{\mathrm{r}}=\text { P.E. }\) \(\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=\mathrm{mgh}\) \(\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=\operatorname{mg}\left(\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\right) \quad\left(\because \mathrm{h}=\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\right).\), given \(\) \(\frac{1}{2} m v^{2}+\frac{1}{2} I \times \frac{v^{2}}{r^{2}}=m g\left(\frac{3 v^{2}}{4 g}\right)\) \(\frac{\mathrm{m}}{2}+\frac{\mathrm{I}}{2 \mathrm{r}^{2}}=\frac{3 \mathrm{~m}}{4}\) \(\frac{\mathrm{I}}{2 \mathrm{r}^{2}} =\frac{3}{4} \mathrm{~m}-\frac{1}{2} \mathrm{~m}\) \(\frac{\mathrm{I}}{2 \mathrm{r}^{2}} =\frac{\mathrm{m}}{4}\) \(\mathrm{I} =\frac{\mathrm{mr}^{2}}{2}\) Hence, the object must be disc.
150391
A small object of uniform density rolls up a curved surface with an initial velocity \(v\). It reaches up to a maximum height of \(\frac{3 v^{2}}{4 g}\) with respect to the initial position. The object is
1 ring
2 solid sphere
3 hollow sphere
4 disc
Explanation:
D Given that, Initial velocity \(=\mathrm{v}\) Height \(_{(\max )}=\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\) If any object rolling on the curved surface then the kinetic energy of object is changes into potential energy. So, by the law of conservation of mechanical energy, Kinetic energy + Rotational kinetic energy \(=\) Potential energy \(\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}=m g h\) \(\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I}\left(\frac{\mathrm{v}}{\mathrm{r}}\right)^2=\mathrm{mg}\left(\frac{3 \mathrm{v}^2}{4 \mathrm{~g}}\right) \quad\left(\begin{array}{ll}\because & \omega=\mathrm{v} / \mathrm{r} \\ \mathrm{h}=\frac{3 \mathrm{v}^2}{4 \mathrm{~g}}\end{array}\right)\) \(\frac{1}{2} I\left(\frac{v^{2}}{r^{2}}\right)=\frac{3 m v^{2}}{4}-\frac{1}{2} m v^{2}\) \(\frac{1}{2} I\left(\frac{v^{2}}{r^{2}}\right) =\frac{3 m^{2}-2 m v^{2}}{4}\) \(\frac{1}{2} I \frac{v^{2}}{r^{2}} =\frac{1}{4} m v^{2}\) \(I =\frac{1}{2} m^{2}\) Hence, object is a disc.
MHT-CET 2013
Rotational Motion
150392
The moment of inertia of two freely rotating bodies \(A\) and \(B\) are \(I_{A}\) and \(I_{B}\), respectively. \(I_{A}>I_{B}\) and their angular moment are equal. If \(K_{A}\) and \(K_{B}\) are their kinetic energies, then
C We know that, The relation between kinetic energy, angular momentum, and moment of intia is \(\mathrm{K}=\frac{\mathrm{L}^{2}}{2 \mathrm{I}}\) The angular momentum is the same for both the bodies, we can write Then, \(\quad \mathrm{K} \propto \frac{1}{\mathrm{I}}\) Kinetic Energy is inversely proportional to the moment of inertia. If \(I_{A}>I_{B}\) Then \(\mathrm{K}_{\mathrm{A}} \lt \mathrm{K}_{\mathrm{B}}\)
MHT-CET 2009
Rotational Motion
150393
A disc of moment of inertia \(\frac{9.8}{\pi^{2}} \mathrm{~kg}-\mathrm{m}^{2}\) is rotating at \(600 \mathrm{rpm}\). If the frequency of rotation changes from \(600 \mathrm{rpm}\) to \(300 \mathrm{rpm}\), then what is the work done ?
1 \(1467 \mathrm{~J}\)
2 \(1452 \mathrm{~J}\)
3 \(1567 \mathrm{~J}\)
4 \(1632 \mathrm{~J}\)
Explanation:
A Using work energy theorem, Work done \(=\) change in rotational kinetic energy Work done \(=\Delta \mathrm{KE}_{\text {rotational }}=\frac{1}{2} \mathrm{I} \omega_{\mathrm{i}}^{2}-\frac{1}{2} \mathrm{I} \omega_{\mathrm{f}}^{2}\) \(\because \quad \omega_{\mathrm{i}}=600 \times \frac{2 \pi}{60}=20 \pi \mathrm{rad} / \mathrm{s}\) Similarly \(\omega_{\mathrm{f}}=10 \pi \mathrm{rad} / \mathrm{s}\) \(I=\frac{9.8}{\pi^{2}} \mathrm{~kg}-\mathrm{m}^{2}\) \(\therefore \quad\) Work Done \(=\frac{1}{2} \mathrm{I}\left[(20 \pi)^{2}-(10 \pi)^{2}\right]\) \(=\frac{1}{2} \times \frac{9.8}{\pi^{2}} \times\left(300 \pi^{2}\right)=1470 \mathrm{~J} \simeq 1467 \mathrm{~J}\) \(=1467 \mathrm{~J}\)
MHT-CET 2004
Rotational Motion
150394
A solid sphere of mass \(M\) and radius \(2 R\) rolls down an inclined plane of height \(h\) without slipping. The speed of its centre of mass when it reaches the bottom is
1 \(\sqrt{\frac{6}{7} \mathrm{gh}}\)
2 \(\sqrt{3 \mathrm{gh}}\)
3 \(\sqrt{\frac{10}{7} \mathrm{gh}}\)
4 \(\sqrt{\frac{4}{3} \mathrm{gh}}\)
Explanation:
C Potential energy of solid Cylinder of mass M at height \(\mathrm{h})=\mathrm{Mgh}\) When the solid sphere rolls down on an inclined plane then it has both rotational and translational kinetic energy. \(\mathrm{K}=\mathrm{K}_{\text {rot }}+\mathrm{K}_{\text {trans }}\) \(\mathrm{K}=\frac{1}{2} \mathrm{I} \omega^{2}+\frac{1}{2} \mathrm{MV}^{2} \quad(\mathrm{~V}=\mathrm{R} \omega\) \(=\frac{1}{2} \times \frac{2}{5} \mathrm{M}(2 \mathrm{R})^{2}\left(\frac{\mathrm{V}}{2 \mathrm{R}}\right)^{2}+\frac{1}{2} \mathrm{MV}^{2} \quad\left(\begin{array}{l}\mathrm{I}=\frac{1}{2} \mathrm{MR}^{2} \\ \text { for solid sphere }\end{array}\right)\) \(=\frac{4}{5} \times \mathrm{MR}^{2} \frac{\mathrm{V}^{2}}{4 \mathrm{R}^{2}}+\frac{1}{2} \mathrm{MV}^{2}\) \(=\frac{1}{5} \mathrm{MV}^{2}+\frac{1}{2} \mathrm{MV}^{2}=\frac{7}{10} \mathrm{MV}^{2}\) Now gain of kinetic energy \(=\) loss in potential energy \(\frac{7}{10} \mathrm{MV}^{2}=\mathrm{Mgh}\) \(\mathrm{V}=\sqrt{\frac{10}{7} \mathrm{gh}}\) (g) Rolling Motion}
JIPMER-2016
Rotational Motion
150395
A small object of uniform density rolls up a curved surface with an initial velocity \(v\). It reaches up to a maximum height of \(\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\) with respect to the initial position. The object is:
1 ring
2 solid sphere
3 hollow sphere
4 disc
Explanation:
D From conservation of energy, \(\text { K.E. })_{\mathrm{t}}+(\text { K.E. })_{\mathrm{r}}=\text { P.E. }\) \(\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=\mathrm{mgh}\) \(\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=\operatorname{mg}\left(\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\right) \quad\left(\because \mathrm{h}=\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\right).\), given \(\) \(\frac{1}{2} m v^{2}+\frac{1}{2} I \times \frac{v^{2}}{r^{2}}=m g\left(\frac{3 v^{2}}{4 g}\right)\) \(\frac{\mathrm{m}}{2}+\frac{\mathrm{I}}{2 \mathrm{r}^{2}}=\frac{3 \mathrm{~m}}{4}\) \(\frac{\mathrm{I}}{2 \mathrm{r}^{2}} =\frac{3}{4} \mathrm{~m}-\frac{1}{2} \mathrm{~m}\) \(\frac{\mathrm{I}}{2 \mathrm{r}^{2}} =\frac{\mathrm{m}}{4}\) \(\mathrm{I} =\frac{\mathrm{mr}^{2}}{2}\) Hence, the object must be disc.
150391
A small object of uniform density rolls up a curved surface with an initial velocity \(v\). It reaches up to a maximum height of \(\frac{3 v^{2}}{4 g}\) with respect to the initial position. The object is
1 ring
2 solid sphere
3 hollow sphere
4 disc
Explanation:
D Given that, Initial velocity \(=\mathrm{v}\) Height \(_{(\max )}=\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\) If any object rolling on the curved surface then the kinetic energy of object is changes into potential energy. So, by the law of conservation of mechanical energy, Kinetic energy + Rotational kinetic energy \(=\) Potential energy \(\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}=m g h\) \(\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I}\left(\frac{\mathrm{v}}{\mathrm{r}}\right)^2=\mathrm{mg}\left(\frac{3 \mathrm{v}^2}{4 \mathrm{~g}}\right) \quad\left(\begin{array}{ll}\because & \omega=\mathrm{v} / \mathrm{r} \\ \mathrm{h}=\frac{3 \mathrm{v}^2}{4 \mathrm{~g}}\end{array}\right)\) \(\frac{1}{2} I\left(\frac{v^{2}}{r^{2}}\right)=\frac{3 m v^{2}}{4}-\frac{1}{2} m v^{2}\) \(\frac{1}{2} I\left(\frac{v^{2}}{r^{2}}\right) =\frac{3 m^{2}-2 m v^{2}}{4}\) \(\frac{1}{2} I \frac{v^{2}}{r^{2}} =\frac{1}{4} m v^{2}\) \(I =\frac{1}{2} m^{2}\) Hence, object is a disc.
MHT-CET 2013
Rotational Motion
150392
The moment of inertia of two freely rotating bodies \(A\) and \(B\) are \(I_{A}\) and \(I_{B}\), respectively. \(I_{A}>I_{B}\) and their angular moment are equal. If \(K_{A}\) and \(K_{B}\) are their kinetic energies, then
C We know that, The relation between kinetic energy, angular momentum, and moment of intia is \(\mathrm{K}=\frac{\mathrm{L}^{2}}{2 \mathrm{I}}\) The angular momentum is the same for both the bodies, we can write Then, \(\quad \mathrm{K} \propto \frac{1}{\mathrm{I}}\) Kinetic Energy is inversely proportional to the moment of inertia. If \(I_{A}>I_{B}\) Then \(\mathrm{K}_{\mathrm{A}} \lt \mathrm{K}_{\mathrm{B}}\)
MHT-CET 2009
Rotational Motion
150393
A disc of moment of inertia \(\frac{9.8}{\pi^{2}} \mathrm{~kg}-\mathrm{m}^{2}\) is rotating at \(600 \mathrm{rpm}\). If the frequency of rotation changes from \(600 \mathrm{rpm}\) to \(300 \mathrm{rpm}\), then what is the work done ?
1 \(1467 \mathrm{~J}\)
2 \(1452 \mathrm{~J}\)
3 \(1567 \mathrm{~J}\)
4 \(1632 \mathrm{~J}\)
Explanation:
A Using work energy theorem, Work done \(=\) change in rotational kinetic energy Work done \(=\Delta \mathrm{KE}_{\text {rotational }}=\frac{1}{2} \mathrm{I} \omega_{\mathrm{i}}^{2}-\frac{1}{2} \mathrm{I} \omega_{\mathrm{f}}^{2}\) \(\because \quad \omega_{\mathrm{i}}=600 \times \frac{2 \pi}{60}=20 \pi \mathrm{rad} / \mathrm{s}\) Similarly \(\omega_{\mathrm{f}}=10 \pi \mathrm{rad} / \mathrm{s}\) \(I=\frac{9.8}{\pi^{2}} \mathrm{~kg}-\mathrm{m}^{2}\) \(\therefore \quad\) Work Done \(=\frac{1}{2} \mathrm{I}\left[(20 \pi)^{2}-(10 \pi)^{2}\right]\) \(=\frac{1}{2} \times \frac{9.8}{\pi^{2}} \times\left(300 \pi^{2}\right)=1470 \mathrm{~J} \simeq 1467 \mathrm{~J}\) \(=1467 \mathrm{~J}\)
MHT-CET 2004
Rotational Motion
150394
A solid sphere of mass \(M\) and radius \(2 R\) rolls down an inclined plane of height \(h\) without slipping. The speed of its centre of mass when it reaches the bottom is
1 \(\sqrt{\frac{6}{7} \mathrm{gh}}\)
2 \(\sqrt{3 \mathrm{gh}}\)
3 \(\sqrt{\frac{10}{7} \mathrm{gh}}\)
4 \(\sqrt{\frac{4}{3} \mathrm{gh}}\)
Explanation:
C Potential energy of solid Cylinder of mass M at height \(\mathrm{h})=\mathrm{Mgh}\) When the solid sphere rolls down on an inclined plane then it has both rotational and translational kinetic energy. \(\mathrm{K}=\mathrm{K}_{\text {rot }}+\mathrm{K}_{\text {trans }}\) \(\mathrm{K}=\frac{1}{2} \mathrm{I} \omega^{2}+\frac{1}{2} \mathrm{MV}^{2} \quad(\mathrm{~V}=\mathrm{R} \omega\) \(=\frac{1}{2} \times \frac{2}{5} \mathrm{M}(2 \mathrm{R})^{2}\left(\frac{\mathrm{V}}{2 \mathrm{R}}\right)^{2}+\frac{1}{2} \mathrm{MV}^{2} \quad\left(\begin{array}{l}\mathrm{I}=\frac{1}{2} \mathrm{MR}^{2} \\ \text { for solid sphere }\end{array}\right)\) \(=\frac{4}{5} \times \mathrm{MR}^{2} \frac{\mathrm{V}^{2}}{4 \mathrm{R}^{2}}+\frac{1}{2} \mathrm{MV}^{2}\) \(=\frac{1}{5} \mathrm{MV}^{2}+\frac{1}{2} \mathrm{MV}^{2}=\frac{7}{10} \mathrm{MV}^{2}\) Now gain of kinetic energy \(=\) loss in potential energy \(\frac{7}{10} \mathrm{MV}^{2}=\mathrm{Mgh}\) \(\mathrm{V}=\sqrt{\frac{10}{7} \mathrm{gh}}\) (g) Rolling Motion}
JIPMER-2016
Rotational Motion
150395
A small object of uniform density rolls up a curved surface with an initial velocity \(v\). It reaches up to a maximum height of \(\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\) with respect to the initial position. The object is:
1 ring
2 solid sphere
3 hollow sphere
4 disc
Explanation:
D From conservation of energy, \(\text { K.E. })_{\mathrm{t}}+(\text { K.E. })_{\mathrm{r}}=\text { P.E. }\) \(\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=\mathrm{mgh}\) \(\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=\operatorname{mg}\left(\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\right) \quad\left(\because \mathrm{h}=\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\right).\), given \(\) \(\frac{1}{2} m v^{2}+\frac{1}{2} I \times \frac{v^{2}}{r^{2}}=m g\left(\frac{3 v^{2}}{4 g}\right)\) \(\frac{\mathrm{m}}{2}+\frac{\mathrm{I}}{2 \mathrm{r}^{2}}=\frac{3 \mathrm{~m}}{4}\) \(\frac{\mathrm{I}}{2 \mathrm{r}^{2}} =\frac{3}{4} \mathrm{~m}-\frac{1}{2} \mathrm{~m}\) \(\frac{\mathrm{I}}{2 \mathrm{r}^{2}} =\frac{\mathrm{m}}{4}\) \(\mathrm{I} =\frac{\mathrm{mr}^{2}}{2}\) Hence, the object must be disc.
150391
A small object of uniform density rolls up a curved surface with an initial velocity \(v\). It reaches up to a maximum height of \(\frac{3 v^{2}}{4 g}\) with respect to the initial position. The object is
1 ring
2 solid sphere
3 hollow sphere
4 disc
Explanation:
D Given that, Initial velocity \(=\mathrm{v}\) Height \(_{(\max )}=\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\) If any object rolling on the curved surface then the kinetic energy of object is changes into potential energy. So, by the law of conservation of mechanical energy, Kinetic energy + Rotational kinetic energy \(=\) Potential energy \(\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}=m g h\) \(\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I}\left(\frac{\mathrm{v}}{\mathrm{r}}\right)^2=\mathrm{mg}\left(\frac{3 \mathrm{v}^2}{4 \mathrm{~g}}\right) \quad\left(\begin{array}{ll}\because & \omega=\mathrm{v} / \mathrm{r} \\ \mathrm{h}=\frac{3 \mathrm{v}^2}{4 \mathrm{~g}}\end{array}\right)\) \(\frac{1}{2} I\left(\frac{v^{2}}{r^{2}}\right)=\frac{3 m v^{2}}{4}-\frac{1}{2} m v^{2}\) \(\frac{1}{2} I\left(\frac{v^{2}}{r^{2}}\right) =\frac{3 m^{2}-2 m v^{2}}{4}\) \(\frac{1}{2} I \frac{v^{2}}{r^{2}} =\frac{1}{4} m v^{2}\) \(I =\frac{1}{2} m^{2}\) Hence, object is a disc.
MHT-CET 2013
Rotational Motion
150392
The moment of inertia of two freely rotating bodies \(A\) and \(B\) are \(I_{A}\) and \(I_{B}\), respectively. \(I_{A}>I_{B}\) and their angular moment are equal. If \(K_{A}\) and \(K_{B}\) are their kinetic energies, then
C We know that, The relation between kinetic energy, angular momentum, and moment of intia is \(\mathrm{K}=\frac{\mathrm{L}^{2}}{2 \mathrm{I}}\) The angular momentum is the same for both the bodies, we can write Then, \(\quad \mathrm{K} \propto \frac{1}{\mathrm{I}}\) Kinetic Energy is inversely proportional to the moment of inertia. If \(I_{A}>I_{B}\) Then \(\mathrm{K}_{\mathrm{A}} \lt \mathrm{K}_{\mathrm{B}}\)
MHT-CET 2009
Rotational Motion
150393
A disc of moment of inertia \(\frac{9.8}{\pi^{2}} \mathrm{~kg}-\mathrm{m}^{2}\) is rotating at \(600 \mathrm{rpm}\). If the frequency of rotation changes from \(600 \mathrm{rpm}\) to \(300 \mathrm{rpm}\), then what is the work done ?
1 \(1467 \mathrm{~J}\)
2 \(1452 \mathrm{~J}\)
3 \(1567 \mathrm{~J}\)
4 \(1632 \mathrm{~J}\)
Explanation:
A Using work energy theorem, Work done \(=\) change in rotational kinetic energy Work done \(=\Delta \mathrm{KE}_{\text {rotational }}=\frac{1}{2} \mathrm{I} \omega_{\mathrm{i}}^{2}-\frac{1}{2} \mathrm{I} \omega_{\mathrm{f}}^{2}\) \(\because \quad \omega_{\mathrm{i}}=600 \times \frac{2 \pi}{60}=20 \pi \mathrm{rad} / \mathrm{s}\) Similarly \(\omega_{\mathrm{f}}=10 \pi \mathrm{rad} / \mathrm{s}\) \(I=\frac{9.8}{\pi^{2}} \mathrm{~kg}-\mathrm{m}^{2}\) \(\therefore \quad\) Work Done \(=\frac{1}{2} \mathrm{I}\left[(20 \pi)^{2}-(10 \pi)^{2}\right]\) \(=\frac{1}{2} \times \frac{9.8}{\pi^{2}} \times\left(300 \pi^{2}\right)=1470 \mathrm{~J} \simeq 1467 \mathrm{~J}\) \(=1467 \mathrm{~J}\)
MHT-CET 2004
Rotational Motion
150394
A solid sphere of mass \(M\) and radius \(2 R\) rolls down an inclined plane of height \(h\) without slipping. The speed of its centre of mass when it reaches the bottom is
1 \(\sqrt{\frac{6}{7} \mathrm{gh}}\)
2 \(\sqrt{3 \mathrm{gh}}\)
3 \(\sqrt{\frac{10}{7} \mathrm{gh}}\)
4 \(\sqrt{\frac{4}{3} \mathrm{gh}}\)
Explanation:
C Potential energy of solid Cylinder of mass M at height \(\mathrm{h})=\mathrm{Mgh}\) When the solid sphere rolls down on an inclined plane then it has both rotational and translational kinetic energy. \(\mathrm{K}=\mathrm{K}_{\text {rot }}+\mathrm{K}_{\text {trans }}\) \(\mathrm{K}=\frac{1}{2} \mathrm{I} \omega^{2}+\frac{1}{2} \mathrm{MV}^{2} \quad(\mathrm{~V}=\mathrm{R} \omega\) \(=\frac{1}{2} \times \frac{2}{5} \mathrm{M}(2 \mathrm{R})^{2}\left(\frac{\mathrm{V}}{2 \mathrm{R}}\right)^{2}+\frac{1}{2} \mathrm{MV}^{2} \quad\left(\begin{array}{l}\mathrm{I}=\frac{1}{2} \mathrm{MR}^{2} \\ \text { for solid sphere }\end{array}\right)\) \(=\frac{4}{5} \times \mathrm{MR}^{2} \frac{\mathrm{V}^{2}}{4 \mathrm{R}^{2}}+\frac{1}{2} \mathrm{MV}^{2}\) \(=\frac{1}{5} \mathrm{MV}^{2}+\frac{1}{2} \mathrm{MV}^{2}=\frac{7}{10} \mathrm{MV}^{2}\) Now gain of kinetic energy \(=\) loss in potential energy \(\frac{7}{10} \mathrm{MV}^{2}=\mathrm{Mgh}\) \(\mathrm{V}=\sqrt{\frac{10}{7} \mathrm{gh}}\) (g) Rolling Motion}
JIPMER-2016
Rotational Motion
150395
A small object of uniform density rolls up a curved surface with an initial velocity \(v\). It reaches up to a maximum height of \(\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\) with respect to the initial position. The object is:
1 ring
2 solid sphere
3 hollow sphere
4 disc
Explanation:
D From conservation of energy, \(\text { K.E. })_{\mathrm{t}}+(\text { K.E. })_{\mathrm{r}}=\text { P.E. }\) \(\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=\mathrm{mgh}\) \(\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=\operatorname{mg}\left(\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\right) \quad\left(\because \mathrm{h}=\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\right).\), given \(\) \(\frac{1}{2} m v^{2}+\frac{1}{2} I \times \frac{v^{2}}{r^{2}}=m g\left(\frac{3 v^{2}}{4 g}\right)\) \(\frac{\mathrm{m}}{2}+\frac{\mathrm{I}}{2 \mathrm{r}^{2}}=\frac{3 \mathrm{~m}}{4}\) \(\frac{\mathrm{I}}{2 \mathrm{r}^{2}} =\frac{3}{4} \mathrm{~m}-\frac{1}{2} \mathrm{~m}\) \(\frac{\mathrm{I}}{2 \mathrm{r}^{2}} =\frac{\mathrm{m}}{4}\) \(\mathrm{I} =\frac{\mathrm{mr}^{2}}{2}\) Hence, the object must be disc.
150391
A small object of uniform density rolls up a curved surface with an initial velocity \(v\). It reaches up to a maximum height of \(\frac{3 v^{2}}{4 g}\) with respect to the initial position. The object is
1 ring
2 solid sphere
3 hollow sphere
4 disc
Explanation:
D Given that, Initial velocity \(=\mathrm{v}\) Height \(_{(\max )}=\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\) If any object rolling on the curved surface then the kinetic energy of object is changes into potential energy. So, by the law of conservation of mechanical energy, Kinetic energy + Rotational kinetic energy \(=\) Potential energy \(\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}=m g h\) \(\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I}\left(\frac{\mathrm{v}}{\mathrm{r}}\right)^2=\mathrm{mg}\left(\frac{3 \mathrm{v}^2}{4 \mathrm{~g}}\right) \quad\left(\begin{array}{ll}\because & \omega=\mathrm{v} / \mathrm{r} \\ \mathrm{h}=\frac{3 \mathrm{v}^2}{4 \mathrm{~g}}\end{array}\right)\) \(\frac{1}{2} I\left(\frac{v^{2}}{r^{2}}\right)=\frac{3 m v^{2}}{4}-\frac{1}{2} m v^{2}\) \(\frac{1}{2} I\left(\frac{v^{2}}{r^{2}}\right) =\frac{3 m^{2}-2 m v^{2}}{4}\) \(\frac{1}{2} I \frac{v^{2}}{r^{2}} =\frac{1}{4} m v^{2}\) \(I =\frac{1}{2} m^{2}\) Hence, object is a disc.
MHT-CET 2013
Rotational Motion
150392
The moment of inertia of two freely rotating bodies \(A\) and \(B\) are \(I_{A}\) and \(I_{B}\), respectively. \(I_{A}>I_{B}\) and their angular moment are equal. If \(K_{A}\) and \(K_{B}\) are their kinetic energies, then
C We know that, The relation between kinetic energy, angular momentum, and moment of intia is \(\mathrm{K}=\frac{\mathrm{L}^{2}}{2 \mathrm{I}}\) The angular momentum is the same for both the bodies, we can write Then, \(\quad \mathrm{K} \propto \frac{1}{\mathrm{I}}\) Kinetic Energy is inversely proportional to the moment of inertia. If \(I_{A}>I_{B}\) Then \(\mathrm{K}_{\mathrm{A}} \lt \mathrm{K}_{\mathrm{B}}\)
MHT-CET 2009
Rotational Motion
150393
A disc of moment of inertia \(\frac{9.8}{\pi^{2}} \mathrm{~kg}-\mathrm{m}^{2}\) is rotating at \(600 \mathrm{rpm}\). If the frequency of rotation changes from \(600 \mathrm{rpm}\) to \(300 \mathrm{rpm}\), then what is the work done ?
1 \(1467 \mathrm{~J}\)
2 \(1452 \mathrm{~J}\)
3 \(1567 \mathrm{~J}\)
4 \(1632 \mathrm{~J}\)
Explanation:
A Using work energy theorem, Work done \(=\) change in rotational kinetic energy Work done \(=\Delta \mathrm{KE}_{\text {rotational }}=\frac{1}{2} \mathrm{I} \omega_{\mathrm{i}}^{2}-\frac{1}{2} \mathrm{I} \omega_{\mathrm{f}}^{2}\) \(\because \quad \omega_{\mathrm{i}}=600 \times \frac{2 \pi}{60}=20 \pi \mathrm{rad} / \mathrm{s}\) Similarly \(\omega_{\mathrm{f}}=10 \pi \mathrm{rad} / \mathrm{s}\) \(I=\frac{9.8}{\pi^{2}} \mathrm{~kg}-\mathrm{m}^{2}\) \(\therefore \quad\) Work Done \(=\frac{1}{2} \mathrm{I}\left[(20 \pi)^{2}-(10 \pi)^{2}\right]\) \(=\frac{1}{2} \times \frac{9.8}{\pi^{2}} \times\left(300 \pi^{2}\right)=1470 \mathrm{~J} \simeq 1467 \mathrm{~J}\) \(=1467 \mathrm{~J}\)
MHT-CET 2004
Rotational Motion
150394
A solid sphere of mass \(M\) and radius \(2 R\) rolls down an inclined plane of height \(h\) without slipping. The speed of its centre of mass when it reaches the bottom is
1 \(\sqrt{\frac{6}{7} \mathrm{gh}}\)
2 \(\sqrt{3 \mathrm{gh}}\)
3 \(\sqrt{\frac{10}{7} \mathrm{gh}}\)
4 \(\sqrt{\frac{4}{3} \mathrm{gh}}\)
Explanation:
C Potential energy of solid Cylinder of mass M at height \(\mathrm{h})=\mathrm{Mgh}\) When the solid sphere rolls down on an inclined plane then it has both rotational and translational kinetic energy. \(\mathrm{K}=\mathrm{K}_{\text {rot }}+\mathrm{K}_{\text {trans }}\) \(\mathrm{K}=\frac{1}{2} \mathrm{I} \omega^{2}+\frac{1}{2} \mathrm{MV}^{2} \quad(\mathrm{~V}=\mathrm{R} \omega\) \(=\frac{1}{2} \times \frac{2}{5} \mathrm{M}(2 \mathrm{R})^{2}\left(\frac{\mathrm{V}}{2 \mathrm{R}}\right)^{2}+\frac{1}{2} \mathrm{MV}^{2} \quad\left(\begin{array}{l}\mathrm{I}=\frac{1}{2} \mathrm{MR}^{2} \\ \text { for solid sphere }\end{array}\right)\) \(=\frac{4}{5} \times \mathrm{MR}^{2} \frac{\mathrm{V}^{2}}{4 \mathrm{R}^{2}}+\frac{1}{2} \mathrm{MV}^{2}\) \(=\frac{1}{5} \mathrm{MV}^{2}+\frac{1}{2} \mathrm{MV}^{2}=\frac{7}{10} \mathrm{MV}^{2}\) Now gain of kinetic energy \(=\) loss in potential energy \(\frac{7}{10} \mathrm{MV}^{2}=\mathrm{Mgh}\) \(\mathrm{V}=\sqrt{\frac{10}{7} \mathrm{gh}}\) (g) Rolling Motion}
JIPMER-2016
Rotational Motion
150395
A small object of uniform density rolls up a curved surface with an initial velocity \(v\). It reaches up to a maximum height of \(\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\) with respect to the initial position. The object is:
1 ring
2 solid sphere
3 hollow sphere
4 disc
Explanation:
D From conservation of energy, \(\text { K.E. })_{\mathrm{t}}+(\text { K.E. })_{\mathrm{r}}=\text { P.E. }\) \(\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=\mathrm{mgh}\) \(\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=\operatorname{mg}\left(\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\right) \quad\left(\because \mathrm{h}=\frac{3 \mathrm{v}^{2}}{4 \mathrm{~g}}\right).\), given \(\) \(\frac{1}{2} m v^{2}+\frac{1}{2} I \times \frac{v^{2}}{r^{2}}=m g\left(\frac{3 v^{2}}{4 g}\right)\) \(\frac{\mathrm{m}}{2}+\frac{\mathrm{I}}{2 \mathrm{r}^{2}}=\frac{3 \mathrm{~m}}{4}\) \(\frac{\mathrm{I}}{2 \mathrm{r}^{2}} =\frac{3}{4} \mathrm{~m}-\frac{1}{2} \mathrm{~m}\) \(\frac{\mathrm{I}}{2 \mathrm{r}^{2}} =\frac{\mathrm{m}}{4}\) \(\mathrm{I} =\frac{\mathrm{mr}^{2}}{2}\) Hence, the object must be disc.
