150436
A body rolls down an inclined plane. If its kinetic energy of rotation is \(40 \%\) of its kinetic energy of translation motion, then the body is
1 Hollow cylinder
2 Ring
3 Solid disc
4 Solid sphere
5 Hollow sphere
Explanation:
D Given, According to question, \(\mathrm{K} . \mathrm{E})_{\text {rotation }}=40 \%(\mathrm{~K} . \mathrm{E} .)_{\text {translation }}\) \(\frac{1}{2} \mathrm{I} \omega^{2}=\frac{40}{100} \times \frac{1}{2} \mathrm{mv}^{2}\) \(\because \quad \mathrm{v}=\mathrm{r} \omega, \omega=\mathrm{v} / \mathrm{r}\) \(\frac{1}{2} \mathrm{mk}^{2}\left(\frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}\right)=0.4 \times \frac{1}{2} \mathrm{mv}^{2}\) \(\frac{\mathrm{k}^{2}}{\mathrm{r}^{2}}=\frac{2}{5}\) \(\mathrm{k}^{2}=\frac{2}{5} \mathrm{r}^{2}\) \(\mathrm{I}=\mathrm{mk}^{2}=\frac{2}{5} \mathrm{mr}^{2}\) As we know that, The moment of inertia of a solid sphere \((I)=\frac{2}{5} \mathrm{mr}^{2}\) So, the body is a solid sphere.
Kerala CEE 2005
Rotational Motion
150437
The total energy of a solid sphere of mass \(300 \mathrm{~g}\) which rolls without slipping with a constant velocity of \(5 \mathrm{~ms}^{-1}\) along a straight line is
1 \(5.25 \mathrm{~J}\)
2 \(3.25 \mathrm{~J}\)
3 \(0.25 \mathrm{~J}\)
4 \(1.25 \mathrm{~J}\)
5 \(0.625 \mathrm{~J}\)
Explanation:
A Given that, Mass of solid sphere \(\mathrm{m})=300 \mathrm{~g}=300 \times 10^{-3} \mathrm{~kg}\) constant velocity \(\mathrm{v})=5 \mathrm{~ms}^{-1}\) We know that, \(\text { kinetic energy (K.E.) }=\frac{1}{2} I \omega^{2}\) For solid sphere moment of inertia \(\mathrm{I}=\frac{2}{5} \mathrm{mr}^{2}\) \(\text { K.E. }=\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \omega^{2}\) Total kinetic energy \(=(\text { K.E. })_{\text {rotation }}+(\text { K.E. })_{\text {translational }}\) \(\text { K.E. })_{\text {total }}=\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \omega^{2}+\frac{1}{2} \mathrm{mr}^{2} \omega^{2}\) \(\text { K.E. })_{\text {total }}=\frac{1}{5} \mathrm{mr}^{2} \omega^{2}+\frac{1}{2} \mathrm{mr}^{2} \omega^{2}\) \(\text { K.E. })_{\text {total }}=\frac{7}{10} \mathrm{mr}^{2} \omega^{2}\) \(\text { K.E. })_{\text {total }}=\frac{7}{10} \mathrm{mv}^{2} \quad(\because \mathrm{v}=\mathrm{r} \omega\) \(\text { K.E. })_{\text {total }}=\frac{7}{10} \times 300 \times 10^{-3} \times 5 \times 5\) \(\text { K.E. })_{\text {total }}=5.25 \mathrm{~J}\)
Kerala CEE- 2014
Rotational Motion
150438
A sphere of mass \(m\) and radius \(r\) rolls on a horizontal plane without slipping with the speed \(u\). Now, if it rolls up vertically, then maximum height it would be attain will be
1 \(\frac{3 u^{2}}{4 g}\)
2 \(\frac{5 \mathrm{u}^{2}}{2 \mathrm{~g}}\)
3 \(\frac{7 \mathrm{u}^{2}}{10 \mathrm{~g}}\)
4 \(\frac{u^{2}}{2 g}\)
5 \(\frac{11 u^{2}}{9 g}\)
Explanation:
C Given that, Mass of sphere \(=m\) Radius of sphere \(=r\) Speed \(=\mathrm{u}\) The rolling sphere has both rotational and translational energy \(\therefore \quad\) K.E. \(=\frac{1}{2} \mathrm{mu}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}\) Moment of inertia of solid sphere (I) \(=\frac{2}{5} \mathrm{mr}^{2}\) K.E. \(=\frac{1}{2} m u^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \omega^{2}\) K.E. \(=\frac{1}{2} m u^{2}+\frac{1}{5} m u^{2} \quad\{\because u=r \omega\}\) \(\mathrm{K} . \mathrm{E}=\frac{5 \mathrm{mu}^{2}+2 \mathrm{mu}^{2}}{10}\) K.E. \(=\frac{7}{10} \mathrm{mu}^{2}\) From conservation of energy principle K.E. \(=\) potential energy (P.E.) \(\frac{7}{10} \mathrm{mu}^{2}=\mathrm{mgh}\) \(\mathrm{h}=\frac{7 \mathrm{u}^{2}}{10 \mathrm{~g}}\)
Kerala CEE 2007
Rotational Motion
150439
A solid sphere is rolling without slipping on a horizontal surface. The ratio of its rotational kinetic energy to its translational kinetic energy is
1 \(2 / 9\)
2 \(2 / 7\)
3 \(2 / 5\)
4 \(7 / 2\)
Explanation:
C Given that, Where \(\mathrm{m}=\) mass of the body \(v=\) velocity of the body \(I=\) moment of inertia \(\omega=\) Angular velocity The Rotational kinetic energy, \(\text { K.E. })_{\text {rotational }}=\frac{1}{2} \mathrm{I} \omega^{2}\) Where, \(\mathrm{I}=\) Moment of inertia \(\omega=\text { angular velocity }\) For Solid sphere moment of inertia \((I)=\frac{2}{5} \mathrm{mr}^{2}\) \(\omega=\frac{\mathrm{v}}{\mathrm{r}} \quad \text { or } \quad \omega^{2}=\frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}\) \(\therefore \quad\) (K.E. \()_{\text {rotational }}=\frac{1}{2} \times \frac{2}{5} \mathrm{mr}^{2} \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}\) \(\text { K.E. })_{\text {rotational }}=\frac{1}{5} \mathrm{mv}^{2}\) The kinetic energy of the solid sphere \(\text { K.E. })_{\text {translational }}=\frac{1}{2} \mathrm{mv}^{2}\) From equation (i) \ (ii) we get, \(\frac{(\mathrm{K} . \mathrm{E} .)_{\text {rotational }}}{(\mathrm{K} . \mathrm{E} .)_{\text {translational }}}=\frac{1 / 5 \mathrm{mv}^{2}}{1 / 2 \mathrm{mv}^{2}}=\frac{2}{5}\) \(\frac{(\mathrm{K} . \mathrm{E} .)_{\text {rotational }}}{(\mathrm{K} . \mathrm{E} .)_{\text {translational }}}=\frac{2}{5}\)
UPSEE - 2008
Rotational Motion
150440
A small disc of radius \(2 \mathrm{~cm}\) is cut from a disc of radius \(6 \mathrm{~cm}\). If the distance between their centre is \(3.2 \mathrm{~cm}\), what is the shift in the centre of mass of the disc?
1 \(0.4 \mathrm{~cm}\)
2 \(2.4 \mathrm{~cm}\)
3 \(1.8 \mathrm{~cm}\)
4 \(1.2 \mathrm{~cm}\)
Explanation:
A Let, to the \(\mathrm{m}=\) mass per unit area \(\mathrm{m}_{1}=\) Mass of removed disc \(=(2)^{2} \pi \mathrm{m}=4 \pi \mathrm{m}\) \(\mathrm{m}=\) Mass of disc \(=\pi \mathrm{R}^{2} \mathrm{~m}=\pi(6)^{2} \mathrm{~m}=36 \pi \mathrm{m}\) \(\mathrm{m}_{2}=\) Mass of remaining portion \(=\mathrm{m}-\mathrm{m}_{1}\) \(=36 \pi \mathrm{m}-4 \pi \mathrm{m}\) \(=32 \pi \mathrm{m}\) Center of mass of disc at origin Centre of mass of removed disc \(=r_{1}=3.2 \mathrm{~cm}\) (given) Centre of mass of remaining disc \(=r_{2}\) Hence, \(\mathrm{m} \times 0=\mathrm{m}_{1} \mathrm{r}_{1}+\mathrm{m}_{2} \mathrm{r}_{2}\) \(0=\mathrm{m}_{1} \mathrm{r}_{1}+\mathrm{m}_{2} \mathrm{r}_{2}\) \(\mathrm{r}_{2}=\frac{-\mathrm{m}_{1} \mathrm{r}_{1}}{\mathrm{~m}_{2}}=\frac{-4 \pi \mathrm{m} \times 3.2}{32 \pi \mathrm{m}}\) \(\mathrm{r}_{2}=-0.4 \mathrm{~cm}=0.4 \mathrm{~cm}\) \(-\mathrm{v}_{\mathrm{e}}\) sign indicate the shift from the center in opposite side.
150436
A body rolls down an inclined plane. If its kinetic energy of rotation is \(40 \%\) of its kinetic energy of translation motion, then the body is
1 Hollow cylinder
2 Ring
3 Solid disc
4 Solid sphere
5 Hollow sphere
Explanation:
D Given, According to question, \(\mathrm{K} . \mathrm{E})_{\text {rotation }}=40 \%(\mathrm{~K} . \mathrm{E} .)_{\text {translation }}\) \(\frac{1}{2} \mathrm{I} \omega^{2}=\frac{40}{100} \times \frac{1}{2} \mathrm{mv}^{2}\) \(\because \quad \mathrm{v}=\mathrm{r} \omega, \omega=\mathrm{v} / \mathrm{r}\) \(\frac{1}{2} \mathrm{mk}^{2}\left(\frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}\right)=0.4 \times \frac{1}{2} \mathrm{mv}^{2}\) \(\frac{\mathrm{k}^{2}}{\mathrm{r}^{2}}=\frac{2}{5}\) \(\mathrm{k}^{2}=\frac{2}{5} \mathrm{r}^{2}\) \(\mathrm{I}=\mathrm{mk}^{2}=\frac{2}{5} \mathrm{mr}^{2}\) As we know that, The moment of inertia of a solid sphere \((I)=\frac{2}{5} \mathrm{mr}^{2}\) So, the body is a solid sphere.
Kerala CEE 2005
Rotational Motion
150437
The total energy of a solid sphere of mass \(300 \mathrm{~g}\) which rolls without slipping with a constant velocity of \(5 \mathrm{~ms}^{-1}\) along a straight line is
1 \(5.25 \mathrm{~J}\)
2 \(3.25 \mathrm{~J}\)
3 \(0.25 \mathrm{~J}\)
4 \(1.25 \mathrm{~J}\)
5 \(0.625 \mathrm{~J}\)
Explanation:
A Given that, Mass of solid sphere \(\mathrm{m})=300 \mathrm{~g}=300 \times 10^{-3} \mathrm{~kg}\) constant velocity \(\mathrm{v})=5 \mathrm{~ms}^{-1}\) We know that, \(\text { kinetic energy (K.E.) }=\frac{1}{2} I \omega^{2}\) For solid sphere moment of inertia \(\mathrm{I}=\frac{2}{5} \mathrm{mr}^{2}\) \(\text { K.E. }=\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \omega^{2}\) Total kinetic energy \(=(\text { K.E. })_{\text {rotation }}+(\text { K.E. })_{\text {translational }}\) \(\text { K.E. })_{\text {total }}=\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \omega^{2}+\frac{1}{2} \mathrm{mr}^{2} \omega^{2}\) \(\text { K.E. })_{\text {total }}=\frac{1}{5} \mathrm{mr}^{2} \omega^{2}+\frac{1}{2} \mathrm{mr}^{2} \omega^{2}\) \(\text { K.E. })_{\text {total }}=\frac{7}{10} \mathrm{mr}^{2} \omega^{2}\) \(\text { K.E. })_{\text {total }}=\frac{7}{10} \mathrm{mv}^{2} \quad(\because \mathrm{v}=\mathrm{r} \omega\) \(\text { K.E. })_{\text {total }}=\frac{7}{10} \times 300 \times 10^{-3} \times 5 \times 5\) \(\text { K.E. })_{\text {total }}=5.25 \mathrm{~J}\)
Kerala CEE- 2014
Rotational Motion
150438
A sphere of mass \(m\) and radius \(r\) rolls on a horizontal plane without slipping with the speed \(u\). Now, if it rolls up vertically, then maximum height it would be attain will be
1 \(\frac{3 u^{2}}{4 g}\)
2 \(\frac{5 \mathrm{u}^{2}}{2 \mathrm{~g}}\)
3 \(\frac{7 \mathrm{u}^{2}}{10 \mathrm{~g}}\)
4 \(\frac{u^{2}}{2 g}\)
5 \(\frac{11 u^{2}}{9 g}\)
Explanation:
C Given that, Mass of sphere \(=m\) Radius of sphere \(=r\) Speed \(=\mathrm{u}\) The rolling sphere has both rotational and translational energy \(\therefore \quad\) K.E. \(=\frac{1}{2} \mathrm{mu}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}\) Moment of inertia of solid sphere (I) \(=\frac{2}{5} \mathrm{mr}^{2}\) K.E. \(=\frac{1}{2} m u^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \omega^{2}\) K.E. \(=\frac{1}{2} m u^{2}+\frac{1}{5} m u^{2} \quad\{\because u=r \omega\}\) \(\mathrm{K} . \mathrm{E}=\frac{5 \mathrm{mu}^{2}+2 \mathrm{mu}^{2}}{10}\) K.E. \(=\frac{7}{10} \mathrm{mu}^{2}\) From conservation of energy principle K.E. \(=\) potential energy (P.E.) \(\frac{7}{10} \mathrm{mu}^{2}=\mathrm{mgh}\) \(\mathrm{h}=\frac{7 \mathrm{u}^{2}}{10 \mathrm{~g}}\)
Kerala CEE 2007
Rotational Motion
150439
A solid sphere is rolling without slipping on a horizontal surface. The ratio of its rotational kinetic energy to its translational kinetic energy is
1 \(2 / 9\)
2 \(2 / 7\)
3 \(2 / 5\)
4 \(7 / 2\)
Explanation:
C Given that, Where \(\mathrm{m}=\) mass of the body \(v=\) velocity of the body \(I=\) moment of inertia \(\omega=\) Angular velocity The Rotational kinetic energy, \(\text { K.E. })_{\text {rotational }}=\frac{1}{2} \mathrm{I} \omega^{2}\) Where, \(\mathrm{I}=\) Moment of inertia \(\omega=\text { angular velocity }\) For Solid sphere moment of inertia \((I)=\frac{2}{5} \mathrm{mr}^{2}\) \(\omega=\frac{\mathrm{v}}{\mathrm{r}} \quad \text { or } \quad \omega^{2}=\frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}\) \(\therefore \quad\) (K.E. \()_{\text {rotational }}=\frac{1}{2} \times \frac{2}{5} \mathrm{mr}^{2} \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}\) \(\text { K.E. })_{\text {rotational }}=\frac{1}{5} \mathrm{mv}^{2}\) The kinetic energy of the solid sphere \(\text { K.E. })_{\text {translational }}=\frac{1}{2} \mathrm{mv}^{2}\) From equation (i) \ (ii) we get, \(\frac{(\mathrm{K} . \mathrm{E} .)_{\text {rotational }}}{(\mathrm{K} . \mathrm{E} .)_{\text {translational }}}=\frac{1 / 5 \mathrm{mv}^{2}}{1 / 2 \mathrm{mv}^{2}}=\frac{2}{5}\) \(\frac{(\mathrm{K} . \mathrm{E} .)_{\text {rotational }}}{(\mathrm{K} . \mathrm{E} .)_{\text {translational }}}=\frac{2}{5}\)
UPSEE - 2008
Rotational Motion
150440
A small disc of radius \(2 \mathrm{~cm}\) is cut from a disc of radius \(6 \mathrm{~cm}\). If the distance between their centre is \(3.2 \mathrm{~cm}\), what is the shift in the centre of mass of the disc?
1 \(0.4 \mathrm{~cm}\)
2 \(2.4 \mathrm{~cm}\)
3 \(1.8 \mathrm{~cm}\)
4 \(1.2 \mathrm{~cm}\)
Explanation:
A Let, to the \(\mathrm{m}=\) mass per unit area \(\mathrm{m}_{1}=\) Mass of removed disc \(=(2)^{2} \pi \mathrm{m}=4 \pi \mathrm{m}\) \(\mathrm{m}=\) Mass of disc \(=\pi \mathrm{R}^{2} \mathrm{~m}=\pi(6)^{2} \mathrm{~m}=36 \pi \mathrm{m}\) \(\mathrm{m}_{2}=\) Mass of remaining portion \(=\mathrm{m}-\mathrm{m}_{1}\) \(=36 \pi \mathrm{m}-4 \pi \mathrm{m}\) \(=32 \pi \mathrm{m}\) Center of mass of disc at origin Centre of mass of removed disc \(=r_{1}=3.2 \mathrm{~cm}\) (given) Centre of mass of remaining disc \(=r_{2}\) Hence, \(\mathrm{m} \times 0=\mathrm{m}_{1} \mathrm{r}_{1}+\mathrm{m}_{2} \mathrm{r}_{2}\) \(0=\mathrm{m}_{1} \mathrm{r}_{1}+\mathrm{m}_{2} \mathrm{r}_{2}\) \(\mathrm{r}_{2}=\frac{-\mathrm{m}_{1} \mathrm{r}_{1}}{\mathrm{~m}_{2}}=\frac{-4 \pi \mathrm{m} \times 3.2}{32 \pi \mathrm{m}}\) \(\mathrm{r}_{2}=-0.4 \mathrm{~cm}=0.4 \mathrm{~cm}\) \(-\mathrm{v}_{\mathrm{e}}\) sign indicate the shift from the center in opposite side.
150436
A body rolls down an inclined plane. If its kinetic energy of rotation is \(40 \%\) of its kinetic energy of translation motion, then the body is
1 Hollow cylinder
2 Ring
3 Solid disc
4 Solid sphere
5 Hollow sphere
Explanation:
D Given, According to question, \(\mathrm{K} . \mathrm{E})_{\text {rotation }}=40 \%(\mathrm{~K} . \mathrm{E} .)_{\text {translation }}\) \(\frac{1}{2} \mathrm{I} \omega^{2}=\frac{40}{100} \times \frac{1}{2} \mathrm{mv}^{2}\) \(\because \quad \mathrm{v}=\mathrm{r} \omega, \omega=\mathrm{v} / \mathrm{r}\) \(\frac{1}{2} \mathrm{mk}^{2}\left(\frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}\right)=0.4 \times \frac{1}{2} \mathrm{mv}^{2}\) \(\frac{\mathrm{k}^{2}}{\mathrm{r}^{2}}=\frac{2}{5}\) \(\mathrm{k}^{2}=\frac{2}{5} \mathrm{r}^{2}\) \(\mathrm{I}=\mathrm{mk}^{2}=\frac{2}{5} \mathrm{mr}^{2}\) As we know that, The moment of inertia of a solid sphere \((I)=\frac{2}{5} \mathrm{mr}^{2}\) So, the body is a solid sphere.
Kerala CEE 2005
Rotational Motion
150437
The total energy of a solid sphere of mass \(300 \mathrm{~g}\) which rolls without slipping with a constant velocity of \(5 \mathrm{~ms}^{-1}\) along a straight line is
1 \(5.25 \mathrm{~J}\)
2 \(3.25 \mathrm{~J}\)
3 \(0.25 \mathrm{~J}\)
4 \(1.25 \mathrm{~J}\)
5 \(0.625 \mathrm{~J}\)
Explanation:
A Given that, Mass of solid sphere \(\mathrm{m})=300 \mathrm{~g}=300 \times 10^{-3} \mathrm{~kg}\) constant velocity \(\mathrm{v})=5 \mathrm{~ms}^{-1}\) We know that, \(\text { kinetic energy (K.E.) }=\frac{1}{2} I \omega^{2}\) For solid sphere moment of inertia \(\mathrm{I}=\frac{2}{5} \mathrm{mr}^{2}\) \(\text { K.E. }=\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \omega^{2}\) Total kinetic energy \(=(\text { K.E. })_{\text {rotation }}+(\text { K.E. })_{\text {translational }}\) \(\text { K.E. })_{\text {total }}=\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \omega^{2}+\frac{1}{2} \mathrm{mr}^{2} \omega^{2}\) \(\text { K.E. })_{\text {total }}=\frac{1}{5} \mathrm{mr}^{2} \omega^{2}+\frac{1}{2} \mathrm{mr}^{2} \omega^{2}\) \(\text { K.E. })_{\text {total }}=\frac{7}{10} \mathrm{mr}^{2} \omega^{2}\) \(\text { K.E. })_{\text {total }}=\frac{7}{10} \mathrm{mv}^{2} \quad(\because \mathrm{v}=\mathrm{r} \omega\) \(\text { K.E. })_{\text {total }}=\frac{7}{10} \times 300 \times 10^{-3} \times 5 \times 5\) \(\text { K.E. })_{\text {total }}=5.25 \mathrm{~J}\)
Kerala CEE- 2014
Rotational Motion
150438
A sphere of mass \(m\) and radius \(r\) rolls on a horizontal plane without slipping with the speed \(u\). Now, if it rolls up vertically, then maximum height it would be attain will be
1 \(\frac{3 u^{2}}{4 g}\)
2 \(\frac{5 \mathrm{u}^{2}}{2 \mathrm{~g}}\)
3 \(\frac{7 \mathrm{u}^{2}}{10 \mathrm{~g}}\)
4 \(\frac{u^{2}}{2 g}\)
5 \(\frac{11 u^{2}}{9 g}\)
Explanation:
C Given that, Mass of sphere \(=m\) Radius of sphere \(=r\) Speed \(=\mathrm{u}\) The rolling sphere has both rotational and translational energy \(\therefore \quad\) K.E. \(=\frac{1}{2} \mathrm{mu}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}\) Moment of inertia of solid sphere (I) \(=\frac{2}{5} \mathrm{mr}^{2}\) K.E. \(=\frac{1}{2} m u^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \omega^{2}\) K.E. \(=\frac{1}{2} m u^{2}+\frac{1}{5} m u^{2} \quad\{\because u=r \omega\}\) \(\mathrm{K} . \mathrm{E}=\frac{5 \mathrm{mu}^{2}+2 \mathrm{mu}^{2}}{10}\) K.E. \(=\frac{7}{10} \mathrm{mu}^{2}\) From conservation of energy principle K.E. \(=\) potential energy (P.E.) \(\frac{7}{10} \mathrm{mu}^{2}=\mathrm{mgh}\) \(\mathrm{h}=\frac{7 \mathrm{u}^{2}}{10 \mathrm{~g}}\)
Kerala CEE 2007
Rotational Motion
150439
A solid sphere is rolling without slipping on a horizontal surface. The ratio of its rotational kinetic energy to its translational kinetic energy is
1 \(2 / 9\)
2 \(2 / 7\)
3 \(2 / 5\)
4 \(7 / 2\)
Explanation:
C Given that, Where \(\mathrm{m}=\) mass of the body \(v=\) velocity of the body \(I=\) moment of inertia \(\omega=\) Angular velocity The Rotational kinetic energy, \(\text { K.E. })_{\text {rotational }}=\frac{1}{2} \mathrm{I} \omega^{2}\) Where, \(\mathrm{I}=\) Moment of inertia \(\omega=\text { angular velocity }\) For Solid sphere moment of inertia \((I)=\frac{2}{5} \mathrm{mr}^{2}\) \(\omega=\frac{\mathrm{v}}{\mathrm{r}} \quad \text { or } \quad \omega^{2}=\frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}\) \(\therefore \quad\) (K.E. \()_{\text {rotational }}=\frac{1}{2} \times \frac{2}{5} \mathrm{mr}^{2} \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}\) \(\text { K.E. })_{\text {rotational }}=\frac{1}{5} \mathrm{mv}^{2}\) The kinetic energy of the solid sphere \(\text { K.E. })_{\text {translational }}=\frac{1}{2} \mathrm{mv}^{2}\) From equation (i) \ (ii) we get, \(\frac{(\mathrm{K} . \mathrm{E} .)_{\text {rotational }}}{(\mathrm{K} . \mathrm{E} .)_{\text {translational }}}=\frac{1 / 5 \mathrm{mv}^{2}}{1 / 2 \mathrm{mv}^{2}}=\frac{2}{5}\) \(\frac{(\mathrm{K} . \mathrm{E} .)_{\text {rotational }}}{(\mathrm{K} . \mathrm{E} .)_{\text {translational }}}=\frac{2}{5}\)
UPSEE - 2008
Rotational Motion
150440
A small disc of radius \(2 \mathrm{~cm}\) is cut from a disc of radius \(6 \mathrm{~cm}\). If the distance between their centre is \(3.2 \mathrm{~cm}\), what is the shift in the centre of mass of the disc?
1 \(0.4 \mathrm{~cm}\)
2 \(2.4 \mathrm{~cm}\)
3 \(1.8 \mathrm{~cm}\)
4 \(1.2 \mathrm{~cm}\)
Explanation:
A Let, to the \(\mathrm{m}=\) mass per unit area \(\mathrm{m}_{1}=\) Mass of removed disc \(=(2)^{2} \pi \mathrm{m}=4 \pi \mathrm{m}\) \(\mathrm{m}=\) Mass of disc \(=\pi \mathrm{R}^{2} \mathrm{~m}=\pi(6)^{2} \mathrm{~m}=36 \pi \mathrm{m}\) \(\mathrm{m}_{2}=\) Mass of remaining portion \(=\mathrm{m}-\mathrm{m}_{1}\) \(=36 \pi \mathrm{m}-4 \pi \mathrm{m}\) \(=32 \pi \mathrm{m}\) Center of mass of disc at origin Centre of mass of removed disc \(=r_{1}=3.2 \mathrm{~cm}\) (given) Centre of mass of remaining disc \(=r_{2}\) Hence, \(\mathrm{m} \times 0=\mathrm{m}_{1} \mathrm{r}_{1}+\mathrm{m}_{2} \mathrm{r}_{2}\) \(0=\mathrm{m}_{1} \mathrm{r}_{1}+\mathrm{m}_{2} \mathrm{r}_{2}\) \(\mathrm{r}_{2}=\frac{-\mathrm{m}_{1} \mathrm{r}_{1}}{\mathrm{~m}_{2}}=\frac{-4 \pi \mathrm{m} \times 3.2}{32 \pi \mathrm{m}}\) \(\mathrm{r}_{2}=-0.4 \mathrm{~cm}=0.4 \mathrm{~cm}\) \(-\mathrm{v}_{\mathrm{e}}\) sign indicate the shift from the center in opposite side.
150436
A body rolls down an inclined plane. If its kinetic energy of rotation is \(40 \%\) of its kinetic energy of translation motion, then the body is
1 Hollow cylinder
2 Ring
3 Solid disc
4 Solid sphere
5 Hollow sphere
Explanation:
D Given, According to question, \(\mathrm{K} . \mathrm{E})_{\text {rotation }}=40 \%(\mathrm{~K} . \mathrm{E} .)_{\text {translation }}\) \(\frac{1}{2} \mathrm{I} \omega^{2}=\frac{40}{100} \times \frac{1}{2} \mathrm{mv}^{2}\) \(\because \quad \mathrm{v}=\mathrm{r} \omega, \omega=\mathrm{v} / \mathrm{r}\) \(\frac{1}{2} \mathrm{mk}^{2}\left(\frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}\right)=0.4 \times \frac{1}{2} \mathrm{mv}^{2}\) \(\frac{\mathrm{k}^{2}}{\mathrm{r}^{2}}=\frac{2}{5}\) \(\mathrm{k}^{2}=\frac{2}{5} \mathrm{r}^{2}\) \(\mathrm{I}=\mathrm{mk}^{2}=\frac{2}{5} \mathrm{mr}^{2}\) As we know that, The moment of inertia of a solid sphere \((I)=\frac{2}{5} \mathrm{mr}^{2}\) So, the body is a solid sphere.
Kerala CEE 2005
Rotational Motion
150437
The total energy of a solid sphere of mass \(300 \mathrm{~g}\) which rolls without slipping with a constant velocity of \(5 \mathrm{~ms}^{-1}\) along a straight line is
1 \(5.25 \mathrm{~J}\)
2 \(3.25 \mathrm{~J}\)
3 \(0.25 \mathrm{~J}\)
4 \(1.25 \mathrm{~J}\)
5 \(0.625 \mathrm{~J}\)
Explanation:
A Given that, Mass of solid sphere \(\mathrm{m})=300 \mathrm{~g}=300 \times 10^{-3} \mathrm{~kg}\) constant velocity \(\mathrm{v})=5 \mathrm{~ms}^{-1}\) We know that, \(\text { kinetic energy (K.E.) }=\frac{1}{2} I \omega^{2}\) For solid sphere moment of inertia \(\mathrm{I}=\frac{2}{5} \mathrm{mr}^{2}\) \(\text { K.E. }=\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \omega^{2}\) Total kinetic energy \(=(\text { K.E. })_{\text {rotation }}+(\text { K.E. })_{\text {translational }}\) \(\text { K.E. })_{\text {total }}=\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \omega^{2}+\frac{1}{2} \mathrm{mr}^{2} \omega^{2}\) \(\text { K.E. })_{\text {total }}=\frac{1}{5} \mathrm{mr}^{2} \omega^{2}+\frac{1}{2} \mathrm{mr}^{2} \omega^{2}\) \(\text { K.E. })_{\text {total }}=\frac{7}{10} \mathrm{mr}^{2} \omega^{2}\) \(\text { K.E. })_{\text {total }}=\frac{7}{10} \mathrm{mv}^{2} \quad(\because \mathrm{v}=\mathrm{r} \omega\) \(\text { K.E. })_{\text {total }}=\frac{7}{10} \times 300 \times 10^{-3} \times 5 \times 5\) \(\text { K.E. })_{\text {total }}=5.25 \mathrm{~J}\)
Kerala CEE- 2014
Rotational Motion
150438
A sphere of mass \(m\) and radius \(r\) rolls on a horizontal plane without slipping with the speed \(u\). Now, if it rolls up vertically, then maximum height it would be attain will be
1 \(\frac{3 u^{2}}{4 g}\)
2 \(\frac{5 \mathrm{u}^{2}}{2 \mathrm{~g}}\)
3 \(\frac{7 \mathrm{u}^{2}}{10 \mathrm{~g}}\)
4 \(\frac{u^{2}}{2 g}\)
5 \(\frac{11 u^{2}}{9 g}\)
Explanation:
C Given that, Mass of sphere \(=m\) Radius of sphere \(=r\) Speed \(=\mathrm{u}\) The rolling sphere has both rotational and translational energy \(\therefore \quad\) K.E. \(=\frac{1}{2} \mathrm{mu}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}\) Moment of inertia of solid sphere (I) \(=\frac{2}{5} \mathrm{mr}^{2}\) K.E. \(=\frac{1}{2} m u^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \omega^{2}\) K.E. \(=\frac{1}{2} m u^{2}+\frac{1}{5} m u^{2} \quad\{\because u=r \omega\}\) \(\mathrm{K} . \mathrm{E}=\frac{5 \mathrm{mu}^{2}+2 \mathrm{mu}^{2}}{10}\) K.E. \(=\frac{7}{10} \mathrm{mu}^{2}\) From conservation of energy principle K.E. \(=\) potential energy (P.E.) \(\frac{7}{10} \mathrm{mu}^{2}=\mathrm{mgh}\) \(\mathrm{h}=\frac{7 \mathrm{u}^{2}}{10 \mathrm{~g}}\)
Kerala CEE 2007
Rotational Motion
150439
A solid sphere is rolling without slipping on a horizontal surface. The ratio of its rotational kinetic energy to its translational kinetic energy is
1 \(2 / 9\)
2 \(2 / 7\)
3 \(2 / 5\)
4 \(7 / 2\)
Explanation:
C Given that, Where \(\mathrm{m}=\) mass of the body \(v=\) velocity of the body \(I=\) moment of inertia \(\omega=\) Angular velocity The Rotational kinetic energy, \(\text { K.E. })_{\text {rotational }}=\frac{1}{2} \mathrm{I} \omega^{2}\) Where, \(\mathrm{I}=\) Moment of inertia \(\omega=\text { angular velocity }\) For Solid sphere moment of inertia \((I)=\frac{2}{5} \mathrm{mr}^{2}\) \(\omega=\frac{\mathrm{v}}{\mathrm{r}} \quad \text { or } \quad \omega^{2}=\frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}\) \(\therefore \quad\) (K.E. \()_{\text {rotational }}=\frac{1}{2} \times \frac{2}{5} \mathrm{mr}^{2} \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}\) \(\text { K.E. })_{\text {rotational }}=\frac{1}{5} \mathrm{mv}^{2}\) The kinetic energy of the solid sphere \(\text { K.E. })_{\text {translational }}=\frac{1}{2} \mathrm{mv}^{2}\) From equation (i) \ (ii) we get, \(\frac{(\mathrm{K} . \mathrm{E} .)_{\text {rotational }}}{(\mathrm{K} . \mathrm{E} .)_{\text {translational }}}=\frac{1 / 5 \mathrm{mv}^{2}}{1 / 2 \mathrm{mv}^{2}}=\frac{2}{5}\) \(\frac{(\mathrm{K} . \mathrm{E} .)_{\text {rotational }}}{(\mathrm{K} . \mathrm{E} .)_{\text {translational }}}=\frac{2}{5}\)
UPSEE - 2008
Rotational Motion
150440
A small disc of radius \(2 \mathrm{~cm}\) is cut from a disc of radius \(6 \mathrm{~cm}\). If the distance between their centre is \(3.2 \mathrm{~cm}\), what is the shift in the centre of mass of the disc?
1 \(0.4 \mathrm{~cm}\)
2 \(2.4 \mathrm{~cm}\)
3 \(1.8 \mathrm{~cm}\)
4 \(1.2 \mathrm{~cm}\)
Explanation:
A Let, to the \(\mathrm{m}=\) mass per unit area \(\mathrm{m}_{1}=\) Mass of removed disc \(=(2)^{2} \pi \mathrm{m}=4 \pi \mathrm{m}\) \(\mathrm{m}=\) Mass of disc \(=\pi \mathrm{R}^{2} \mathrm{~m}=\pi(6)^{2} \mathrm{~m}=36 \pi \mathrm{m}\) \(\mathrm{m}_{2}=\) Mass of remaining portion \(=\mathrm{m}-\mathrm{m}_{1}\) \(=36 \pi \mathrm{m}-4 \pi \mathrm{m}\) \(=32 \pi \mathrm{m}\) Center of mass of disc at origin Centre of mass of removed disc \(=r_{1}=3.2 \mathrm{~cm}\) (given) Centre of mass of remaining disc \(=r_{2}\) Hence, \(\mathrm{m} \times 0=\mathrm{m}_{1} \mathrm{r}_{1}+\mathrm{m}_{2} \mathrm{r}_{2}\) \(0=\mathrm{m}_{1} \mathrm{r}_{1}+\mathrm{m}_{2} \mathrm{r}_{2}\) \(\mathrm{r}_{2}=\frac{-\mathrm{m}_{1} \mathrm{r}_{1}}{\mathrm{~m}_{2}}=\frac{-4 \pi \mathrm{m} \times 3.2}{32 \pi \mathrm{m}}\) \(\mathrm{r}_{2}=-0.4 \mathrm{~cm}=0.4 \mathrm{~cm}\) \(-\mathrm{v}_{\mathrm{e}}\) sign indicate the shift from the center in opposite side.
150436
A body rolls down an inclined plane. If its kinetic energy of rotation is \(40 \%\) of its kinetic energy of translation motion, then the body is
1 Hollow cylinder
2 Ring
3 Solid disc
4 Solid sphere
5 Hollow sphere
Explanation:
D Given, According to question, \(\mathrm{K} . \mathrm{E})_{\text {rotation }}=40 \%(\mathrm{~K} . \mathrm{E} .)_{\text {translation }}\) \(\frac{1}{2} \mathrm{I} \omega^{2}=\frac{40}{100} \times \frac{1}{2} \mathrm{mv}^{2}\) \(\because \quad \mathrm{v}=\mathrm{r} \omega, \omega=\mathrm{v} / \mathrm{r}\) \(\frac{1}{2} \mathrm{mk}^{2}\left(\frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}\right)=0.4 \times \frac{1}{2} \mathrm{mv}^{2}\) \(\frac{\mathrm{k}^{2}}{\mathrm{r}^{2}}=\frac{2}{5}\) \(\mathrm{k}^{2}=\frac{2}{5} \mathrm{r}^{2}\) \(\mathrm{I}=\mathrm{mk}^{2}=\frac{2}{5} \mathrm{mr}^{2}\) As we know that, The moment of inertia of a solid sphere \((I)=\frac{2}{5} \mathrm{mr}^{2}\) So, the body is a solid sphere.
Kerala CEE 2005
Rotational Motion
150437
The total energy of a solid sphere of mass \(300 \mathrm{~g}\) which rolls without slipping with a constant velocity of \(5 \mathrm{~ms}^{-1}\) along a straight line is
1 \(5.25 \mathrm{~J}\)
2 \(3.25 \mathrm{~J}\)
3 \(0.25 \mathrm{~J}\)
4 \(1.25 \mathrm{~J}\)
5 \(0.625 \mathrm{~J}\)
Explanation:
A Given that, Mass of solid sphere \(\mathrm{m})=300 \mathrm{~g}=300 \times 10^{-3} \mathrm{~kg}\) constant velocity \(\mathrm{v})=5 \mathrm{~ms}^{-1}\) We know that, \(\text { kinetic energy (K.E.) }=\frac{1}{2} I \omega^{2}\) For solid sphere moment of inertia \(\mathrm{I}=\frac{2}{5} \mathrm{mr}^{2}\) \(\text { K.E. }=\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \omega^{2}\) Total kinetic energy \(=(\text { K.E. })_{\text {rotation }}+(\text { K.E. })_{\text {translational }}\) \(\text { K.E. })_{\text {total }}=\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \omega^{2}+\frac{1}{2} \mathrm{mr}^{2} \omega^{2}\) \(\text { K.E. })_{\text {total }}=\frac{1}{5} \mathrm{mr}^{2} \omega^{2}+\frac{1}{2} \mathrm{mr}^{2} \omega^{2}\) \(\text { K.E. })_{\text {total }}=\frac{7}{10} \mathrm{mr}^{2} \omega^{2}\) \(\text { K.E. })_{\text {total }}=\frac{7}{10} \mathrm{mv}^{2} \quad(\because \mathrm{v}=\mathrm{r} \omega\) \(\text { K.E. })_{\text {total }}=\frac{7}{10} \times 300 \times 10^{-3} \times 5 \times 5\) \(\text { K.E. })_{\text {total }}=5.25 \mathrm{~J}\)
Kerala CEE- 2014
Rotational Motion
150438
A sphere of mass \(m\) and radius \(r\) rolls on a horizontal plane without slipping with the speed \(u\). Now, if it rolls up vertically, then maximum height it would be attain will be
1 \(\frac{3 u^{2}}{4 g}\)
2 \(\frac{5 \mathrm{u}^{2}}{2 \mathrm{~g}}\)
3 \(\frac{7 \mathrm{u}^{2}}{10 \mathrm{~g}}\)
4 \(\frac{u^{2}}{2 g}\)
5 \(\frac{11 u^{2}}{9 g}\)
Explanation:
C Given that, Mass of sphere \(=m\) Radius of sphere \(=r\) Speed \(=\mathrm{u}\) The rolling sphere has both rotational and translational energy \(\therefore \quad\) K.E. \(=\frac{1}{2} \mathrm{mu}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}\) Moment of inertia of solid sphere (I) \(=\frac{2}{5} \mathrm{mr}^{2}\) K.E. \(=\frac{1}{2} m u^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right) \omega^{2}\) K.E. \(=\frac{1}{2} m u^{2}+\frac{1}{5} m u^{2} \quad\{\because u=r \omega\}\) \(\mathrm{K} . \mathrm{E}=\frac{5 \mathrm{mu}^{2}+2 \mathrm{mu}^{2}}{10}\) K.E. \(=\frac{7}{10} \mathrm{mu}^{2}\) From conservation of energy principle K.E. \(=\) potential energy (P.E.) \(\frac{7}{10} \mathrm{mu}^{2}=\mathrm{mgh}\) \(\mathrm{h}=\frac{7 \mathrm{u}^{2}}{10 \mathrm{~g}}\)
Kerala CEE 2007
Rotational Motion
150439
A solid sphere is rolling without slipping on a horizontal surface. The ratio of its rotational kinetic energy to its translational kinetic energy is
1 \(2 / 9\)
2 \(2 / 7\)
3 \(2 / 5\)
4 \(7 / 2\)
Explanation:
C Given that, Where \(\mathrm{m}=\) mass of the body \(v=\) velocity of the body \(I=\) moment of inertia \(\omega=\) Angular velocity The Rotational kinetic energy, \(\text { K.E. })_{\text {rotational }}=\frac{1}{2} \mathrm{I} \omega^{2}\) Where, \(\mathrm{I}=\) Moment of inertia \(\omega=\text { angular velocity }\) For Solid sphere moment of inertia \((I)=\frac{2}{5} \mathrm{mr}^{2}\) \(\omega=\frac{\mathrm{v}}{\mathrm{r}} \quad \text { or } \quad \omega^{2}=\frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}\) \(\therefore \quad\) (K.E. \()_{\text {rotational }}=\frac{1}{2} \times \frac{2}{5} \mathrm{mr}^{2} \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}\) \(\text { K.E. })_{\text {rotational }}=\frac{1}{5} \mathrm{mv}^{2}\) The kinetic energy of the solid sphere \(\text { K.E. })_{\text {translational }}=\frac{1}{2} \mathrm{mv}^{2}\) From equation (i) \ (ii) we get, \(\frac{(\mathrm{K} . \mathrm{E} .)_{\text {rotational }}}{(\mathrm{K} . \mathrm{E} .)_{\text {translational }}}=\frac{1 / 5 \mathrm{mv}^{2}}{1 / 2 \mathrm{mv}^{2}}=\frac{2}{5}\) \(\frac{(\mathrm{K} . \mathrm{E} .)_{\text {rotational }}}{(\mathrm{K} . \mathrm{E} .)_{\text {translational }}}=\frac{2}{5}\)
UPSEE - 2008
Rotational Motion
150440
A small disc of radius \(2 \mathrm{~cm}\) is cut from a disc of radius \(6 \mathrm{~cm}\). If the distance between their centre is \(3.2 \mathrm{~cm}\), what is the shift in the centre of mass of the disc?
1 \(0.4 \mathrm{~cm}\)
2 \(2.4 \mathrm{~cm}\)
3 \(1.8 \mathrm{~cm}\)
4 \(1.2 \mathrm{~cm}\)
Explanation:
A Let, to the \(\mathrm{m}=\) mass per unit area \(\mathrm{m}_{1}=\) Mass of removed disc \(=(2)^{2} \pi \mathrm{m}=4 \pi \mathrm{m}\) \(\mathrm{m}=\) Mass of disc \(=\pi \mathrm{R}^{2} \mathrm{~m}=\pi(6)^{2} \mathrm{~m}=36 \pi \mathrm{m}\) \(\mathrm{m}_{2}=\) Mass of remaining portion \(=\mathrm{m}-\mathrm{m}_{1}\) \(=36 \pi \mathrm{m}-4 \pi \mathrm{m}\) \(=32 \pi \mathrm{m}\) Center of mass of disc at origin Centre of mass of removed disc \(=r_{1}=3.2 \mathrm{~cm}\) (given) Centre of mass of remaining disc \(=r_{2}\) Hence, \(\mathrm{m} \times 0=\mathrm{m}_{1} \mathrm{r}_{1}+\mathrm{m}_{2} \mathrm{r}_{2}\) \(0=\mathrm{m}_{1} \mathrm{r}_{1}+\mathrm{m}_{2} \mathrm{r}_{2}\) \(\mathrm{r}_{2}=\frac{-\mathrm{m}_{1} \mathrm{r}_{1}}{\mathrm{~m}_{2}}=\frac{-4 \pi \mathrm{m} \times 3.2}{32 \pi \mathrm{m}}\) \(\mathrm{r}_{2}=-0.4 \mathrm{~cm}=0.4 \mathrm{~cm}\) \(-\mathrm{v}_{\mathrm{e}}\) sign indicate the shift from the center in opposite side.
