NEET Test Series from KOTA - 10 Papers In MS WORD
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Rotational Motion
150409
If mass, speed and radius of the circular path of the particle are increased by \(100 \%\), then the necessary force required to maintain the circular path will have to be increased by
1 \(100 \%\)
2 \(250 \%\)
3 \(300 \%\)
4 \(400 \%\)
Explanation:
C To the given data, mass \(=\mathrm{m}\) radius \(=\mathrm{r}\) velocity \(=\mathrm{v}\) \(\mathrm{F}=\) centripetal force We know that, applying the force \(\mathrm{F}_{\mathrm{c}}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) Now to the increment is \(100 \%\) in mass velocity radius \(\mathrm{m}=\mathrm{m}+\mathrm{m}=2 \mathrm{~m}\) \(\mathrm{v}=\mathrm{v}+\mathrm{v}=2 \mathrm{v}\) \(\mathrm{r}=\mathrm{r}+\mathrm{r}=2 \mathrm{r}\) Now hence, force is \(\mathrm{F}^{\prime}=\frac{(\mathrm{m}+\mathrm{m})(\mathrm{v}+\mathrm{v})^{2}}{\mathrm{r}+\mathrm{r}}\) \(\mathrm{F}^{\prime}=\frac{2 \mathrm{~m} \times 4 \mathrm{v}^{2}}{2 \mathrm{r}}=4\left(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\right)\) \(\mathrm{F}^{\prime}=\mathrm{F} \times 4\) Now the \(\%\) change in force \(=\frac{F^{\prime}-F}{F} \times 100=\frac{4 F-F}{F} \times 100=(4-1) \times 100\) \(=3 \times 100=300 \%\)
AP EAMCET (17.09.2020) Shift-I
Rotational Motion
150410
The speed of a uniform solid sphere after rolling down from rest without slipping along a fixed inclined plane of vertical height \(h\) is
1 \(\sqrt{\frac{10 g h}{7}}\)
2 \(\sqrt{g h}\)
3 \(\sqrt{\frac{6 g h}{5}}\)
4 \(\sqrt{\frac{4 g h}{3}}\)
Explanation:
A Using Work Energy Theorem, \(\mathrm{mgh}=\frac{1}{2} \mathrm{I} \omega^{2}+\frac{1}{2} \mathrm{mv}^{2}\) For solid sphere, moment of inertia \(\mathrm{I}=\frac{2}{5} \mathrm{mr}^{2}\) and \(\omega=\frac{\mathrm{v}}{\mathrm{r}}\) \(\mathrm{mgh}=\frac{1}{2} \times \frac{2}{5} \mathrm{mr}^{2} \times\left(\frac{\mathrm{v}}{\mathrm{r}}\right)^{2}+\frac{1}{2} \mathrm{mv}^{2}\) \(\mathrm{mgh}=\frac{1}{5} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{mv}^{2}\) \(\mathrm{mgh}=\frac{7}{10} \mathrm{mv}^{2}\) \(\mathrm{v}^{2}=\frac{10 \mathrm{gh}}{7}\) \(\mathrm{v}=\sqrt{\frac{10 \mathrm{gh}}{7}}\)
AP EAMCET (17.09.2020) Shift-II
Rotational Motion
150411
The centre of a wheel rolling on a plane surface move with a speed \(v_{0}\). A particle on the rim of the wheel at the same level as the centre will be moving at a speed
1 0
2 \(\mathrm{v}_{0}\)
3 \(\sqrt{2} \mathrm{v}_{0}\)
4 \(2 \mathrm{v}_{0}\)
Explanation:
C Given that, The velocity of the particle \(=\mathrm{v}\) The speed of the centre of wheel \(=\mathrm{v}_{0}\) Particle is on the rim of the wheel For Pure rolling \(\mathrm{v}_{0}=\mathrm{R} \omega\) At \(\mathrm{P}, \mathrm{v}=\mathrm{R} \omega\) Therefore we have \(v=\sqrt{\left(R^{2}+R^{2}\right)} \omega\) \(\mathrm{v} =\sqrt{2} \mathrm{R} \omega\) \(\mathrm{v} =\sqrt{2} \mathrm{v}_{0} \quad\{\text { from }(\mathrm{i})\}\)
AP EAMCET (21.09.2020) Shift-II
Rotational Motion
150412
A solid cylinder of mass \(M\) and radius \(R\) rolls down an inclined plane of length \(L\) and height \(h\), without slipping. Find the speed of its centre of mass when the cylinder reaches its bottom.
1 \(\sqrt{2 \mathrm{gh}}\)
2 \(\sqrt{\frac{3 g h}{4}}\)
3 \(\sqrt{\frac{4 \mathrm{gh}}{3}}\)
4 \(\sqrt{4 \mathrm{gh}}\)
Explanation:
C The given situation is shown in above figure when cylinder reaches at bottom, then its potential energy is converted into its rotational and linear kinetic energy of its centre of mass. Hence, By using work energy theorem, \(\mathrm{Mgh}=\frac{1}{2} \mathrm{I} \omega^{2}+\frac{1}{2} \mathrm{Mv}_{\mathrm{cm}}^{2}\) \(\mathrm{Mgh}=\frac{1}{2} \times \frac{\mathrm{MR}^{2}}{2} \times\left(\frac{\mathrm{v}_{\mathrm{cm}}}{\mathrm{R}}\right)^{2}+\frac{1}{2} \mathrm{Mv}_{\mathrm{cm}}^{2}\) \(=\frac{M v_{\mathrm{cm}}^2}{4}+\frac{\mathrm{Mv}_{\mathrm{cm}}^2}{2} \quad\left\{\begin{array}{l}\because \mathrm{I}=\frac{\mathrm{Mv}^2}{2} \text { and } \\ \omega=\frac{\mathrm{v}_{\mathrm{cm}}}{\mathrm{R}}\end{array}\right\}\) \(\mathrm{Mgh}=\frac{3}{4} \mathrm{Mv}_{\mathrm{cm}}^{2} \mathrm{v}_{\mathrm{cm}}^{2}=\frac{4 \mathrm{gh}}{3}\) \(\begin{aligned} & \mathrm{Mgh}=\frac{3}{4} \mathrm{Mv}_{\mathrm{cm}}^2 \\ & \mathrm{v}_{\mathrm{cm}}^2=\frac{4 \mathrm{gh}}{3} \\ & \mathrm{v}_{\mathrm{cm}}=\sqrt{\frac{4 \mathrm{gh}}{3}}\end{aligned}\)
150409
If mass, speed and radius of the circular path of the particle are increased by \(100 \%\), then the necessary force required to maintain the circular path will have to be increased by
1 \(100 \%\)
2 \(250 \%\)
3 \(300 \%\)
4 \(400 \%\)
Explanation:
C To the given data, mass \(=\mathrm{m}\) radius \(=\mathrm{r}\) velocity \(=\mathrm{v}\) \(\mathrm{F}=\) centripetal force We know that, applying the force \(\mathrm{F}_{\mathrm{c}}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) Now to the increment is \(100 \%\) in mass velocity radius \(\mathrm{m}=\mathrm{m}+\mathrm{m}=2 \mathrm{~m}\) \(\mathrm{v}=\mathrm{v}+\mathrm{v}=2 \mathrm{v}\) \(\mathrm{r}=\mathrm{r}+\mathrm{r}=2 \mathrm{r}\) Now hence, force is \(\mathrm{F}^{\prime}=\frac{(\mathrm{m}+\mathrm{m})(\mathrm{v}+\mathrm{v})^{2}}{\mathrm{r}+\mathrm{r}}\) \(\mathrm{F}^{\prime}=\frac{2 \mathrm{~m} \times 4 \mathrm{v}^{2}}{2 \mathrm{r}}=4\left(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\right)\) \(\mathrm{F}^{\prime}=\mathrm{F} \times 4\) Now the \(\%\) change in force \(=\frac{F^{\prime}-F}{F} \times 100=\frac{4 F-F}{F} \times 100=(4-1) \times 100\) \(=3 \times 100=300 \%\)
AP EAMCET (17.09.2020) Shift-I
Rotational Motion
150410
The speed of a uniform solid sphere after rolling down from rest without slipping along a fixed inclined plane of vertical height \(h\) is
1 \(\sqrt{\frac{10 g h}{7}}\)
2 \(\sqrt{g h}\)
3 \(\sqrt{\frac{6 g h}{5}}\)
4 \(\sqrt{\frac{4 g h}{3}}\)
Explanation:
A Using Work Energy Theorem, \(\mathrm{mgh}=\frac{1}{2} \mathrm{I} \omega^{2}+\frac{1}{2} \mathrm{mv}^{2}\) For solid sphere, moment of inertia \(\mathrm{I}=\frac{2}{5} \mathrm{mr}^{2}\) and \(\omega=\frac{\mathrm{v}}{\mathrm{r}}\) \(\mathrm{mgh}=\frac{1}{2} \times \frac{2}{5} \mathrm{mr}^{2} \times\left(\frac{\mathrm{v}}{\mathrm{r}}\right)^{2}+\frac{1}{2} \mathrm{mv}^{2}\) \(\mathrm{mgh}=\frac{1}{5} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{mv}^{2}\) \(\mathrm{mgh}=\frac{7}{10} \mathrm{mv}^{2}\) \(\mathrm{v}^{2}=\frac{10 \mathrm{gh}}{7}\) \(\mathrm{v}=\sqrt{\frac{10 \mathrm{gh}}{7}}\)
AP EAMCET (17.09.2020) Shift-II
Rotational Motion
150411
The centre of a wheel rolling on a plane surface move with a speed \(v_{0}\). A particle on the rim of the wheel at the same level as the centre will be moving at a speed
1 0
2 \(\mathrm{v}_{0}\)
3 \(\sqrt{2} \mathrm{v}_{0}\)
4 \(2 \mathrm{v}_{0}\)
Explanation:
C Given that, The velocity of the particle \(=\mathrm{v}\) The speed of the centre of wheel \(=\mathrm{v}_{0}\) Particle is on the rim of the wheel For Pure rolling \(\mathrm{v}_{0}=\mathrm{R} \omega\) At \(\mathrm{P}, \mathrm{v}=\mathrm{R} \omega\) Therefore we have \(v=\sqrt{\left(R^{2}+R^{2}\right)} \omega\) \(\mathrm{v} =\sqrt{2} \mathrm{R} \omega\) \(\mathrm{v} =\sqrt{2} \mathrm{v}_{0} \quad\{\text { from }(\mathrm{i})\}\)
AP EAMCET (21.09.2020) Shift-II
Rotational Motion
150412
A solid cylinder of mass \(M\) and radius \(R\) rolls down an inclined plane of length \(L\) and height \(h\), without slipping. Find the speed of its centre of mass when the cylinder reaches its bottom.
1 \(\sqrt{2 \mathrm{gh}}\)
2 \(\sqrt{\frac{3 g h}{4}}\)
3 \(\sqrt{\frac{4 \mathrm{gh}}{3}}\)
4 \(\sqrt{4 \mathrm{gh}}\)
Explanation:
C The given situation is shown in above figure when cylinder reaches at bottom, then its potential energy is converted into its rotational and linear kinetic energy of its centre of mass. Hence, By using work energy theorem, \(\mathrm{Mgh}=\frac{1}{2} \mathrm{I} \omega^{2}+\frac{1}{2} \mathrm{Mv}_{\mathrm{cm}}^{2}\) \(\mathrm{Mgh}=\frac{1}{2} \times \frac{\mathrm{MR}^{2}}{2} \times\left(\frac{\mathrm{v}_{\mathrm{cm}}}{\mathrm{R}}\right)^{2}+\frac{1}{2} \mathrm{Mv}_{\mathrm{cm}}^{2}\) \(=\frac{M v_{\mathrm{cm}}^2}{4}+\frac{\mathrm{Mv}_{\mathrm{cm}}^2}{2} \quad\left\{\begin{array}{l}\because \mathrm{I}=\frac{\mathrm{Mv}^2}{2} \text { and } \\ \omega=\frac{\mathrm{v}_{\mathrm{cm}}}{\mathrm{R}}\end{array}\right\}\) \(\mathrm{Mgh}=\frac{3}{4} \mathrm{Mv}_{\mathrm{cm}}^{2} \mathrm{v}_{\mathrm{cm}}^{2}=\frac{4 \mathrm{gh}}{3}\) \(\begin{aligned} & \mathrm{Mgh}=\frac{3}{4} \mathrm{Mv}_{\mathrm{cm}}^2 \\ & \mathrm{v}_{\mathrm{cm}}^2=\frac{4 \mathrm{gh}}{3} \\ & \mathrm{v}_{\mathrm{cm}}=\sqrt{\frac{4 \mathrm{gh}}{3}}\end{aligned}\)
150409
If mass, speed and radius of the circular path of the particle are increased by \(100 \%\), then the necessary force required to maintain the circular path will have to be increased by
1 \(100 \%\)
2 \(250 \%\)
3 \(300 \%\)
4 \(400 \%\)
Explanation:
C To the given data, mass \(=\mathrm{m}\) radius \(=\mathrm{r}\) velocity \(=\mathrm{v}\) \(\mathrm{F}=\) centripetal force We know that, applying the force \(\mathrm{F}_{\mathrm{c}}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) Now to the increment is \(100 \%\) in mass velocity radius \(\mathrm{m}=\mathrm{m}+\mathrm{m}=2 \mathrm{~m}\) \(\mathrm{v}=\mathrm{v}+\mathrm{v}=2 \mathrm{v}\) \(\mathrm{r}=\mathrm{r}+\mathrm{r}=2 \mathrm{r}\) Now hence, force is \(\mathrm{F}^{\prime}=\frac{(\mathrm{m}+\mathrm{m})(\mathrm{v}+\mathrm{v})^{2}}{\mathrm{r}+\mathrm{r}}\) \(\mathrm{F}^{\prime}=\frac{2 \mathrm{~m} \times 4 \mathrm{v}^{2}}{2 \mathrm{r}}=4\left(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\right)\) \(\mathrm{F}^{\prime}=\mathrm{F} \times 4\) Now the \(\%\) change in force \(=\frac{F^{\prime}-F}{F} \times 100=\frac{4 F-F}{F} \times 100=(4-1) \times 100\) \(=3 \times 100=300 \%\)
AP EAMCET (17.09.2020) Shift-I
Rotational Motion
150410
The speed of a uniform solid sphere after rolling down from rest without slipping along a fixed inclined plane of vertical height \(h\) is
1 \(\sqrt{\frac{10 g h}{7}}\)
2 \(\sqrt{g h}\)
3 \(\sqrt{\frac{6 g h}{5}}\)
4 \(\sqrt{\frac{4 g h}{3}}\)
Explanation:
A Using Work Energy Theorem, \(\mathrm{mgh}=\frac{1}{2} \mathrm{I} \omega^{2}+\frac{1}{2} \mathrm{mv}^{2}\) For solid sphere, moment of inertia \(\mathrm{I}=\frac{2}{5} \mathrm{mr}^{2}\) and \(\omega=\frac{\mathrm{v}}{\mathrm{r}}\) \(\mathrm{mgh}=\frac{1}{2} \times \frac{2}{5} \mathrm{mr}^{2} \times\left(\frac{\mathrm{v}}{\mathrm{r}}\right)^{2}+\frac{1}{2} \mathrm{mv}^{2}\) \(\mathrm{mgh}=\frac{1}{5} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{mv}^{2}\) \(\mathrm{mgh}=\frac{7}{10} \mathrm{mv}^{2}\) \(\mathrm{v}^{2}=\frac{10 \mathrm{gh}}{7}\) \(\mathrm{v}=\sqrt{\frac{10 \mathrm{gh}}{7}}\)
AP EAMCET (17.09.2020) Shift-II
Rotational Motion
150411
The centre of a wheel rolling on a plane surface move with a speed \(v_{0}\). A particle on the rim of the wheel at the same level as the centre will be moving at a speed
1 0
2 \(\mathrm{v}_{0}\)
3 \(\sqrt{2} \mathrm{v}_{0}\)
4 \(2 \mathrm{v}_{0}\)
Explanation:
C Given that, The velocity of the particle \(=\mathrm{v}\) The speed of the centre of wheel \(=\mathrm{v}_{0}\) Particle is on the rim of the wheel For Pure rolling \(\mathrm{v}_{0}=\mathrm{R} \omega\) At \(\mathrm{P}, \mathrm{v}=\mathrm{R} \omega\) Therefore we have \(v=\sqrt{\left(R^{2}+R^{2}\right)} \omega\) \(\mathrm{v} =\sqrt{2} \mathrm{R} \omega\) \(\mathrm{v} =\sqrt{2} \mathrm{v}_{0} \quad\{\text { from }(\mathrm{i})\}\)
AP EAMCET (21.09.2020) Shift-II
Rotational Motion
150412
A solid cylinder of mass \(M\) and radius \(R\) rolls down an inclined plane of length \(L\) and height \(h\), without slipping. Find the speed of its centre of mass when the cylinder reaches its bottom.
1 \(\sqrt{2 \mathrm{gh}}\)
2 \(\sqrt{\frac{3 g h}{4}}\)
3 \(\sqrt{\frac{4 \mathrm{gh}}{3}}\)
4 \(\sqrt{4 \mathrm{gh}}\)
Explanation:
C The given situation is shown in above figure when cylinder reaches at bottom, then its potential energy is converted into its rotational and linear kinetic energy of its centre of mass. Hence, By using work energy theorem, \(\mathrm{Mgh}=\frac{1}{2} \mathrm{I} \omega^{2}+\frac{1}{2} \mathrm{Mv}_{\mathrm{cm}}^{2}\) \(\mathrm{Mgh}=\frac{1}{2} \times \frac{\mathrm{MR}^{2}}{2} \times\left(\frac{\mathrm{v}_{\mathrm{cm}}}{\mathrm{R}}\right)^{2}+\frac{1}{2} \mathrm{Mv}_{\mathrm{cm}}^{2}\) \(=\frac{M v_{\mathrm{cm}}^2}{4}+\frac{\mathrm{Mv}_{\mathrm{cm}}^2}{2} \quad\left\{\begin{array}{l}\because \mathrm{I}=\frac{\mathrm{Mv}^2}{2} \text { and } \\ \omega=\frac{\mathrm{v}_{\mathrm{cm}}}{\mathrm{R}}\end{array}\right\}\) \(\mathrm{Mgh}=\frac{3}{4} \mathrm{Mv}_{\mathrm{cm}}^{2} \mathrm{v}_{\mathrm{cm}}^{2}=\frac{4 \mathrm{gh}}{3}\) \(\begin{aligned} & \mathrm{Mgh}=\frac{3}{4} \mathrm{Mv}_{\mathrm{cm}}^2 \\ & \mathrm{v}_{\mathrm{cm}}^2=\frac{4 \mathrm{gh}}{3} \\ & \mathrm{v}_{\mathrm{cm}}=\sqrt{\frac{4 \mathrm{gh}}{3}}\end{aligned}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Rotational Motion
150409
If mass, speed and radius of the circular path of the particle are increased by \(100 \%\), then the necessary force required to maintain the circular path will have to be increased by
1 \(100 \%\)
2 \(250 \%\)
3 \(300 \%\)
4 \(400 \%\)
Explanation:
C To the given data, mass \(=\mathrm{m}\) radius \(=\mathrm{r}\) velocity \(=\mathrm{v}\) \(\mathrm{F}=\) centripetal force We know that, applying the force \(\mathrm{F}_{\mathrm{c}}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) Now to the increment is \(100 \%\) in mass velocity radius \(\mathrm{m}=\mathrm{m}+\mathrm{m}=2 \mathrm{~m}\) \(\mathrm{v}=\mathrm{v}+\mathrm{v}=2 \mathrm{v}\) \(\mathrm{r}=\mathrm{r}+\mathrm{r}=2 \mathrm{r}\) Now hence, force is \(\mathrm{F}^{\prime}=\frac{(\mathrm{m}+\mathrm{m})(\mathrm{v}+\mathrm{v})^{2}}{\mathrm{r}+\mathrm{r}}\) \(\mathrm{F}^{\prime}=\frac{2 \mathrm{~m} \times 4 \mathrm{v}^{2}}{2 \mathrm{r}}=4\left(\frac{\mathrm{mv}^{2}}{\mathrm{r}}\right)\) \(\mathrm{F}^{\prime}=\mathrm{F} \times 4\) Now the \(\%\) change in force \(=\frac{F^{\prime}-F}{F} \times 100=\frac{4 F-F}{F} \times 100=(4-1) \times 100\) \(=3 \times 100=300 \%\)
AP EAMCET (17.09.2020) Shift-I
Rotational Motion
150410
The speed of a uniform solid sphere after rolling down from rest without slipping along a fixed inclined plane of vertical height \(h\) is
1 \(\sqrt{\frac{10 g h}{7}}\)
2 \(\sqrt{g h}\)
3 \(\sqrt{\frac{6 g h}{5}}\)
4 \(\sqrt{\frac{4 g h}{3}}\)
Explanation:
A Using Work Energy Theorem, \(\mathrm{mgh}=\frac{1}{2} \mathrm{I} \omega^{2}+\frac{1}{2} \mathrm{mv}^{2}\) For solid sphere, moment of inertia \(\mathrm{I}=\frac{2}{5} \mathrm{mr}^{2}\) and \(\omega=\frac{\mathrm{v}}{\mathrm{r}}\) \(\mathrm{mgh}=\frac{1}{2} \times \frac{2}{5} \mathrm{mr}^{2} \times\left(\frac{\mathrm{v}}{\mathrm{r}}\right)^{2}+\frac{1}{2} \mathrm{mv}^{2}\) \(\mathrm{mgh}=\frac{1}{5} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{mv}^{2}\) \(\mathrm{mgh}=\frac{7}{10} \mathrm{mv}^{2}\) \(\mathrm{v}^{2}=\frac{10 \mathrm{gh}}{7}\) \(\mathrm{v}=\sqrt{\frac{10 \mathrm{gh}}{7}}\)
AP EAMCET (17.09.2020) Shift-II
Rotational Motion
150411
The centre of a wheel rolling on a plane surface move with a speed \(v_{0}\). A particle on the rim of the wheel at the same level as the centre will be moving at a speed
1 0
2 \(\mathrm{v}_{0}\)
3 \(\sqrt{2} \mathrm{v}_{0}\)
4 \(2 \mathrm{v}_{0}\)
Explanation:
C Given that, The velocity of the particle \(=\mathrm{v}\) The speed of the centre of wheel \(=\mathrm{v}_{0}\) Particle is on the rim of the wheel For Pure rolling \(\mathrm{v}_{0}=\mathrm{R} \omega\) At \(\mathrm{P}, \mathrm{v}=\mathrm{R} \omega\) Therefore we have \(v=\sqrt{\left(R^{2}+R^{2}\right)} \omega\) \(\mathrm{v} =\sqrt{2} \mathrm{R} \omega\) \(\mathrm{v} =\sqrt{2} \mathrm{v}_{0} \quad\{\text { from }(\mathrm{i})\}\)
AP EAMCET (21.09.2020) Shift-II
Rotational Motion
150412
A solid cylinder of mass \(M\) and radius \(R\) rolls down an inclined plane of length \(L\) and height \(h\), without slipping. Find the speed of its centre of mass when the cylinder reaches its bottom.
1 \(\sqrt{2 \mathrm{gh}}\)
2 \(\sqrt{\frac{3 g h}{4}}\)
3 \(\sqrt{\frac{4 \mathrm{gh}}{3}}\)
4 \(\sqrt{4 \mathrm{gh}}\)
Explanation:
C The given situation is shown in above figure when cylinder reaches at bottom, then its potential energy is converted into its rotational and linear kinetic energy of its centre of mass. Hence, By using work energy theorem, \(\mathrm{Mgh}=\frac{1}{2} \mathrm{I} \omega^{2}+\frac{1}{2} \mathrm{Mv}_{\mathrm{cm}}^{2}\) \(\mathrm{Mgh}=\frac{1}{2} \times \frac{\mathrm{MR}^{2}}{2} \times\left(\frac{\mathrm{v}_{\mathrm{cm}}}{\mathrm{R}}\right)^{2}+\frac{1}{2} \mathrm{Mv}_{\mathrm{cm}}^{2}\) \(=\frac{M v_{\mathrm{cm}}^2}{4}+\frac{\mathrm{Mv}_{\mathrm{cm}}^2}{2} \quad\left\{\begin{array}{l}\because \mathrm{I}=\frac{\mathrm{Mv}^2}{2} \text { and } \\ \omega=\frac{\mathrm{v}_{\mathrm{cm}}}{\mathrm{R}}\end{array}\right\}\) \(\mathrm{Mgh}=\frac{3}{4} \mathrm{Mv}_{\mathrm{cm}}^{2} \mathrm{v}_{\mathrm{cm}}^{2}=\frac{4 \mathrm{gh}}{3}\) \(\begin{aligned} & \mathrm{Mgh}=\frac{3}{4} \mathrm{Mv}_{\mathrm{cm}}^2 \\ & \mathrm{v}_{\mathrm{cm}}^2=\frac{4 \mathrm{gh}}{3} \\ & \mathrm{v}_{\mathrm{cm}}=\sqrt{\frac{4 \mathrm{gh}}{3}}\end{aligned}\)
