149743
A loaded spring gun of mass \(M\) fires a bullet of mass \(m\) with a velocity \(v\) at an angle of elevation \(\theta\). The gun is initially at rest on a horizontal frictionless surface. After firing, the center of mass of the gun-bullet system
1 moves with a velocity \(\frac{\mathrm{v}(\mathrm{M}-\mathrm{m})}{\mathrm{M}+\mathrm{m}}\) in the horizontal direction
2 moves with velocity \(\frac{m v \cos \theta}{(M+m)}\)
3 moves with a velocity \(v\left(\frac{m}{M+m}\right)\)
4 moves with velocity \(\left(\frac{m v \sin \theta}{M+m}\right)\)
Explanation:
D Initially, The Gun and bullet both are at rest. Since there is no applied external force in the horizontal direction, the \(\mathrm{CM}\) of the gun and bullet will not have any acceleration. From the principle of conservationof momentum, the \(\mathrm{CM}\) will not move in the horizontal direction. After firing, Centre of mass along \(x\)-direction, we get \(\mathrm{v}(\mathrm{x})=\mathrm{v} \cos \theta(\mathrm{x})=0\) Centre of mass along \(y\)-direction \(\mathrm{v}_{\mathrm{y}}=\left(\frac{\mathrm{mv} \sin \theta}{\mathrm{m}+\mathrm{M}}\right)\)
TS EAMCET(Medical)-2015
Rotational Motion
149744
A small disc of radius \(2 \mathrm{~cm}\) is cut from a disc of radius \(6 \mathrm{~cm}\). If the distance between their centers 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 Given, Radius of complete disc (a) \(=6 \mathrm{~cm}\) Radius of small disc \((b)=2 \mathrm{~cm}\) Distance \(\left(\mathrm{x}_{1}\right)=3.2 \mathrm{~cm}\) \(\vec{x}_{\mathrm{cm}}=\frac{\mathrm{M} \times(0)+(-\mathrm{m}) \times\left(\mathrm{x}_{1}\right)}{\mathrm{M}-\mathrm{m}}, \quad \overrightarrow{\mathrm{y}}_{\mathrm{cm}}=\frac{\mathrm{M} \times 0+(-\mathrm{m}) \times(0)}{\mathrm{M}-\mathrm{m}}\) \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}}=\frac{-\mathrm{mx}_{1}}{\mathrm{M}-\mathrm{m}} \quad, \quad \overrightarrow{\mathrm{y}}_{\mathrm{cm}}=0\) So, position of new centre of mass is \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}}=\frac{-\mathrm{mx}_{1}}{\mathrm{M}-\mathrm{m}}\) Let, surface charge density of disc is \(\sigma\) So, surface charge density \(=\frac{\text { mass }}{\text { per unit area }}\) \(\sigma=\frac{\mathrm{M}}{\pi \mathrm{a}^{2}}\) \(\mathrm{M}=\sigma \pi \mathrm{a}^{2}\) Similarly, \(\sigma=\frac{m}{\pi b^{2}} \quad\) (a \& b are radius respectively) \(\mathrm{m}=\sigma \pi \mathrm{b}^{2}\) Putting value of \(M\) and \(m\) in equation (i), we get \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}}=\frac{-\sigma \pi \mathrm{b}^{2} \mathrm{x}_{1}}{\sigma \pi \mathrm{a}^{2}-\sigma \pi \mathrm{b}^{2}}\) \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}}=\frac{-\mathrm{b}^{2} \mathrm{x}_{1}}{\mathrm{a}^{2}-\mathrm{b}^{2}}\) \(=\frac{-(2)^{2} \times 3.2}{(6)^{2}-(2)^{2}}\) \(=-\frac{12.8}{36-4}=\frac{-12.8}{32}\) \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}} =-0.4 \mathrm{~cm}\) So, the shift in centre of mass of disc is \(0.4 \mathrm{~cm}\) in negative \(\mathrm{x}\) - direction.
JIPMER-2008
Rotational Motion
149745
A circular plate of uniform thickness has a diameter of \(56 \mathrm{~cm}\). A circular portion of diameter \(42 \mathrm{~cm}\) is removed from one edge of the plate as shown in the figure below. The position of centre of mass of the remaining portion is
1 \(7 \mathrm{~cm}\) to the left of the centre of plate
2 \(8 \mathrm{~cm}\) to the left of the centre of plate
3 \(9 \mathrm{~cm}\) to the left of the centre of plate
4 \(10 \mathrm{~cm}\) to the left of the centre of plate
Explanation:
C Let, \(\quad \mathrm{M}_{0}=\) mass of circular plate. \(\mathrm{M}_{0}=\rho \pi \mathrm{r}^{2}=\rho \pi(28)^{2}\) \(\mathrm{M}_{\mathrm{r}}=\) removed mass of circular plate \(\mathrm{M}_{\mathrm{r}}=\rho \pi(\mathrm{r})^{2}=\rho \pi(21)^{2}\) mass of remaining portion \(=M_{0}-M_{r}\) \(=\rho \pi(28)^{2}-\rho \pi(21)^{2}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\left[\rho \pi(28)^{2} \times 0\right]-\left[\rho \pi(21)^{2} \times 7\right]}{\rho \pi\left[(28)^{2}-(21)^{2}\right]}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{-(21)^{2} \times 7}{(28)^{2}-(21)^{2}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{-3087}{784-441}\) \(\mathrm{x}_{\mathrm{cm}}=-9 \mathrm{~cm}\) negative sign denote \(9 \mathrm{~cm}\) left to the centre of plate.
Assam CEE-2016
Rotational Motion
149746
A marble and a cube of equal mass are initially at rest. Now the marble rolls and cube slides down a frictionless ramp. When they arrive at the bottom, the ratio of speed of the cube with respect to the centre of mass and speed of the marble is
1 \(7: 5\)
2 \(\sqrt{7}: \sqrt{5}\)
3 \(5: 3\)
4 \(\sqrt{5}: \sqrt{3}\)
Explanation:
B For cube- \((\mathrm{KE})_{\mathrm{A}}+(\mathrm{PE})_{\mathrm{A}}=(\mathrm{KE})_{\mathrm{B}}+(\mathrm{PE})_{\mathrm{B}}\) \(0+\mathrm{mgh}=\frac{1}{2} \mathrm{mV}_{\mathrm{c}}^{2}+0\) \(\mathrm{mgh}=\frac{1}{2} \mathrm{mV}_{\mathrm{c}}^{2}\) \(\mathrm{~V}_{\mathrm{c}}=\sqrt{2 \mathrm{gh}}\) For marble- \((\mathrm{KE})_{\mathrm{A}}+(\mathrm{PE})_{\mathrm{A}}=(\mathrm{KE})_{\mathrm{B}}+(\mathrm{PE})_{\mathrm{B}}\) \(0+\mathrm{mgh}=\frac{1}{2} \mathrm{mV}_{\mathrm{m}}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}+0\) \(\mathrm{mgh}=\frac{1}{2} \mathrm{mV}_{\mathrm{m}}^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right)\left(\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{r}}\right)^{2}+0 \quad[\because \mathrm{v}=\mathrm{r} \omega]\) \(\mathrm{mgh}=\frac{1}{2} \mathrm{mV}_{\mathrm{m}}^{2}+\frac{2}{10} \mathrm{mV}_{\mathrm{m}}^{2}\) \(\mathrm{gh} =\frac{1}{2} \mathrm{~V}_{\mathrm{m}}^{2}+\frac{1}{5} \mathrm{~V}_{\mathrm{m}}^{2}\) \(\mathrm{gh} =\frac{7 \mathrm{~V}_{\mathrm{m}}^{2}}{10}\) \(\mathrm{~V}_{\mathrm{m}}^{2} =\frac{10}{7} \mathrm{gh}\) \(\mathrm{V}_{\mathrm{m}}=\sqrt{\frac{10}{7} \mathrm{gh}}\) Dividing equation (i) and (ii), we get - \(\frac{\mathrm{V}_{\mathrm{c}}}{\mathrm{V}_{\mathrm{m}}}=\sqrt{\frac{2 \mathrm{gh}}{\frac{10}{7} \mathrm{gh}}}\) \(\frac{\mathrm{V}_{\mathrm{c}}}{\mathrm{V}_{\mathrm{m}}}=\sqrt{\frac{7}{5}}\) \(\mathrm{~V}_{\mathrm{c}}: \mathrm{V}_{\mathrm{m}}=\sqrt{7}: \sqrt{5}\)
149743
A loaded spring gun of mass \(M\) fires a bullet of mass \(m\) with a velocity \(v\) at an angle of elevation \(\theta\). The gun is initially at rest on a horizontal frictionless surface. After firing, the center of mass of the gun-bullet system
1 moves with a velocity \(\frac{\mathrm{v}(\mathrm{M}-\mathrm{m})}{\mathrm{M}+\mathrm{m}}\) in the horizontal direction
2 moves with velocity \(\frac{m v \cos \theta}{(M+m)}\)
3 moves with a velocity \(v\left(\frac{m}{M+m}\right)\)
4 moves with velocity \(\left(\frac{m v \sin \theta}{M+m}\right)\)
Explanation:
D Initially, The Gun and bullet both are at rest. Since there is no applied external force in the horizontal direction, the \(\mathrm{CM}\) of the gun and bullet will not have any acceleration. From the principle of conservationof momentum, the \(\mathrm{CM}\) will not move in the horizontal direction. After firing, Centre of mass along \(x\)-direction, we get \(\mathrm{v}(\mathrm{x})=\mathrm{v} \cos \theta(\mathrm{x})=0\) Centre of mass along \(y\)-direction \(\mathrm{v}_{\mathrm{y}}=\left(\frac{\mathrm{mv} \sin \theta}{\mathrm{m}+\mathrm{M}}\right)\)
TS EAMCET(Medical)-2015
Rotational Motion
149744
A small disc of radius \(2 \mathrm{~cm}\) is cut from a disc of radius \(6 \mathrm{~cm}\). If the distance between their centers 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 Given, Radius of complete disc (a) \(=6 \mathrm{~cm}\) Radius of small disc \((b)=2 \mathrm{~cm}\) Distance \(\left(\mathrm{x}_{1}\right)=3.2 \mathrm{~cm}\) \(\vec{x}_{\mathrm{cm}}=\frac{\mathrm{M} \times(0)+(-\mathrm{m}) \times\left(\mathrm{x}_{1}\right)}{\mathrm{M}-\mathrm{m}}, \quad \overrightarrow{\mathrm{y}}_{\mathrm{cm}}=\frac{\mathrm{M} \times 0+(-\mathrm{m}) \times(0)}{\mathrm{M}-\mathrm{m}}\) \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}}=\frac{-\mathrm{mx}_{1}}{\mathrm{M}-\mathrm{m}} \quad, \quad \overrightarrow{\mathrm{y}}_{\mathrm{cm}}=0\) So, position of new centre of mass is \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}}=\frac{-\mathrm{mx}_{1}}{\mathrm{M}-\mathrm{m}}\) Let, surface charge density of disc is \(\sigma\) So, surface charge density \(=\frac{\text { mass }}{\text { per unit area }}\) \(\sigma=\frac{\mathrm{M}}{\pi \mathrm{a}^{2}}\) \(\mathrm{M}=\sigma \pi \mathrm{a}^{2}\) Similarly, \(\sigma=\frac{m}{\pi b^{2}} \quad\) (a \& b are radius respectively) \(\mathrm{m}=\sigma \pi \mathrm{b}^{2}\) Putting value of \(M\) and \(m\) in equation (i), we get \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}}=\frac{-\sigma \pi \mathrm{b}^{2} \mathrm{x}_{1}}{\sigma \pi \mathrm{a}^{2}-\sigma \pi \mathrm{b}^{2}}\) \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}}=\frac{-\mathrm{b}^{2} \mathrm{x}_{1}}{\mathrm{a}^{2}-\mathrm{b}^{2}}\) \(=\frac{-(2)^{2} \times 3.2}{(6)^{2}-(2)^{2}}\) \(=-\frac{12.8}{36-4}=\frac{-12.8}{32}\) \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}} =-0.4 \mathrm{~cm}\) So, the shift in centre of mass of disc is \(0.4 \mathrm{~cm}\) in negative \(\mathrm{x}\) - direction.
JIPMER-2008
Rotational Motion
149745
A circular plate of uniform thickness has a diameter of \(56 \mathrm{~cm}\). A circular portion of diameter \(42 \mathrm{~cm}\) is removed from one edge of the plate as shown in the figure below. The position of centre of mass of the remaining portion is
1 \(7 \mathrm{~cm}\) to the left of the centre of plate
2 \(8 \mathrm{~cm}\) to the left of the centre of plate
3 \(9 \mathrm{~cm}\) to the left of the centre of plate
4 \(10 \mathrm{~cm}\) to the left of the centre of plate
Explanation:
C Let, \(\quad \mathrm{M}_{0}=\) mass of circular plate. \(\mathrm{M}_{0}=\rho \pi \mathrm{r}^{2}=\rho \pi(28)^{2}\) \(\mathrm{M}_{\mathrm{r}}=\) removed mass of circular plate \(\mathrm{M}_{\mathrm{r}}=\rho \pi(\mathrm{r})^{2}=\rho \pi(21)^{2}\) mass of remaining portion \(=M_{0}-M_{r}\) \(=\rho \pi(28)^{2}-\rho \pi(21)^{2}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\left[\rho \pi(28)^{2} \times 0\right]-\left[\rho \pi(21)^{2} \times 7\right]}{\rho \pi\left[(28)^{2}-(21)^{2}\right]}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{-(21)^{2} \times 7}{(28)^{2}-(21)^{2}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{-3087}{784-441}\) \(\mathrm{x}_{\mathrm{cm}}=-9 \mathrm{~cm}\) negative sign denote \(9 \mathrm{~cm}\) left to the centre of plate.
Assam CEE-2016
Rotational Motion
149746
A marble and a cube of equal mass are initially at rest. Now the marble rolls and cube slides down a frictionless ramp. When they arrive at the bottom, the ratio of speed of the cube with respect to the centre of mass and speed of the marble is
1 \(7: 5\)
2 \(\sqrt{7}: \sqrt{5}\)
3 \(5: 3\)
4 \(\sqrt{5}: \sqrt{3}\)
Explanation:
B For cube- \((\mathrm{KE})_{\mathrm{A}}+(\mathrm{PE})_{\mathrm{A}}=(\mathrm{KE})_{\mathrm{B}}+(\mathrm{PE})_{\mathrm{B}}\) \(0+\mathrm{mgh}=\frac{1}{2} \mathrm{mV}_{\mathrm{c}}^{2}+0\) \(\mathrm{mgh}=\frac{1}{2} \mathrm{mV}_{\mathrm{c}}^{2}\) \(\mathrm{~V}_{\mathrm{c}}=\sqrt{2 \mathrm{gh}}\) For marble- \((\mathrm{KE})_{\mathrm{A}}+(\mathrm{PE})_{\mathrm{A}}=(\mathrm{KE})_{\mathrm{B}}+(\mathrm{PE})_{\mathrm{B}}\) \(0+\mathrm{mgh}=\frac{1}{2} \mathrm{mV}_{\mathrm{m}}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}+0\) \(\mathrm{mgh}=\frac{1}{2} \mathrm{mV}_{\mathrm{m}}^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right)\left(\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{r}}\right)^{2}+0 \quad[\because \mathrm{v}=\mathrm{r} \omega]\) \(\mathrm{mgh}=\frac{1}{2} \mathrm{mV}_{\mathrm{m}}^{2}+\frac{2}{10} \mathrm{mV}_{\mathrm{m}}^{2}\) \(\mathrm{gh} =\frac{1}{2} \mathrm{~V}_{\mathrm{m}}^{2}+\frac{1}{5} \mathrm{~V}_{\mathrm{m}}^{2}\) \(\mathrm{gh} =\frac{7 \mathrm{~V}_{\mathrm{m}}^{2}}{10}\) \(\mathrm{~V}_{\mathrm{m}}^{2} =\frac{10}{7} \mathrm{gh}\) \(\mathrm{V}_{\mathrm{m}}=\sqrt{\frac{10}{7} \mathrm{gh}}\) Dividing equation (i) and (ii), we get - \(\frac{\mathrm{V}_{\mathrm{c}}}{\mathrm{V}_{\mathrm{m}}}=\sqrt{\frac{2 \mathrm{gh}}{\frac{10}{7} \mathrm{gh}}}\) \(\frac{\mathrm{V}_{\mathrm{c}}}{\mathrm{V}_{\mathrm{m}}}=\sqrt{\frac{7}{5}}\) \(\mathrm{~V}_{\mathrm{c}}: \mathrm{V}_{\mathrm{m}}=\sqrt{7}: \sqrt{5}\)
149743
A loaded spring gun of mass \(M\) fires a bullet of mass \(m\) with a velocity \(v\) at an angle of elevation \(\theta\). The gun is initially at rest on a horizontal frictionless surface. After firing, the center of mass of the gun-bullet system
1 moves with a velocity \(\frac{\mathrm{v}(\mathrm{M}-\mathrm{m})}{\mathrm{M}+\mathrm{m}}\) in the horizontal direction
2 moves with velocity \(\frac{m v \cos \theta}{(M+m)}\)
3 moves with a velocity \(v\left(\frac{m}{M+m}\right)\)
4 moves with velocity \(\left(\frac{m v \sin \theta}{M+m}\right)\)
Explanation:
D Initially, The Gun and bullet both are at rest. Since there is no applied external force in the horizontal direction, the \(\mathrm{CM}\) of the gun and bullet will not have any acceleration. From the principle of conservationof momentum, the \(\mathrm{CM}\) will not move in the horizontal direction. After firing, Centre of mass along \(x\)-direction, we get \(\mathrm{v}(\mathrm{x})=\mathrm{v} \cos \theta(\mathrm{x})=0\) Centre of mass along \(y\)-direction \(\mathrm{v}_{\mathrm{y}}=\left(\frac{\mathrm{mv} \sin \theta}{\mathrm{m}+\mathrm{M}}\right)\)
TS EAMCET(Medical)-2015
Rotational Motion
149744
A small disc of radius \(2 \mathrm{~cm}\) is cut from a disc of radius \(6 \mathrm{~cm}\). If the distance between their centers 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 Given, Radius of complete disc (a) \(=6 \mathrm{~cm}\) Radius of small disc \((b)=2 \mathrm{~cm}\) Distance \(\left(\mathrm{x}_{1}\right)=3.2 \mathrm{~cm}\) \(\vec{x}_{\mathrm{cm}}=\frac{\mathrm{M} \times(0)+(-\mathrm{m}) \times\left(\mathrm{x}_{1}\right)}{\mathrm{M}-\mathrm{m}}, \quad \overrightarrow{\mathrm{y}}_{\mathrm{cm}}=\frac{\mathrm{M} \times 0+(-\mathrm{m}) \times(0)}{\mathrm{M}-\mathrm{m}}\) \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}}=\frac{-\mathrm{mx}_{1}}{\mathrm{M}-\mathrm{m}} \quad, \quad \overrightarrow{\mathrm{y}}_{\mathrm{cm}}=0\) So, position of new centre of mass is \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}}=\frac{-\mathrm{mx}_{1}}{\mathrm{M}-\mathrm{m}}\) Let, surface charge density of disc is \(\sigma\) So, surface charge density \(=\frac{\text { mass }}{\text { per unit area }}\) \(\sigma=\frac{\mathrm{M}}{\pi \mathrm{a}^{2}}\) \(\mathrm{M}=\sigma \pi \mathrm{a}^{2}\) Similarly, \(\sigma=\frac{m}{\pi b^{2}} \quad\) (a \& b are radius respectively) \(\mathrm{m}=\sigma \pi \mathrm{b}^{2}\) Putting value of \(M\) and \(m\) in equation (i), we get \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}}=\frac{-\sigma \pi \mathrm{b}^{2} \mathrm{x}_{1}}{\sigma \pi \mathrm{a}^{2}-\sigma \pi \mathrm{b}^{2}}\) \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}}=\frac{-\mathrm{b}^{2} \mathrm{x}_{1}}{\mathrm{a}^{2}-\mathrm{b}^{2}}\) \(=\frac{-(2)^{2} \times 3.2}{(6)^{2}-(2)^{2}}\) \(=-\frac{12.8}{36-4}=\frac{-12.8}{32}\) \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}} =-0.4 \mathrm{~cm}\) So, the shift in centre of mass of disc is \(0.4 \mathrm{~cm}\) in negative \(\mathrm{x}\) - direction.
JIPMER-2008
Rotational Motion
149745
A circular plate of uniform thickness has a diameter of \(56 \mathrm{~cm}\). A circular portion of diameter \(42 \mathrm{~cm}\) is removed from one edge of the plate as shown in the figure below. The position of centre of mass of the remaining portion is
1 \(7 \mathrm{~cm}\) to the left of the centre of plate
2 \(8 \mathrm{~cm}\) to the left of the centre of plate
3 \(9 \mathrm{~cm}\) to the left of the centre of plate
4 \(10 \mathrm{~cm}\) to the left of the centre of plate
Explanation:
C Let, \(\quad \mathrm{M}_{0}=\) mass of circular plate. \(\mathrm{M}_{0}=\rho \pi \mathrm{r}^{2}=\rho \pi(28)^{2}\) \(\mathrm{M}_{\mathrm{r}}=\) removed mass of circular plate \(\mathrm{M}_{\mathrm{r}}=\rho \pi(\mathrm{r})^{2}=\rho \pi(21)^{2}\) mass of remaining portion \(=M_{0}-M_{r}\) \(=\rho \pi(28)^{2}-\rho \pi(21)^{2}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\left[\rho \pi(28)^{2} \times 0\right]-\left[\rho \pi(21)^{2} \times 7\right]}{\rho \pi\left[(28)^{2}-(21)^{2}\right]}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{-(21)^{2} \times 7}{(28)^{2}-(21)^{2}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{-3087}{784-441}\) \(\mathrm{x}_{\mathrm{cm}}=-9 \mathrm{~cm}\) negative sign denote \(9 \mathrm{~cm}\) left to the centre of plate.
Assam CEE-2016
Rotational Motion
149746
A marble and a cube of equal mass are initially at rest. Now the marble rolls and cube slides down a frictionless ramp. When they arrive at the bottom, the ratio of speed of the cube with respect to the centre of mass and speed of the marble is
1 \(7: 5\)
2 \(\sqrt{7}: \sqrt{5}\)
3 \(5: 3\)
4 \(\sqrt{5}: \sqrt{3}\)
Explanation:
B For cube- \((\mathrm{KE})_{\mathrm{A}}+(\mathrm{PE})_{\mathrm{A}}=(\mathrm{KE})_{\mathrm{B}}+(\mathrm{PE})_{\mathrm{B}}\) \(0+\mathrm{mgh}=\frac{1}{2} \mathrm{mV}_{\mathrm{c}}^{2}+0\) \(\mathrm{mgh}=\frac{1}{2} \mathrm{mV}_{\mathrm{c}}^{2}\) \(\mathrm{~V}_{\mathrm{c}}=\sqrt{2 \mathrm{gh}}\) For marble- \((\mathrm{KE})_{\mathrm{A}}+(\mathrm{PE})_{\mathrm{A}}=(\mathrm{KE})_{\mathrm{B}}+(\mathrm{PE})_{\mathrm{B}}\) \(0+\mathrm{mgh}=\frac{1}{2} \mathrm{mV}_{\mathrm{m}}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}+0\) \(\mathrm{mgh}=\frac{1}{2} \mathrm{mV}_{\mathrm{m}}^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right)\left(\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{r}}\right)^{2}+0 \quad[\because \mathrm{v}=\mathrm{r} \omega]\) \(\mathrm{mgh}=\frac{1}{2} \mathrm{mV}_{\mathrm{m}}^{2}+\frac{2}{10} \mathrm{mV}_{\mathrm{m}}^{2}\) \(\mathrm{gh} =\frac{1}{2} \mathrm{~V}_{\mathrm{m}}^{2}+\frac{1}{5} \mathrm{~V}_{\mathrm{m}}^{2}\) \(\mathrm{gh} =\frac{7 \mathrm{~V}_{\mathrm{m}}^{2}}{10}\) \(\mathrm{~V}_{\mathrm{m}}^{2} =\frac{10}{7} \mathrm{gh}\) \(\mathrm{V}_{\mathrm{m}}=\sqrt{\frac{10}{7} \mathrm{gh}}\) Dividing equation (i) and (ii), we get - \(\frac{\mathrm{V}_{\mathrm{c}}}{\mathrm{V}_{\mathrm{m}}}=\sqrt{\frac{2 \mathrm{gh}}{\frac{10}{7} \mathrm{gh}}}\) \(\frac{\mathrm{V}_{\mathrm{c}}}{\mathrm{V}_{\mathrm{m}}}=\sqrt{\frac{7}{5}}\) \(\mathrm{~V}_{\mathrm{c}}: \mathrm{V}_{\mathrm{m}}=\sqrt{7}: \sqrt{5}\)
149743
A loaded spring gun of mass \(M\) fires a bullet of mass \(m\) with a velocity \(v\) at an angle of elevation \(\theta\). The gun is initially at rest on a horizontal frictionless surface. After firing, the center of mass of the gun-bullet system
1 moves with a velocity \(\frac{\mathrm{v}(\mathrm{M}-\mathrm{m})}{\mathrm{M}+\mathrm{m}}\) in the horizontal direction
2 moves with velocity \(\frac{m v \cos \theta}{(M+m)}\)
3 moves with a velocity \(v\left(\frac{m}{M+m}\right)\)
4 moves with velocity \(\left(\frac{m v \sin \theta}{M+m}\right)\)
Explanation:
D Initially, The Gun and bullet both are at rest. Since there is no applied external force in the horizontal direction, the \(\mathrm{CM}\) of the gun and bullet will not have any acceleration. From the principle of conservationof momentum, the \(\mathrm{CM}\) will not move in the horizontal direction. After firing, Centre of mass along \(x\)-direction, we get \(\mathrm{v}(\mathrm{x})=\mathrm{v} \cos \theta(\mathrm{x})=0\) Centre of mass along \(y\)-direction \(\mathrm{v}_{\mathrm{y}}=\left(\frac{\mathrm{mv} \sin \theta}{\mathrm{m}+\mathrm{M}}\right)\)
TS EAMCET(Medical)-2015
Rotational Motion
149744
A small disc of radius \(2 \mathrm{~cm}\) is cut from a disc of radius \(6 \mathrm{~cm}\). If the distance between their centers 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 Given, Radius of complete disc (a) \(=6 \mathrm{~cm}\) Radius of small disc \((b)=2 \mathrm{~cm}\) Distance \(\left(\mathrm{x}_{1}\right)=3.2 \mathrm{~cm}\) \(\vec{x}_{\mathrm{cm}}=\frac{\mathrm{M} \times(0)+(-\mathrm{m}) \times\left(\mathrm{x}_{1}\right)}{\mathrm{M}-\mathrm{m}}, \quad \overrightarrow{\mathrm{y}}_{\mathrm{cm}}=\frac{\mathrm{M} \times 0+(-\mathrm{m}) \times(0)}{\mathrm{M}-\mathrm{m}}\) \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}}=\frac{-\mathrm{mx}_{1}}{\mathrm{M}-\mathrm{m}} \quad, \quad \overrightarrow{\mathrm{y}}_{\mathrm{cm}}=0\) So, position of new centre of mass is \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}}=\frac{-\mathrm{mx}_{1}}{\mathrm{M}-\mathrm{m}}\) Let, surface charge density of disc is \(\sigma\) So, surface charge density \(=\frac{\text { mass }}{\text { per unit area }}\) \(\sigma=\frac{\mathrm{M}}{\pi \mathrm{a}^{2}}\) \(\mathrm{M}=\sigma \pi \mathrm{a}^{2}\) Similarly, \(\sigma=\frac{m}{\pi b^{2}} \quad\) (a \& b are radius respectively) \(\mathrm{m}=\sigma \pi \mathrm{b}^{2}\) Putting value of \(M\) and \(m\) in equation (i), we get \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}}=\frac{-\sigma \pi \mathrm{b}^{2} \mathrm{x}_{1}}{\sigma \pi \mathrm{a}^{2}-\sigma \pi \mathrm{b}^{2}}\) \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}}=\frac{-\mathrm{b}^{2} \mathrm{x}_{1}}{\mathrm{a}^{2}-\mathrm{b}^{2}}\) \(=\frac{-(2)^{2} \times 3.2}{(6)^{2}-(2)^{2}}\) \(=-\frac{12.8}{36-4}=\frac{-12.8}{32}\) \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}} =-0.4 \mathrm{~cm}\) So, the shift in centre of mass of disc is \(0.4 \mathrm{~cm}\) in negative \(\mathrm{x}\) - direction.
JIPMER-2008
Rotational Motion
149745
A circular plate of uniform thickness has a diameter of \(56 \mathrm{~cm}\). A circular portion of diameter \(42 \mathrm{~cm}\) is removed from one edge of the plate as shown in the figure below. The position of centre of mass of the remaining portion is
1 \(7 \mathrm{~cm}\) to the left of the centre of plate
2 \(8 \mathrm{~cm}\) to the left of the centre of plate
3 \(9 \mathrm{~cm}\) to the left of the centre of plate
4 \(10 \mathrm{~cm}\) to the left of the centre of plate
Explanation:
C Let, \(\quad \mathrm{M}_{0}=\) mass of circular plate. \(\mathrm{M}_{0}=\rho \pi \mathrm{r}^{2}=\rho \pi(28)^{2}\) \(\mathrm{M}_{\mathrm{r}}=\) removed mass of circular plate \(\mathrm{M}_{\mathrm{r}}=\rho \pi(\mathrm{r})^{2}=\rho \pi(21)^{2}\) mass of remaining portion \(=M_{0}-M_{r}\) \(=\rho \pi(28)^{2}-\rho \pi(21)^{2}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{\left[\rho \pi(28)^{2} \times 0\right]-\left[\rho \pi(21)^{2} \times 7\right]}{\rho \pi\left[(28)^{2}-(21)^{2}\right]}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{-(21)^{2} \times 7}{(28)^{2}-(21)^{2}}\) \(\mathrm{x}_{\mathrm{cm}}=\frac{-3087}{784-441}\) \(\mathrm{x}_{\mathrm{cm}}=-9 \mathrm{~cm}\) negative sign denote \(9 \mathrm{~cm}\) left to the centre of plate.
Assam CEE-2016
Rotational Motion
149746
A marble and a cube of equal mass are initially at rest. Now the marble rolls and cube slides down a frictionless ramp. When they arrive at the bottom, the ratio of speed of the cube with respect to the centre of mass and speed of the marble is
1 \(7: 5\)
2 \(\sqrt{7}: \sqrt{5}\)
3 \(5: 3\)
4 \(\sqrt{5}: \sqrt{3}\)
Explanation:
B For cube- \((\mathrm{KE})_{\mathrm{A}}+(\mathrm{PE})_{\mathrm{A}}=(\mathrm{KE})_{\mathrm{B}}+(\mathrm{PE})_{\mathrm{B}}\) \(0+\mathrm{mgh}=\frac{1}{2} \mathrm{mV}_{\mathrm{c}}^{2}+0\) \(\mathrm{mgh}=\frac{1}{2} \mathrm{mV}_{\mathrm{c}}^{2}\) \(\mathrm{~V}_{\mathrm{c}}=\sqrt{2 \mathrm{gh}}\) For marble- \((\mathrm{KE})_{\mathrm{A}}+(\mathrm{PE})_{\mathrm{A}}=(\mathrm{KE})_{\mathrm{B}}+(\mathrm{PE})_{\mathrm{B}}\) \(0+\mathrm{mgh}=\frac{1}{2} \mathrm{mV}_{\mathrm{m}}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}+0\) \(\mathrm{mgh}=\frac{1}{2} \mathrm{mV}_{\mathrm{m}}^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{mr}^{2}\right)\left(\frac{\mathrm{V}_{\mathrm{m}}}{\mathrm{r}}\right)^{2}+0 \quad[\because \mathrm{v}=\mathrm{r} \omega]\) \(\mathrm{mgh}=\frac{1}{2} \mathrm{mV}_{\mathrm{m}}^{2}+\frac{2}{10} \mathrm{mV}_{\mathrm{m}}^{2}\) \(\mathrm{gh} =\frac{1}{2} \mathrm{~V}_{\mathrm{m}}^{2}+\frac{1}{5} \mathrm{~V}_{\mathrm{m}}^{2}\) \(\mathrm{gh} =\frac{7 \mathrm{~V}_{\mathrm{m}}^{2}}{10}\) \(\mathrm{~V}_{\mathrm{m}}^{2} =\frac{10}{7} \mathrm{gh}\) \(\mathrm{V}_{\mathrm{m}}=\sqrt{\frac{10}{7} \mathrm{gh}}\) Dividing equation (i) and (ii), we get - \(\frac{\mathrm{V}_{\mathrm{c}}}{\mathrm{V}_{\mathrm{m}}}=\sqrt{\frac{2 \mathrm{gh}}{\frac{10}{7} \mathrm{gh}}}\) \(\frac{\mathrm{V}_{\mathrm{c}}}{\mathrm{V}_{\mathrm{m}}}=\sqrt{\frac{7}{5}}\) \(\mathrm{~V}_{\mathrm{c}}: \mathrm{V}_{\mathrm{m}}=\sqrt{7}: \sqrt{5}\)
