NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Rotational Motion
149747
A body of mass \(m_{1}=4 \mathrm{~kg}\) moves at \(5 \hat{i} \mathrm{~m} / \mathrm{s}\) and another body of mass \(m_{2}=2 \mathrm{~kg}\) moves at \(10 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s}\). The kinetic energy of centre of mass is
1 \(\frac{200}{3} \mathrm{~J}\)
2 \(\frac{500}{3} \mathrm{~J}\)
3 \(\frac{400}{3} \mathrm{~J}\)
4 \(\frac{800}{3} \mathrm{~J}\)
Explanation:
C Given that, \(\mathrm{m}_{1}=4 \mathrm{~kg}, \mathrm{~m}_{2}=2 \mathrm{~kg}\) \(\frac{\mathrm{dr}_{1}}{\mathrm{dt}}=5 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}, \frac{\mathrm{dr}_{2}}{\mathrm{dt}}=10 \hat{\mathrm{i} \mathrm{m} / \mathrm{s}}\) Velocity of centre of mass \(=\) \(\mathrm{v}_{\mathrm{cm}}=\frac{\mathrm{m}_{1} \frac{\mathrm{dr}_{1}}{\mathrm{dt}}+\mathrm{m}_{2} \frac{\mathrm{dr}_{2}}{\mathrm{dt}}}{\mathrm{m}_{1}+\mathrm{m}_{2}}=\frac{(4 \times 5) \mathrm{i}+(2 \times 10) \mathrm{i}}{4+2}\) \(\mathrm{v}_{\mathrm{cm}}=\frac{40}{6} \hat{\mathrm{i}}=\frac{20}{3} \hat{\mathrm{i}}\) The kinetic energy of the centre of the mass of system- \(\mathrm{K} . \mathrm{E}_{\mathrm{cm}} =\frac{1}{2} \mathrm{mv}_{\mathrm{cm}}^{2}\) \(=\frac{1}{2}(4+2) \times \frac{20}{3} \times \frac{20}{3}\) \(\mathrm{~K} . \mathrm{E}_{\mathrm{cm}} =\frac{400}{3} \mathrm{~J}\)
AP EAMCET -2010
Rotational Motion
149750
Two masses of 6 and 2 units are at positions \((6 \hat{\mathbf{i}}-7 \hat{\mathbf{j}})\) and \((2 \hat{\mathbf{i}}+\mathbf{5} \hat{\mathbf{j}}-8 \hat{\mathbf{k}})\) respectively. The co-ordinates of the centre-of-mass (c.m.) are
1 \((2,-5,3)\)
2 \((5,-5,-3)\)
3 \((5,-4,-2)\)
4 \((5,-4,-4)\)
Explanation:
C Given, \(\mathrm{m}_{1}=6 \text { unit }\) \(\mathrm{m}_{2}=2 \text { unit }\) \(\overrightarrow{\mathrm{r}}_{1}=(6 \hat{\mathrm{i}}-7 \hat{\mathrm{j}})\) \(\overrightarrow{\mathrm{r}}_{2}=(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-8 \hat{\mathrm{k}})\) Co-ordinate of the centre of mass \(\left(\vec{x}_{\mathrm{cm}}\right)=\frac{\mathrm{m}_{1} \overrightarrow{\mathrm{r}}_{1}+\mathrm{m}_{1} \overrightarrow{\mathrm{r}}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}} =\frac{6(6 \hat{\mathrm{i}}-7 \hat{\mathrm{j}})+2 \times(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-8 \hat{\mathrm{k}})}{6+2}\) \(=\frac{36 \hat{\mathrm{i}}-42 \hat{\mathrm{j}}+4 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-16 \hat{\mathrm{k}}}{8}\) \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}} =5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}\)
AMU-2014
Rotational Motion
149751
With \(\mathrm{O}\) as the origin of the coordinate axis, the \(X\) and \(Y\) - coordinates of the centre of mass of the system of particles shown in the figure may be given as
C Here \(\mathrm{m}\) and \(2 \mathrm{~m}\) represent the masses of the particles. The taking moments with respect to \(0 \overrightarrow{\mathrm{x}}_{\mathrm{cm}}\) is \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}} =\frac{\mathrm{m}(-2 \mathrm{~b})+2 \mathrm{~m}(-\mathrm{b})+\mathrm{m} \times 0+2 \mathrm{~m}(\mathrm{~b})}{\mathrm{m}+2 \mathrm{~m}+\mathrm{m}+2 \mathrm{~m}}\) \(=-\frac{\mathrm{b}}{3}\) \(\overrightarrow{\mathrm{y}}_{\mathrm{cm}} =\mathrm{b}\) such that at the \(\overrightarrow{\mathrm{R}}_{\mathrm{cm}}=\left(\frac{-\mathrm{b}}{3}, \mathrm{~b}\right)\)
AMU-2006
Rotational Motion
149752
A straight rod of length \(L\) has one of its ends at the origin and the other at \(x=L\). If the mass per unit length of the rod is given by Ax where \(A\) is a constant, where is its mass centre
1 \(\mathrm{L} / 3\)
2 \(\mathrm{L} / 2\)
3 \(2 \mathrm{~L} / 3\)
4 \(3 \mathrm{~L} / 4\)
Explanation:
C At let to be assume A is a constant \(\mathrm{x}_{\mathrm{c}}=\) coordinate of center of mass At given data, \(\frac{M}{L}=A x\) \(X_{\text {COM }}=\frac{\int x \cdot d m}{\int d m}\) \(=\frac{\int_{0}^{L} x \cdot A x d x}{\left\{\int_{0}^{L} A x d x\right\}} \quad \text { T Total Mass }=\int_{0}^{L} A x d x \text { ) }\) \(\frac{d m}{d x}=A \cdot x\) \(d m=A x \cdot d x\) \(X_{\text {COM }}=\frac{\int_{0}^{L} x^{2} d x}{\int_{0}^{L} x \cdot d x}\) \(X_{\text {COM }}=\frac{\left[\frac{x^{3}}{3}\right]_{0}^{L}}{\left[\frac{x^{2}}{2}\right]_{0}^{L}}\) \(X_{\text {COM }}=\frac{\frac{L^{3}}{L^{2}}}{\frac{2}{2}}\) \(X_{\text {COM }}=\frac{2 L}{3}\)
149747
A body of mass \(m_{1}=4 \mathrm{~kg}\) moves at \(5 \hat{i} \mathrm{~m} / \mathrm{s}\) and another body of mass \(m_{2}=2 \mathrm{~kg}\) moves at \(10 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s}\). The kinetic energy of centre of mass is
1 \(\frac{200}{3} \mathrm{~J}\)
2 \(\frac{500}{3} \mathrm{~J}\)
3 \(\frac{400}{3} \mathrm{~J}\)
4 \(\frac{800}{3} \mathrm{~J}\)
Explanation:
C Given that, \(\mathrm{m}_{1}=4 \mathrm{~kg}, \mathrm{~m}_{2}=2 \mathrm{~kg}\) \(\frac{\mathrm{dr}_{1}}{\mathrm{dt}}=5 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}, \frac{\mathrm{dr}_{2}}{\mathrm{dt}}=10 \hat{\mathrm{i} \mathrm{m} / \mathrm{s}}\) Velocity of centre of mass \(=\) \(\mathrm{v}_{\mathrm{cm}}=\frac{\mathrm{m}_{1} \frac{\mathrm{dr}_{1}}{\mathrm{dt}}+\mathrm{m}_{2} \frac{\mathrm{dr}_{2}}{\mathrm{dt}}}{\mathrm{m}_{1}+\mathrm{m}_{2}}=\frac{(4 \times 5) \mathrm{i}+(2 \times 10) \mathrm{i}}{4+2}\) \(\mathrm{v}_{\mathrm{cm}}=\frac{40}{6} \hat{\mathrm{i}}=\frac{20}{3} \hat{\mathrm{i}}\) The kinetic energy of the centre of the mass of system- \(\mathrm{K} . \mathrm{E}_{\mathrm{cm}} =\frac{1}{2} \mathrm{mv}_{\mathrm{cm}}^{2}\) \(=\frac{1}{2}(4+2) \times \frac{20}{3} \times \frac{20}{3}\) \(\mathrm{~K} . \mathrm{E}_{\mathrm{cm}} =\frac{400}{3} \mathrm{~J}\)
AP EAMCET -2010
Rotational Motion
149750
Two masses of 6 and 2 units are at positions \((6 \hat{\mathbf{i}}-7 \hat{\mathbf{j}})\) and \((2 \hat{\mathbf{i}}+\mathbf{5} \hat{\mathbf{j}}-8 \hat{\mathbf{k}})\) respectively. The co-ordinates of the centre-of-mass (c.m.) are
1 \((2,-5,3)\)
2 \((5,-5,-3)\)
3 \((5,-4,-2)\)
4 \((5,-4,-4)\)
Explanation:
C Given, \(\mathrm{m}_{1}=6 \text { unit }\) \(\mathrm{m}_{2}=2 \text { unit }\) \(\overrightarrow{\mathrm{r}}_{1}=(6 \hat{\mathrm{i}}-7 \hat{\mathrm{j}})\) \(\overrightarrow{\mathrm{r}}_{2}=(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-8 \hat{\mathrm{k}})\) Co-ordinate of the centre of mass \(\left(\vec{x}_{\mathrm{cm}}\right)=\frac{\mathrm{m}_{1} \overrightarrow{\mathrm{r}}_{1}+\mathrm{m}_{1} \overrightarrow{\mathrm{r}}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}} =\frac{6(6 \hat{\mathrm{i}}-7 \hat{\mathrm{j}})+2 \times(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-8 \hat{\mathrm{k}})}{6+2}\) \(=\frac{36 \hat{\mathrm{i}}-42 \hat{\mathrm{j}}+4 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-16 \hat{\mathrm{k}}}{8}\) \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}} =5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}\)
AMU-2014
Rotational Motion
149751
With \(\mathrm{O}\) as the origin of the coordinate axis, the \(X\) and \(Y\) - coordinates of the centre of mass of the system of particles shown in the figure may be given as
C Here \(\mathrm{m}\) and \(2 \mathrm{~m}\) represent the masses of the particles. The taking moments with respect to \(0 \overrightarrow{\mathrm{x}}_{\mathrm{cm}}\) is \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}} =\frac{\mathrm{m}(-2 \mathrm{~b})+2 \mathrm{~m}(-\mathrm{b})+\mathrm{m} \times 0+2 \mathrm{~m}(\mathrm{~b})}{\mathrm{m}+2 \mathrm{~m}+\mathrm{m}+2 \mathrm{~m}}\) \(=-\frac{\mathrm{b}}{3}\) \(\overrightarrow{\mathrm{y}}_{\mathrm{cm}} =\mathrm{b}\) such that at the \(\overrightarrow{\mathrm{R}}_{\mathrm{cm}}=\left(\frac{-\mathrm{b}}{3}, \mathrm{~b}\right)\)
AMU-2006
Rotational Motion
149752
A straight rod of length \(L\) has one of its ends at the origin and the other at \(x=L\). If the mass per unit length of the rod is given by Ax where \(A\) is a constant, where is its mass centre
1 \(\mathrm{L} / 3\)
2 \(\mathrm{L} / 2\)
3 \(2 \mathrm{~L} / 3\)
4 \(3 \mathrm{~L} / 4\)
Explanation:
C At let to be assume A is a constant \(\mathrm{x}_{\mathrm{c}}=\) coordinate of center of mass At given data, \(\frac{M}{L}=A x\) \(X_{\text {COM }}=\frac{\int x \cdot d m}{\int d m}\) \(=\frac{\int_{0}^{L} x \cdot A x d x}{\left\{\int_{0}^{L} A x d x\right\}} \quad \text { T Total Mass }=\int_{0}^{L} A x d x \text { ) }\) \(\frac{d m}{d x}=A \cdot x\) \(d m=A x \cdot d x\) \(X_{\text {COM }}=\frac{\int_{0}^{L} x^{2} d x}{\int_{0}^{L} x \cdot d x}\) \(X_{\text {COM }}=\frac{\left[\frac{x^{3}}{3}\right]_{0}^{L}}{\left[\frac{x^{2}}{2}\right]_{0}^{L}}\) \(X_{\text {COM }}=\frac{\frac{L^{3}}{L^{2}}}{\frac{2}{2}}\) \(X_{\text {COM }}=\frac{2 L}{3}\)
149747
A body of mass \(m_{1}=4 \mathrm{~kg}\) moves at \(5 \hat{i} \mathrm{~m} / \mathrm{s}\) and another body of mass \(m_{2}=2 \mathrm{~kg}\) moves at \(10 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s}\). The kinetic energy of centre of mass is
1 \(\frac{200}{3} \mathrm{~J}\)
2 \(\frac{500}{3} \mathrm{~J}\)
3 \(\frac{400}{3} \mathrm{~J}\)
4 \(\frac{800}{3} \mathrm{~J}\)
Explanation:
C Given that, \(\mathrm{m}_{1}=4 \mathrm{~kg}, \mathrm{~m}_{2}=2 \mathrm{~kg}\) \(\frac{\mathrm{dr}_{1}}{\mathrm{dt}}=5 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}, \frac{\mathrm{dr}_{2}}{\mathrm{dt}}=10 \hat{\mathrm{i} \mathrm{m} / \mathrm{s}}\) Velocity of centre of mass \(=\) \(\mathrm{v}_{\mathrm{cm}}=\frac{\mathrm{m}_{1} \frac{\mathrm{dr}_{1}}{\mathrm{dt}}+\mathrm{m}_{2} \frac{\mathrm{dr}_{2}}{\mathrm{dt}}}{\mathrm{m}_{1}+\mathrm{m}_{2}}=\frac{(4 \times 5) \mathrm{i}+(2 \times 10) \mathrm{i}}{4+2}\) \(\mathrm{v}_{\mathrm{cm}}=\frac{40}{6} \hat{\mathrm{i}}=\frac{20}{3} \hat{\mathrm{i}}\) The kinetic energy of the centre of the mass of system- \(\mathrm{K} . \mathrm{E}_{\mathrm{cm}} =\frac{1}{2} \mathrm{mv}_{\mathrm{cm}}^{2}\) \(=\frac{1}{2}(4+2) \times \frac{20}{3} \times \frac{20}{3}\) \(\mathrm{~K} . \mathrm{E}_{\mathrm{cm}} =\frac{400}{3} \mathrm{~J}\)
AP EAMCET -2010
Rotational Motion
149750
Two masses of 6 and 2 units are at positions \((6 \hat{\mathbf{i}}-7 \hat{\mathbf{j}})\) and \((2 \hat{\mathbf{i}}+\mathbf{5} \hat{\mathbf{j}}-8 \hat{\mathbf{k}})\) respectively. The co-ordinates of the centre-of-mass (c.m.) are
1 \((2,-5,3)\)
2 \((5,-5,-3)\)
3 \((5,-4,-2)\)
4 \((5,-4,-4)\)
Explanation:
C Given, \(\mathrm{m}_{1}=6 \text { unit }\) \(\mathrm{m}_{2}=2 \text { unit }\) \(\overrightarrow{\mathrm{r}}_{1}=(6 \hat{\mathrm{i}}-7 \hat{\mathrm{j}})\) \(\overrightarrow{\mathrm{r}}_{2}=(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-8 \hat{\mathrm{k}})\) Co-ordinate of the centre of mass \(\left(\vec{x}_{\mathrm{cm}}\right)=\frac{\mathrm{m}_{1} \overrightarrow{\mathrm{r}}_{1}+\mathrm{m}_{1} \overrightarrow{\mathrm{r}}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}} =\frac{6(6 \hat{\mathrm{i}}-7 \hat{\mathrm{j}})+2 \times(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-8 \hat{\mathrm{k}})}{6+2}\) \(=\frac{36 \hat{\mathrm{i}}-42 \hat{\mathrm{j}}+4 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-16 \hat{\mathrm{k}}}{8}\) \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}} =5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}\)
AMU-2014
Rotational Motion
149751
With \(\mathrm{O}\) as the origin of the coordinate axis, the \(X\) and \(Y\) - coordinates of the centre of mass of the system of particles shown in the figure may be given as
C Here \(\mathrm{m}\) and \(2 \mathrm{~m}\) represent the masses of the particles. The taking moments with respect to \(0 \overrightarrow{\mathrm{x}}_{\mathrm{cm}}\) is \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}} =\frac{\mathrm{m}(-2 \mathrm{~b})+2 \mathrm{~m}(-\mathrm{b})+\mathrm{m} \times 0+2 \mathrm{~m}(\mathrm{~b})}{\mathrm{m}+2 \mathrm{~m}+\mathrm{m}+2 \mathrm{~m}}\) \(=-\frac{\mathrm{b}}{3}\) \(\overrightarrow{\mathrm{y}}_{\mathrm{cm}} =\mathrm{b}\) such that at the \(\overrightarrow{\mathrm{R}}_{\mathrm{cm}}=\left(\frac{-\mathrm{b}}{3}, \mathrm{~b}\right)\)
AMU-2006
Rotational Motion
149752
A straight rod of length \(L\) has one of its ends at the origin and the other at \(x=L\). If the mass per unit length of the rod is given by Ax where \(A\) is a constant, where is its mass centre
1 \(\mathrm{L} / 3\)
2 \(\mathrm{L} / 2\)
3 \(2 \mathrm{~L} / 3\)
4 \(3 \mathrm{~L} / 4\)
Explanation:
C At let to be assume A is a constant \(\mathrm{x}_{\mathrm{c}}=\) coordinate of center of mass At given data, \(\frac{M}{L}=A x\) \(X_{\text {COM }}=\frac{\int x \cdot d m}{\int d m}\) \(=\frac{\int_{0}^{L} x \cdot A x d x}{\left\{\int_{0}^{L} A x d x\right\}} \quad \text { T Total Mass }=\int_{0}^{L} A x d x \text { ) }\) \(\frac{d m}{d x}=A \cdot x\) \(d m=A x \cdot d x\) \(X_{\text {COM }}=\frac{\int_{0}^{L} x^{2} d x}{\int_{0}^{L} x \cdot d x}\) \(X_{\text {COM }}=\frac{\left[\frac{x^{3}}{3}\right]_{0}^{L}}{\left[\frac{x^{2}}{2}\right]_{0}^{L}}\) \(X_{\text {COM }}=\frac{\frac{L^{3}}{L^{2}}}{\frac{2}{2}}\) \(X_{\text {COM }}=\frac{2 L}{3}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Rotational Motion
149747
A body of mass \(m_{1}=4 \mathrm{~kg}\) moves at \(5 \hat{i} \mathrm{~m} / \mathrm{s}\) and another body of mass \(m_{2}=2 \mathrm{~kg}\) moves at \(10 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s}\). The kinetic energy of centre of mass is
1 \(\frac{200}{3} \mathrm{~J}\)
2 \(\frac{500}{3} \mathrm{~J}\)
3 \(\frac{400}{3} \mathrm{~J}\)
4 \(\frac{800}{3} \mathrm{~J}\)
Explanation:
C Given that, \(\mathrm{m}_{1}=4 \mathrm{~kg}, \mathrm{~m}_{2}=2 \mathrm{~kg}\) \(\frac{\mathrm{dr}_{1}}{\mathrm{dt}}=5 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}, \frac{\mathrm{dr}_{2}}{\mathrm{dt}}=10 \hat{\mathrm{i} \mathrm{m} / \mathrm{s}}\) Velocity of centre of mass \(=\) \(\mathrm{v}_{\mathrm{cm}}=\frac{\mathrm{m}_{1} \frac{\mathrm{dr}_{1}}{\mathrm{dt}}+\mathrm{m}_{2} \frac{\mathrm{dr}_{2}}{\mathrm{dt}}}{\mathrm{m}_{1}+\mathrm{m}_{2}}=\frac{(4 \times 5) \mathrm{i}+(2 \times 10) \mathrm{i}}{4+2}\) \(\mathrm{v}_{\mathrm{cm}}=\frac{40}{6} \hat{\mathrm{i}}=\frac{20}{3} \hat{\mathrm{i}}\) The kinetic energy of the centre of the mass of system- \(\mathrm{K} . \mathrm{E}_{\mathrm{cm}} =\frac{1}{2} \mathrm{mv}_{\mathrm{cm}}^{2}\) \(=\frac{1}{2}(4+2) \times \frac{20}{3} \times \frac{20}{3}\) \(\mathrm{~K} . \mathrm{E}_{\mathrm{cm}} =\frac{400}{3} \mathrm{~J}\)
AP EAMCET -2010
Rotational Motion
149750
Two masses of 6 and 2 units are at positions \((6 \hat{\mathbf{i}}-7 \hat{\mathbf{j}})\) and \((2 \hat{\mathbf{i}}+\mathbf{5} \hat{\mathbf{j}}-8 \hat{\mathbf{k}})\) respectively. The co-ordinates of the centre-of-mass (c.m.) are
1 \((2,-5,3)\)
2 \((5,-5,-3)\)
3 \((5,-4,-2)\)
4 \((5,-4,-4)\)
Explanation:
C Given, \(\mathrm{m}_{1}=6 \text { unit }\) \(\mathrm{m}_{2}=2 \text { unit }\) \(\overrightarrow{\mathrm{r}}_{1}=(6 \hat{\mathrm{i}}-7 \hat{\mathrm{j}})\) \(\overrightarrow{\mathrm{r}}_{2}=(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-8 \hat{\mathrm{k}})\) Co-ordinate of the centre of mass \(\left(\vec{x}_{\mathrm{cm}}\right)=\frac{\mathrm{m}_{1} \overrightarrow{\mathrm{r}}_{1}+\mathrm{m}_{1} \overrightarrow{\mathrm{r}}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}} =\frac{6(6 \hat{\mathrm{i}}-7 \hat{\mathrm{j}})+2 \times(2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-8 \hat{\mathrm{k}})}{6+2}\) \(=\frac{36 \hat{\mathrm{i}}-42 \hat{\mathrm{j}}+4 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-16 \hat{\mathrm{k}}}{8}\) \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}} =5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}\)
AMU-2014
Rotational Motion
149751
With \(\mathrm{O}\) as the origin of the coordinate axis, the \(X\) and \(Y\) - coordinates of the centre of mass of the system of particles shown in the figure may be given as
C Here \(\mathrm{m}\) and \(2 \mathrm{~m}\) represent the masses of the particles. The taking moments with respect to \(0 \overrightarrow{\mathrm{x}}_{\mathrm{cm}}\) is \(\overrightarrow{\mathrm{x}}_{\mathrm{cm}} =\frac{\mathrm{m}(-2 \mathrm{~b})+2 \mathrm{~m}(-\mathrm{b})+\mathrm{m} \times 0+2 \mathrm{~m}(\mathrm{~b})}{\mathrm{m}+2 \mathrm{~m}+\mathrm{m}+2 \mathrm{~m}}\) \(=-\frac{\mathrm{b}}{3}\) \(\overrightarrow{\mathrm{y}}_{\mathrm{cm}} =\mathrm{b}\) such that at the \(\overrightarrow{\mathrm{R}}_{\mathrm{cm}}=\left(\frac{-\mathrm{b}}{3}, \mathrm{~b}\right)\)
AMU-2006
Rotational Motion
149752
A straight rod of length \(L\) has one of its ends at the origin and the other at \(x=L\). If the mass per unit length of the rod is given by Ax where \(A\) is a constant, where is its mass centre
1 \(\mathrm{L} / 3\)
2 \(\mathrm{L} / 2\)
3 \(2 \mathrm{~L} / 3\)
4 \(3 \mathrm{~L} / 4\)
Explanation:
C At let to be assume A is a constant \(\mathrm{x}_{\mathrm{c}}=\) coordinate of center of mass At given data, \(\frac{M}{L}=A x\) \(X_{\text {COM }}=\frac{\int x \cdot d m}{\int d m}\) \(=\frac{\int_{0}^{L} x \cdot A x d x}{\left\{\int_{0}^{L} A x d x\right\}} \quad \text { T Total Mass }=\int_{0}^{L} A x d x \text { ) }\) \(\frac{d m}{d x}=A \cdot x\) \(d m=A x \cdot d x\) \(X_{\text {COM }}=\frac{\int_{0}^{L} x^{2} d x}{\int_{0}^{L} x \cdot d x}\) \(X_{\text {COM }}=\frac{\left[\frac{x^{3}}{3}\right]_{0}^{L}}{\left[\frac{x^{2}}{2}\right]_{0}^{L}}\) \(X_{\text {COM }}=\frac{\frac{L^{3}}{L^{2}}}{\frac{2}{2}}\) \(X_{\text {COM }}=\frac{2 L}{3}\)
