D Given, \(\mathrm{m}=150 \mathrm{gram}=0.15 \mathrm{~kg}\) \(\mathrm{r}=15 \mathrm{~cm}=0.15 \mathrm{~m}\) For circular ring \(\mathrm{I}=\mathrm{mr}^{2}\) \(\mathrm{I}=0.15(0.15)^{2}=15 \times 225 \times 10^{-6}\) \(\mathrm{I}=0.003 \mathrm{~kg} . \mathrm{m}^{2}\)
AP EAMCET-24.09.2020
Rotational Motion
150144
The masses of \(200 \mathrm{~g}\) and \(300 \mathrm{~g}\) are attached to the \(20 \mathrm{~cm}\) and \(70 \mathrm{~cm}\) marks of a light meter rod respectively. The moment of inertia of the system about an axis passing through \(50 \mathrm{~cm}\) mark is
1 \(0.15 \mathrm{~kg} \mathrm{~m}^{2}\)
2 \(0.036 \mathrm{~kg} \mathrm{~m}^{2}\)
3 \(0.3 \mathrm{~kg} \mathrm{~m}^{2}\)
4 zero
Explanation:
B Given, \(\mathrm{m}_{1}=200 \mathrm{~g}=0.2 \mathrm{~kg}\) \(\mathrm{m}_{2}=300 \mathrm{~g}=0.3 \mathrm{~kg}\) \(\mathrm{r}_{1}=20 \mathrm{~cm}=0.2 \mathrm{~m}\) \(\mathrm{r}_{2}=30 \mathrm{~cm}==0.3 \mathrm{~m}\) The moment of inertia about axis passing through \(50 \mathrm{~cm}\) mark, \(\because \mathrm{I}=\sum \mathrm{mr}^{2}\) \(\mathrm{I}_{50}=\mathrm{m}_{1} \mathrm{r}_{1}^{2}+\mathrm{m}_{2} \mathrm{r}_{2}^{2}\) \(=0.2 \times(0.3)^{2}+0.3 \times(0.2)^{2}\) \(=0.2 \times 0.09+0.3 \times 0.04\) \(=0.018+0.012\) \(\mathrm{I}_{50}=0.03 \mathrm{kgm}^{2}\)
COMEDK 2020
Rotational Motion
150145
Four particles each of the mass \(m\) are placed at the corners of a square of side length \(l\), the radius of gyration of the system about an axis perpendicular to the square and passing through its centre is
1 \(\frac{l}{\sqrt{2}}\)
2 \(\frac{l}{2}\)
3 \(l\)
4 \(\sqrt{2} l\)
Explanation:
A The distance of the particle from the centre of square, \(\mathrm{AC}=\sqrt{2} l\) \(\mathrm{OC}=\frac{\sqrt{2} l}{2}=\frac{l}{\sqrt{2}}\) Given, \(M=4 \mathrm{~m}\) MOI of system, \(\mathrm{I}=\sum \mathrm{mr}^{2}\) \(=4 \mathrm{~m}\left(\frac{l}{\sqrt{2}}\right)^{2}\) \(=\frac{4 \mathrm{~m} l^{2}}{2}\) \(\mathrm{I}=2 \mathrm{~m} l^{2}\) And the radius of gyration, \(\mathrm{K}=\sqrt{\frac{\mathrm{I}}{\mathrm{M}}}=\sqrt{\frac{2 \mathrm{~m} l^{2}}{4 \mathrm{~m}}}\) \(\therefore \mathrm{K}=\frac{l}{\sqrt{2}}\)
COMEDK 2020
Rotational Motion
150146
Consider a thin metal strip of mass \(1 \mathrm{~kg}\) and length \(5 \mathrm{~m}\). Calculate its moment of inertia about an axis perpendicular to strip and located at \(100 \mathrm{~cm}\) on strip from one its end. (Assume the breadth as the strip is negligible)
1 \(4.33 \mathrm{~kg}-\mathrm{m}^{2}\)
2 \(4.85 \mathrm{~kg}-\mathrm{m}^{2}\)
3 \(4.11 \mathrm{~kg}-\mathrm{m}^{2}\)
4 \(4.66 \mathrm{~kg}-\mathrm{m}^{2}\)
Explanation:
A The given situation is as shown below, Here, \(\mathrm{O}\) is the point where centre of mass of whole strip exist. The point \(\mathrm{O}^{\prime}\) is \(100 \mathrm{~cm}\) or \(1 \mathrm{~m}\) away from one end. So, its distance from centre of mass or from point \(\mathrm{O}\) is \(1.5 \mathrm{~m}\) \(\therefore\) Moment of inertia about an axis perpendicular to strip at point \(\mathrm{O}^{\prime}\) can be calculated using parallel axis theorem as \(\mathrm{I}^{\prime} =\mathrm{I}_{\mathrm{CM}}+\mathrm{MR}^{2}\) \(=\frac{\mathrm{ML}^{2}}{12}+\mathrm{MR}^{2}\) \(=\frac{1 \times 5 \times 5}{12}+1 \times 1.5 \times 1.5\) \(=2.08+2.25\) \(=4.33 \mathrm{~kg}-\mathrm{m}^{2}\)
D Given, \(\mathrm{m}=150 \mathrm{gram}=0.15 \mathrm{~kg}\) \(\mathrm{r}=15 \mathrm{~cm}=0.15 \mathrm{~m}\) For circular ring \(\mathrm{I}=\mathrm{mr}^{2}\) \(\mathrm{I}=0.15(0.15)^{2}=15 \times 225 \times 10^{-6}\) \(\mathrm{I}=0.003 \mathrm{~kg} . \mathrm{m}^{2}\)
AP EAMCET-24.09.2020
Rotational Motion
150144
The masses of \(200 \mathrm{~g}\) and \(300 \mathrm{~g}\) are attached to the \(20 \mathrm{~cm}\) and \(70 \mathrm{~cm}\) marks of a light meter rod respectively. The moment of inertia of the system about an axis passing through \(50 \mathrm{~cm}\) mark is
1 \(0.15 \mathrm{~kg} \mathrm{~m}^{2}\)
2 \(0.036 \mathrm{~kg} \mathrm{~m}^{2}\)
3 \(0.3 \mathrm{~kg} \mathrm{~m}^{2}\)
4 zero
Explanation:
B Given, \(\mathrm{m}_{1}=200 \mathrm{~g}=0.2 \mathrm{~kg}\) \(\mathrm{m}_{2}=300 \mathrm{~g}=0.3 \mathrm{~kg}\) \(\mathrm{r}_{1}=20 \mathrm{~cm}=0.2 \mathrm{~m}\) \(\mathrm{r}_{2}=30 \mathrm{~cm}==0.3 \mathrm{~m}\) The moment of inertia about axis passing through \(50 \mathrm{~cm}\) mark, \(\because \mathrm{I}=\sum \mathrm{mr}^{2}\) \(\mathrm{I}_{50}=\mathrm{m}_{1} \mathrm{r}_{1}^{2}+\mathrm{m}_{2} \mathrm{r}_{2}^{2}\) \(=0.2 \times(0.3)^{2}+0.3 \times(0.2)^{2}\) \(=0.2 \times 0.09+0.3 \times 0.04\) \(=0.018+0.012\) \(\mathrm{I}_{50}=0.03 \mathrm{kgm}^{2}\)
COMEDK 2020
Rotational Motion
150145
Four particles each of the mass \(m\) are placed at the corners of a square of side length \(l\), the radius of gyration of the system about an axis perpendicular to the square and passing through its centre is
1 \(\frac{l}{\sqrt{2}}\)
2 \(\frac{l}{2}\)
3 \(l\)
4 \(\sqrt{2} l\)
Explanation:
A The distance of the particle from the centre of square, \(\mathrm{AC}=\sqrt{2} l\) \(\mathrm{OC}=\frac{\sqrt{2} l}{2}=\frac{l}{\sqrt{2}}\) Given, \(M=4 \mathrm{~m}\) MOI of system, \(\mathrm{I}=\sum \mathrm{mr}^{2}\) \(=4 \mathrm{~m}\left(\frac{l}{\sqrt{2}}\right)^{2}\) \(=\frac{4 \mathrm{~m} l^{2}}{2}\) \(\mathrm{I}=2 \mathrm{~m} l^{2}\) And the radius of gyration, \(\mathrm{K}=\sqrt{\frac{\mathrm{I}}{\mathrm{M}}}=\sqrt{\frac{2 \mathrm{~m} l^{2}}{4 \mathrm{~m}}}\) \(\therefore \mathrm{K}=\frac{l}{\sqrt{2}}\)
COMEDK 2020
Rotational Motion
150146
Consider a thin metal strip of mass \(1 \mathrm{~kg}\) and length \(5 \mathrm{~m}\). Calculate its moment of inertia about an axis perpendicular to strip and located at \(100 \mathrm{~cm}\) on strip from one its end. (Assume the breadth as the strip is negligible)
1 \(4.33 \mathrm{~kg}-\mathrm{m}^{2}\)
2 \(4.85 \mathrm{~kg}-\mathrm{m}^{2}\)
3 \(4.11 \mathrm{~kg}-\mathrm{m}^{2}\)
4 \(4.66 \mathrm{~kg}-\mathrm{m}^{2}\)
Explanation:
A The given situation is as shown below, Here, \(\mathrm{O}\) is the point where centre of mass of whole strip exist. The point \(\mathrm{O}^{\prime}\) is \(100 \mathrm{~cm}\) or \(1 \mathrm{~m}\) away from one end. So, its distance from centre of mass or from point \(\mathrm{O}\) is \(1.5 \mathrm{~m}\) \(\therefore\) Moment of inertia about an axis perpendicular to strip at point \(\mathrm{O}^{\prime}\) can be calculated using parallel axis theorem as \(\mathrm{I}^{\prime} =\mathrm{I}_{\mathrm{CM}}+\mathrm{MR}^{2}\) \(=\frac{\mathrm{ML}^{2}}{12}+\mathrm{MR}^{2}\) \(=\frac{1 \times 5 \times 5}{12}+1 \times 1.5 \times 1.5\) \(=2.08+2.25\) \(=4.33 \mathrm{~kg}-\mathrm{m}^{2}\)
D Given, \(\mathrm{m}=150 \mathrm{gram}=0.15 \mathrm{~kg}\) \(\mathrm{r}=15 \mathrm{~cm}=0.15 \mathrm{~m}\) For circular ring \(\mathrm{I}=\mathrm{mr}^{2}\) \(\mathrm{I}=0.15(0.15)^{2}=15 \times 225 \times 10^{-6}\) \(\mathrm{I}=0.003 \mathrm{~kg} . \mathrm{m}^{2}\)
AP EAMCET-24.09.2020
Rotational Motion
150144
The masses of \(200 \mathrm{~g}\) and \(300 \mathrm{~g}\) are attached to the \(20 \mathrm{~cm}\) and \(70 \mathrm{~cm}\) marks of a light meter rod respectively. The moment of inertia of the system about an axis passing through \(50 \mathrm{~cm}\) mark is
1 \(0.15 \mathrm{~kg} \mathrm{~m}^{2}\)
2 \(0.036 \mathrm{~kg} \mathrm{~m}^{2}\)
3 \(0.3 \mathrm{~kg} \mathrm{~m}^{2}\)
4 zero
Explanation:
B Given, \(\mathrm{m}_{1}=200 \mathrm{~g}=0.2 \mathrm{~kg}\) \(\mathrm{m}_{2}=300 \mathrm{~g}=0.3 \mathrm{~kg}\) \(\mathrm{r}_{1}=20 \mathrm{~cm}=0.2 \mathrm{~m}\) \(\mathrm{r}_{2}=30 \mathrm{~cm}==0.3 \mathrm{~m}\) The moment of inertia about axis passing through \(50 \mathrm{~cm}\) mark, \(\because \mathrm{I}=\sum \mathrm{mr}^{2}\) \(\mathrm{I}_{50}=\mathrm{m}_{1} \mathrm{r}_{1}^{2}+\mathrm{m}_{2} \mathrm{r}_{2}^{2}\) \(=0.2 \times(0.3)^{2}+0.3 \times(0.2)^{2}\) \(=0.2 \times 0.09+0.3 \times 0.04\) \(=0.018+0.012\) \(\mathrm{I}_{50}=0.03 \mathrm{kgm}^{2}\)
COMEDK 2020
Rotational Motion
150145
Four particles each of the mass \(m\) are placed at the corners of a square of side length \(l\), the radius of gyration of the system about an axis perpendicular to the square and passing through its centre is
1 \(\frac{l}{\sqrt{2}}\)
2 \(\frac{l}{2}\)
3 \(l\)
4 \(\sqrt{2} l\)
Explanation:
A The distance of the particle from the centre of square, \(\mathrm{AC}=\sqrt{2} l\) \(\mathrm{OC}=\frac{\sqrt{2} l}{2}=\frac{l}{\sqrt{2}}\) Given, \(M=4 \mathrm{~m}\) MOI of system, \(\mathrm{I}=\sum \mathrm{mr}^{2}\) \(=4 \mathrm{~m}\left(\frac{l}{\sqrt{2}}\right)^{2}\) \(=\frac{4 \mathrm{~m} l^{2}}{2}\) \(\mathrm{I}=2 \mathrm{~m} l^{2}\) And the radius of gyration, \(\mathrm{K}=\sqrt{\frac{\mathrm{I}}{\mathrm{M}}}=\sqrt{\frac{2 \mathrm{~m} l^{2}}{4 \mathrm{~m}}}\) \(\therefore \mathrm{K}=\frac{l}{\sqrt{2}}\)
COMEDK 2020
Rotational Motion
150146
Consider a thin metal strip of mass \(1 \mathrm{~kg}\) and length \(5 \mathrm{~m}\). Calculate its moment of inertia about an axis perpendicular to strip and located at \(100 \mathrm{~cm}\) on strip from one its end. (Assume the breadth as the strip is negligible)
1 \(4.33 \mathrm{~kg}-\mathrm{m}^{2}\)
2 \(4.85 \mathrm{~kg}-\mathrm{m}^{2}\)
3 \(4.11 \mathrm{~kg}-\mathrm{m}^{2}\)
4 \(4.66 \mathrm{~kg}-\mathrm{m}^{2}\)
Explanation:
A The given situation is as shown below, Here, \(\mathrm{O}\) is the point where centre of mass of whole strip exist. The point \(\mathrm{O}^{\prime}\) is \(100 \mathrm{~cm}\) or \(1 \mathrm{~m}\) away from one end. So, its distance from centre of mass or from point \(\mathrm{O}\) is \(1.5 \mathrm{~m}\) \(\therefore\) Moment of inertia about an axis perpendicular to strip at point \(\mathrm{O}^{\prime}\) can be calculated using parallel axis theorem as \(\mathrm{I}^{\prime} =\mathrm{I}_{\mathrm{CM}}+\mathrm{MR}^{2}\) \(=\frac{\mathrm{ML}^{2}}{12}+\mathrm{MR}^{2}\) \(=\frac{1 \times 5 \times 5}{12}+1 \times 1.5 \times 1.5\) \(=2.08+2.25\) \(=4.33 \mathrm{~kg}-\mathrm{m}^{2}\)
D Given, \(\mathrm{m}=150 \mathrm{gram}=0.15 \mathrm{~kg}\) \(\mathrm{r}=15 \mathrm{~cm}=0.15 \mathrm{~m}\) For circular ring \(\mathrm{I}=\mathrm{mr}^{2}\) \(\mathrm{I}=0.15(0.15)^{2}=15 \times 225 \times 10^{-6}\) \(\mathrm{I}=0.003 \mathrm{~kg} . \mathrm{m}^{2}\)
AP EAMCET-24.09.2020
Rotational Motion
150144
The masses of \(200 \mathrm{~g}\) and \(300 \mathrm{~g}\) are attached to the \(20 \mathrm{~cm}\) and \(70 \mathrm{~cm}\) marks of a light meter rod respectively. The moment of inertia of the system about an axis passing through \(50 \mathrm{~cm}\) mark is
1 \(0.15 \mathrm{~kg} \mathrm{~m}^{2}\)
2 \(0.036 \mathrm{~kg} \mathrm{~m}^{2}\)
3 \(0.3 \mathrm{~kg} \mathrm{~m}^{2}\)
4 zero
Explanation:
B Given, \(\mathrm{m}_{1}=200 \mathrm{~g}=0.2 \mathrm{~kg}\) \(\mathrm{m}_{2}=300 \mathrm{~g}=0.3 \mathrm{~kg}\) \(\mathrm{r}_{1}=20 \mathrm{~cm}=0.2 \mathrm{~m}\) \(\mathrm{r}_{2}=30 \mathrm{~cm}==0.3 \mathrm{~m}\) The moment of inertia about axis passing through \(50 \mathrm{~cm}\) mark, \(\because \mathrm{I}=\sum \mathrm{mr}^{2}\) \(\mathrm{I}_{50}=\mathrm{m}_{1} \mathrm{r}_{1}^{2}+\mathrm{m}_{2} \mathrm{r}_{2}^{2}\) \(=0.2 \times(0.3)^{2}+0.3 \times(0.2)^{2}\) \(=0.2 \times 0.09+0.3 \times 0.04\) \(=0.018+0.012\) \(\mathrm{I}_{50}=0.03 \mathrm{kgm}^{2}\)
COMEDK 2020
Rotational Motion
150145
Four particles each of the mass \(m\) are placed at the corners of a square of side length \(l\), the radius of gyration of the system about an axis perpendicular to the square and passing through its centre is
1 \(\frac{l}{\sqrt{2}}\)
2 \(\frac{l}{2}\)
3 \(l\)
4 \(\sqrt{2} l\)
Explanation:
A The distance of the particle from the centre of square, \(\mathrm{AC}=\sqrt{2} l\) \(\mathrm{OC}=\frac{\sqrt{2} l}{2}=\frac{l}{\sqrt{2}}\) Given, \(M=4 \mathrm{~m}\) MOI of system, \(\mathrm{I}=\sum \mathrm{mr}^{2}\) \(=4 \mathrm{~m}\left(\frac{l}{\sqrt{2}}\right)^{2}\) \(=\frac{4 \mathrm{~m} l^{2}}{2}\) \(\mathrm{I}=2 \mathrm{~m} l^{2}\) And the radius of gyration, \(\mathrm{K}=\sqrt{\frac{\mathrm{I}}{\mathrm{M}}}=\sqrt{\frac{2 \mathrm{~m} l^{2}}{4 \mathrm{~m}}}\) \(\therefore \mathrm{K}=\frac{l}{\sqrt{2}}\)
COMEDK 2020
Rotational Motion
150146
Consider a thin metal strip of mass \(1 \mathrm{~kg}\) and length \(5 \mathrm{~m}\). Calculate its moment of inertia about an axis perpendicular to strip and located at \(100 \mathrm{~cm}\) on strip from one its end. (Assume the breadth as the strip is negligible)
1 \(4.33 \mathrm{~kg}-\mathrm{m}^{2}\)
2 \(4.85 \mathrm{~kg}-\mathrm{m}^{2}\)
3 \(4.11 \mathrm{~kg}-\mathrm{m}^{2}\)
4 \(4.66 \mathrm{~kg}-\mathrm{m}^{2}\)
Explanation:
A The given situation is as shown below, Here, \(\mathrm{O}\) is the point where centre of mass of whole strip exist. The point \(\mathrm{O}^{\prime}\) is \(100 \mathrm{~cm}\) or \(1 \mathrm{~m}\) away from one end. So, its distance from centre of mass or from point \(\mathrm{O}\) is \(1.5 \mathrm{~m}\) \(\therefore\) Moment of inertia about an axis perpendicular to strip at point \(\mathrm{O}^{\prime}\) can be calculated using parallel axis theorem as \(\mathrm{I}^{\prime} =\mathrm{I}_{\mathrm{CM}}+\mathrm{MR}^{2}\) \(=\frac{\mathrm{ML}^{2}}{12}+\mathrm{MR}^{2}\) \(=\frac{1 \times 5 \times 5}{12}+1 \times 1.5 \times 1.5\) \(=2.08+2.25\) \(=4.33 \mathrm{~kg}-\mathrm{m}^{2}\)
