365904
Two identical concentric rings each of mass \(m\) and radius \(R\) are placed perpendicularly. What is the moment of inertia about axis of one of the rings?
1 \((3 / 2) M R^{2}\)
2 \(2 M R^{2}\)
3 \(3 M R^{2}\)
4 \((1 / 4) M R^{2}\)
Explanation:
Since, the axis of two rings are also perpendicular to each other, therefore applying the perpendicular axis theorem \(I=M R^{2}+\dfrac{1}{2} M R^{2}=\dfrac{3}{2} M R^{2}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365905
A wheel comprises a ring of radius \(\mathrm{R}\) and mass \(M\) and three spokes of mass \(m\) each. The moment of inertia of the wheel about its axis is
365906
Figure shows a thin metallic triangular sheet \(ABC\). The mass of the sheet is \(M\). The moment of inertia of the sheet about side \(AC\) is:
1 \(\dfrac{M l^{2}}{12}\)
2 \(\dfrac{M l^{2}}{6}\)
3 \(\dfrac{M l^{2}}{18}\)
4 \(\dfrac{M l^{2}}{4}\)
Explanation:
We can imagine the given triangular plate as a half part of a square of side length \(l\). \(I_{A C}=\dfrac{M l^{2}}{12}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365907
A thin uniform rod has mass \(M\) and length \(L\). The moment of inertia about an axis perpendicular to it and passing through the point at a distance \(\dfrac{L}{3}\) from one of its end, will be
1 \(\dfrac{M L^{2}}{12}\)
2 \(\dfrac{7 M L^{2}}{8}\)
3 \(\dfrac{M L^{2}}{9}\)
4 \(\dfrac{M L^{2}}{3}\)
Explanation:
A point '\(P\)' at a distance \(\dfrac{L}{3}\) from one end of the rod is at a distance of \(d = \frac{L}{2} - \frac{L}{3} = \frac{L}{6}\) from the centre '\(C\)' of rod The moment of inertia of the rod, about an axis passing through centre and perpendicular to it, is \(I_{c}=\dfrac{M L^{2}}{12}\) Using parallel axes theorem, moment of inertia of the rod, about an axis perpendicular to the rod and passing through \(P\), is \(I_{p}=I_{c}+M d^{2}=\dfrac{M L^{2}}{12}+M\left(\dfrac{L}{6}\right)^{2}=\dfrac{M L^{2}}{9}\)
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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365904
Two identical concentric rings each of mass \(m\) and radius \(R\) are placed perpendicularly. What is the moment of inertia about axis of one of the rings?
1 \((3 / 2) M R^{2}\)
2 \(2 M R^{2}\)
3 \(3 M R^{2}\)
4 \((1 / 4) M R^{2}\)
Explanation:
Since, the axis of two rings are also perpendicular to each other, therefore applying the perpendicular axis theorem \(I=M R^{2}+\dfrac{1}{2} M R^{2}=\dfrac{3}{2} M R^{2}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365905
A wheel comprises a ring of radius \(\mathrm{R}\) and mass \(M\) and three spokes of mass \(m\) each. The moment of inertia of the wheel about its axis is
365906
Figure shows a thin metallic triangular sheet \(ABC\). The mass of the sheet is \(M\). The moment of inertia of the sheet about side \(AC\) is:
1 \(\dfrac{M l^{2}}{12}\)
2 \(\dfrac{M l^{2}}{6}\)
3 \(\dfrac{M l^{2}}{18}\)
4 \(\dfrac{M l^{2}}{4}\)
Explanation:
We can imagine the given triangular plate as a half part of a square of side length \(l\). \(I_{A C}=\dfrac{M l^{2}}{12}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365907
A thin uniform rod has mass \(M\) and length \(L\). The moment of inertia about an axis perpendicular to it and passing through the point at a distance \(\dfrac{L}{3}\) from one of its end, will be
1 \(\dfrac{M L^{2}}{12}\)
2 \(\dfrac{7 M L^{2}}{8}\)
3 \(\dfrac{M L^{2}}{9}\)
4 \(\dfrac{M L^{2}}{3}\)
Explanation:
A point '\(P\)' at a distance \(\dfrac{L}{3}\) from one end of the rod is at a distance of \(d = \frac{L}{2} - \frac{L}{3} = \frac{L}{6}\) from the centre '\(C\)' of rod The moment of inertia of the rod, about an axis passing through centre and perpendicular to it, is \(I_{c}=\dfrac{M L^{2}}{12}\) Using parallel axes theorem, moment of inertia of the rod, about an axis perpendicular to the rod and passing through \(P\), is \(I_{p}=I_{c}+M d^{2}=\dfrac{M L^{2}}{12}+M\left(\dfrac{L}{6}\right)^{2}=\dfrac{M L^{2}}{9}\)
365904
Two identical concentric rings each of mass \(m\) and radius \(R\) are placed perpendicularly. What is the moment of inertia about axis of one of the rings?
1 \((3 / 2) M R^{2}\)
2 \(2 M R^{2}\)
3 \(3 M R^{2}\)
4 \((1 / 4) M R^{2}\)
Explanation:
Since, the axis of two rings are also perpendicular to each other, therefore applying the perpendicular axis theorem \(I=M R^{2}+\dfrac{1}{2} M R^{2}=\dfrac{3}{2} M R^{2}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365905
A wheel comprises a ring of radius \(\mathrm{R}\) and mass \(M\) and three spokes of mass \(m\) each. The moment of inertia of the wheel about its axis is
365906
Figure shows a thin metallic triangular sheet \(ABC\). The mass of the sheet is \(M\). The moment of inertia of the sheet about side \(AC\) is:
1 \(\dfrac{M l^{2}}{12}\)
2 \(\dfrac{M l^{2}}{6}\)
3 \(\dfrac{M l^{2}}{18}\)
4 \(\dfrac{M l^{2}}{4}\)
Explanation:
We can imagine the given triangular plate as a half part of a square of side length \(l\). \(I_{A C}=\dfrac{M l^{2}}{12}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365907
A thin uniform rod has mass \(M\) and length \(L\). The moment of inertia about an axis perpendicular to it and passing through the point at a distance \(\dfrac{L}{3}\) from one of its end, will be
1 \(\dfrac{M L^{2}}{12}\)
2 \(\dfrac{7 M L^{2}}{8}\)
3 \(\dfrac{M L^{2}}{9}\)
4 \(\dfrac{M L^{2}}{3}\)
Explanation:
A point '\(P\)' at a distance \(\dfrac{L}{3}\) from one end of the rod is at a distance of \(d = \frac{L}{2} - \frac{L}{3} = \frac{L}{6}\) from the centre '\(C\)' of rod The moment of inertia of the rod, about an axis passing through centre and perpendicular to it, is \(I_{c}=\dfrac{M L^{2}}{12}\) Using parallel axes theorem, moment of inertia of the rod, about an axis perpendicular to the rod and passing through \(P\), is \(I_{p}=I_{c}+M d^{2}=\dfrac{M L^{2}}{12}+M\left(\dfrac{L}{6}\right)^{2}=\dfrac{M L^{2}}{9}\)
365904
Two identical concentric rings each of mass \(m\) and radius \(R\) are placed perpendicularly. What is the moment of inertia about axis of one of the rings?
1 \((3 / 2) M R^{2}\)
2 \(2 M R^{2}\)
3 \(3 M R^{2}\)
4 \((1 / 4) M R^{2}\)
Explanation:
Since, the axis of two rings are also perpendicular to each other, therefore applying the perpendicular axis theorem \(I=M R^{2}+\dfrac{1}{2} M R^{2}=\dfrac{3}{2} M R^{2}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365905
A wheel comprises a ring of radius \(\mathrm{R}\) and mass \(M\) and three spokes of mass \(m\) each. The moment of inertia of the wheel about its axis is
365906
Figure shows a thin metallic triangular sheet \(ABC\). The mass of the sheet is \(M\). The moment of inertia of the sheet about side \(AC\) is:
1 \(\dfrac{M l^{2}}{12}\)
2 \(\dfrac{M l^{2}}{6}\)
3 \(\dfrac{M l^{2}}{18}\)
4 \(\dfrac{M l^{2}}{4}\)
Explanation:
We can imagine the given triangular plate as a half part of a square of side length \(l\). \(I_{A C}=\dfrac{M l^{2}}{12}\)
PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
365907
A thin uniform rod has mass \(M\) and length \(L\). The moment of inertia about an axis perpendicular to it and passing through the point at a distance \(\dfrac{L}{3}\) from one of its end, will be
1 \(\dfrac{M L^{2}}{12}\)
2 \(\dfrac{7 M L^{2}}{8}\)
3 \(\dfrac{M L^{2}}{9}\)
4 \(\dfrac{M L^{2}}{3}\)
Explanation:
A point '\(P\)' at a distance \(\dfrac{L}{3}\) from one end of the rod is at a distance of \(d = \frac{L}{2} - \frac{L}{3} = \frac{L}{6}\) from the centre '\(C\)' of rod The moment of inertia of the rod, about an axis passing through centre and perpendicular to it, is \(I_{c}=\dfrac{M L^{2}}{12}\) Using parallel axes theorem, moment of inertia of the rod, about an axis perpendicular to the rod and passing through \(P\), is \(I_{p}=I_{c}+M d^{2}=\dfrac{M L^{2}}{12}+M\left(\dfrac{L}{6}\right)^{2}=\dfrac{M L^{2}}{9}\)
