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 \(I_{p}=I_{c}+M d^{2}=\dfrac{M L^{2}}{12}+M\left(\dfrac{L}{6}\right)^{2}=\dfrac{M L^{2}}{9}\)
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 \(I_{p}=I_{c}+M d^{2}=\dfrac{M L^{2}}{12}+M\left(\dfrac{L}{6}\right)^{2}=\dfrac{M L^{2}}{9}\)
