Unit 5: Torque and Rotational Dynamics
Weightage : 10-15%
The rod shown above can pivot about the point \(x=0\) and rotates in a plane perpendicular to the page. Its linear density, \(\lambda(x)\), increases with \(x\) such that
\(\lambda(x)=kx\),
where \(k\) is a positive constant. Determine the rod’s moment of inertia in terms of its length \(L\) and its total mass \(M\).
Correct Answer: \(\boxed{\mathrm{D}}\)
Consider an infinitesimal element of the rod of length \(dx\) at a distance \(x\) from the pivot.
Its mass is
\(dm=\lambda(x)\,dx=kx\,dx\)
The moment of inertia is
\(I=\int x^2\,dm\)
\(I=\int_0^L x^2(kx\,dx)=k\int_0^L x^3\,dx=\frac{1}{4}kL^4\)
The total mass of the rod is
\(M=\int_0^L dm=\int_0^L kx\,dx=\frac{1}{2}kL^2\)
Hence,
\(k=\frac{2M}{L^2}\)
Substituting into the expression for \(I\),
\(I=\frac{1}{4}\left(\frac{2M}{L^2}\right)L^4=\frac{1}{2}ML^2\)
Since the mass density increases linearly with distance from the pivot, more mass is located farther from the axis, giving a larger moment of inertia than that of a uniform rod about one end.
Therefore, the correct answer is (D).
A uniform stick has length \(L\). The moment of inertia about the center of the stick is \(I_0\). A particle of mass \(M\) is attached to one end of the stick. The moment of inertia of the combined system about the center of the stick is
Correct Answer: \(\boxed{\mathrm{A}}\)
The total moment of inertia is the sum of the stick’s moment of inertia and that of the attached particle.
The particle is located at one end of the stick, a distance
\(r=\frac{L}{2}\)
from the center of the stick (the axis of rotation).
The particle’s moment of inertia is
\(I_{\mathrm{particle}}=Mr^2=M\left(\frac{L}{2}\right)^2=\frac{1}{4}ML^2\)
Therefore, the total moment of inertia is
\(I_{\mathrm{total}}=I_0+\frac{1}{4}ML^2\)
Thus, attaching the particle increases the rotational inertia by an amount proportional to the square of its distance from the axis.
Therefore, the correct answer is (A).
A \(5\)-kilogram sphere is connected to a \(10\)-kilogram sphere by a rigid rod of negligible mass, as shown above. The sphere-rod combination can be pivoted about an axis that is perpendicular to the plane of the page and that passes through one of the five lettered points.
Through which point should the axis pass for the moment of inertia of the sphere-rod combination about this axis to be greatest?
Correct Answer: \( \boxed{\mathrm{E}} \)
The moment of inertia about any axis is given by the parallel-axis theorem:
\( I = I_{\mathrm{CM}} + MD^{2} \)
where \(I_{\mathrm{CM}}\) is the moment of inertia about the center of mass, \(M\) is the total mass, and \(D\) is the distance between the axis and the center of mass.
Since the \(10\,\mathrm{kg}\) sphere is heavier than the \(5\,\mathrm{kg}\) sphere, the center of mass of the system lies closer to the \(10\,\mathrm{kg}\) sphere, at point B.
The farther the axis is from the center of mass, the larger the value of \(MD^{2}\), and hence the larger the total moment of inertia.
Of the five points shown, point E is farthest from the center of mass.
Therefore, the moment of inertia is greatest when the axis passes through (E).
