Unit 6: Energy and Momentum of Rotating Systems
Weightage : 10-15%
A cylinder rotates with constant angular acceleration about a fixed axis. The cylinder’s moment of inertia about the axis is \(4\,\mathrm{kg\cdot m^2}\). At time \(t=0\), the cylinder is at rest. At time \(t=2\,\mathrm{s}\), its angular velocity is \(1\,\mathrm{rad/s}\). What is the angular momentum of the cylinder at time \(t=2\,\mathrm{s}\)?
Correct Answer: \(\boxed{\mathrm{D}}\)
The angular momentum of a rotating rigid body is given by
\(L=I\omega\)
Substitute the given values:
\(I=4\,\mathrm{kg\cdot m^2}\)
\(\omega=1\,\mathrm{rad/s}\)
Therefore,
\(L=(4\,\mathrm{kg\cdot m^2})(1\,\mathrm{rad/s})=4\,\mathrm{kg\cdot m^2/s}\)
The radius of the cylinder is not needed because the moment of inertia is already provided.
Therefore, the correct answer is (D).
A particle of mass \(m\) moves with a constant speed \(v\) along the dashed line \(y=a\). When the \(x\)-coordinate of the particle is \(x_0\), the magnitude of the angular momentum of the particle with respect to the origin of the coordinate system is
Correct Answer: \(\boxed{\mathrm{B}}\)
The magnitude of the angular momentum of a particle about a point is
\(L=|\vec{r}\times\vec{p}|=mvr_{\perp}\)
where \(r_{\perp}\) is the perpendicular distance from the origin to the particle’s line of motion.
The particle moves along the horizontal line
\(y=a\),
so the perpendicular distance from the origin to this line is simply
\(r_{\perp}=a\)
Therefore,
\(L=mvr_{\perp}=mv(a)=mva\)
Notice that the angular momentum depends only on the perpendicular distance from the origin to the line of motion and is independent of the particle’s \(x\)-coordinate \(x_0\).
Therefore, the correct answer is (B).
The rigid body shown in the diagram above consists of a vertical support post and two horizontal crossbars with spheres attached. The masses of the spheres and the lengths of the crossbars are indicated in the diagram. The body rotates about a vertical axis along the support post with constant angular speed \(\omega\). If the masses of the support post and the crossbars are negligible, what is the ratio of the angular momentum of the two upper spheres to that of the two lower spheres?
Correct Answer: \(\boxed{\mathrm{E}}\)
The angular momentum of a rotating rigid body is
\(L=I\omega\)
Since all spheres rotate with the same angular speed \(\omega\), the ratio of their angular momenta is equal to the ratio of their moments of inertia:
\(\frac{L_{\mathrm{upper}}}{L_{\mathrm{lower}}}=\frac{I_{\mathrm{upper}}}{I_{\mathrm{lower}}}\)
For the two upper spheres:
\(I_{\mathrm{upper}}=2(mL^2)\)
since there are two spheres of mass \(m\), each at a distance \(L\) from the axis.
For the two lower spheres:
\(I_{\mathrm{lower}}=2(2m)(2L)^2=16mL^2\)
Therefore,
$ \dfrac{L_{\mathrm{upper}}}{L_{\mathrm{lower}}} =\dfrac{2mL^2}{16mL^2} =\dfrac{1}{8} $
Thus, the angular momentum of the two upper spheres is one-eighth that of the two lower spheres.
Therefore, the correct answer is (E).
