= Solution
\b[Definitions]. The <torque> of a <force> $\mathbf F$ about an origin is $\boldsymbol\tau=\mathbf r\times\mathbf F$, where $\mathbf r$ runs from that origin to the force's point of application. Its component about a fixed axis changes <angular momentum> about that axis. The <moment of inertia> about that axis is
$$
\boxed{I=\int s^2\,dm,}
$$
where $s$ is perpendicular distance to the axis. It measures resistance to <angular acceleration>; for fixed-axis rigid rotation, $\tau_{\rm net}=I\alpha$. The units are $\mathrm{N\,m}$ for <torque> and $\mathrm{kg\,m^2}$ for <moment of inertia>.
\b[Mass and axial inertia]. Slice the shaft perpendicular to its common axis. A slice of radius $r(z)$ has <mass> $dm=\rho\pi r(z)^2dz$ and <moment of inertia> $dI=\tfrac12r(z)^2dm$, using the <moment of inertia of a uniform disc>. Thus the <axial moment of inertia of a solid of revolution> uses the second and fourth powers of its radius:
$$
M=\rho\pi\int r(z)^2dz,\qquad
I=\frac{\rho\pi}{2}\int r(z)^4dz.
$$
For the tapered section, take $z$ from its narrow end toward its wide end, so $r(z)=a(1+z/l)$ for $0\leq z\leq l$. Its contributions are
$$
M_{\rm taper}=\rho\pi a^2l\int_0^1(1+u)^2du
=\frac73\rho\pi a^2l,
\qquad
I_{\rm taper}=\frac{\rho\pi a^4l}{2}\int_0^1(1+u)^4du
=\frac{31}{10}\rho\pi a^4l.
$$
The two cylinders contribute $4\rho\pi a^2l$ and $\rho\pi a^2l$ to <mass>, and $8\rho\pi a^4l$ and $\tfrac12\rho\pi a^4l$ to <moment of inertia>. Adding the three sections gives
$$
\boxed{M=\left(4+\frac73+1\right)\rho\pi a^2l
=\frac{22}{3}\rho\pi a^2l,}
$$
$$
I=\left(8+\frac{31}{10}+\frac12\right)\rho\pi a^4l
=\frac{58}{5}\rho\pi a^4l,
\qquad
\boxed{I=\frac{58/5}{22/3}Ma^2=\frac{87}{55}Ma^2.}
$$
All slice rotation axes coincide with the shaft axis, so the moments add directly.
\b[Frictional torque]. The normal load on the narrow circular end is $Mg$. With the stipulated uniform <pressure>, $p_N=Mg/(\pi a^2)$. A contact annulus of radius $s$ and width $ds$ has area $dA=2\pi s\,ds$, normal <force> $p_NdA$ and tangential <kinetic friction> $dF=\mu p_NdA$. Multiplying by its lever arm $s$ and integrating gives the <uniform-pressure friction torque on a circular contact>:
$$
G_f=\int_0^a s\,\mu\frac{Mg}{\pi a^2}\,2\pi s\,ds
=\frac{2\mu Mg}{a^2}\frac{a^3}{3},
\qquad
\boxed{G_f=\frac{2M\mu ga}{3}.}
$$
Here $G_f$ is the positive magnitude; its <torque> opposes the <angular velocity>.
\b[Steady speed and stopping rotations]. Mechanical <power> in rotation is $P=\tau\omega$. At steady <angular velocity>, the driving and frictional torques balance, so <constant-power rotation against dry friction> gives
$$
\boxed{\omega_*=\frac{P}{G_f}=\frac{3P}{2M\mu ga}.}
$$
Using $g=9.81\,\mathrm{m\,s^{-2}}$ and $a=0.100\,\mathrm m$,
$$
I=0.0316364\,\mathrm{kg\,m^2},\qquad
G_f=0.3924\,\mathrm{N\,m},\qquad
\boxed{\omega_*=25.4842\,\mathrm{rad\,s^{-1}}\simeq25.5\,\mathrm{rad\,s^{-1}}.}
$$
Once the motor is removed, the speed-independent <kinetic friction> gives constant <angular acceleration> $-G_f/I$. Equivalently, the initial <rotational kinetic energy> is dissipated as frictional work. The <stopping angle under constant resisting torque> is therefore
$$
G_f\Delta\phi=\frac12I\omega_*^2,\qquad
\Delta\phi=\frac{I\omega_*^2}{2G_f}=26.1800\,\mathrm{rad}.
$$
The total <angular displacement> corresponds to
$$
N=\frac{\Delta\phi}{2\pi}=\frac{IP^2}{4\pi G_f^3}=4.16668\ \text{revolutions}.
$$
Hence the shaft completes
$$
\boxed{\lfloor N\rfloor=4\ \text{complete rotations},}
$$
and then another $0.16668$ of a rotation. Its stopping time is $I\omega_*/G_f=2.05461\,\mathrm s$. Using the rounded value $g=9.8\,\mathrm{m\,s^{-2}}$ changes the fractional rotation slightly but leaves four complete rotations.
Back to article page
