91Ó°ÊÓ

The lengthof road$1.20m$

Mass of the rod $M=6.40kg$

Mass of the ball$m=1.06kg$

The initial angular velocity${\mathrm{Ó\neg }}_0=39.0\text{rev}/\text{s}$

Time to come to rest $t=32.0s$

Constant angular acceleration kinematics can be used to compute the angular acceleration

Formula:

$\mathrm{Ó\neg }={\mathrm{Ó\neg }}_0+\mathrm{Î\pm }t$

(a)

$\mathrm{Ó\neg }={\mathrm{Ó\neg }}_0+\mathrm{Î\pm }t$Using this equation,$\mathrm{Ó\neg }=0$

$\begin{array}{rcl}\mathrm{Î\pm }& =& \frac{{\mathrm{Ó\neg }}_0}{t}\\ & =& \frac{39.0rev/s}{32.0s}\\ & =& -1.22rev/s^2\\ & =& -7.66rad/s^2\\ & & \end{array}$

(b)

$\mathrm{Ï„}=l\mathrm{Î\pm }$,$\mathrm{Ï„}$ is the torque and Iis the rotational inertia. The contribution of the rod to

$l=\frac{M{\mathrm{ℓ}}^2}{12}$, l is the length and M being mass of the rod. The contribution of each ball is$=m\left[\frac{ \mathrm{ℓ}   }{2}\right]{ }^2$where, m is the mass of the ball. The total rotational inertia is

$\begin{array}{rcl}l& =& \frac{M{\mathrm{â„“}}^2}{12}+2\frac{m{\mathrm{â„“}}^2}{4}\\ & =& \frac{\left[6.40kg\right]{\left[1.20m\right]}^2}{12}+2\frac{\left[1.06kg\right]{\left[1.20m\right]}^2}{4}\\ & =& 1.53kg.m^2\end{array}$

$\begin{array}{rcl}\mathrm{Ï„}& =& l\mathrm{Î\pm }\\ & =& \left[1.53kg.m^2\right]\left[-7.66rad/s^2\right]\\ & =& -11.7N.m\\ & & \end{array}$

The retarding torque is $-11.7N.m$

(c)

$\begin{array}{rcl}Ki& =& \frac{1}{2}I (\mathrm{Ó\neg }0{)}^2\\ & =& \frac{1}{2}\left[1.53kg.m^2\right]{\left[2\mathrm{Ï€}× (39rad/s)\right]}^2\\ & =& 4.59×{10}^{4}J\\ & & \end{array}$

Thetotal energy transferred from mechanical energy to thermal energy by frictionis $4.59×{10}^{4}J$

(d)

$\begin{array}{rcl}\mathrm{θ}& =& {\mathrm{Ó\neg }}_{0}t+\frac{1}{2}\mathrm{Î\pm }{t}^2\\ & & \text{=}\left[2\mathrm{Ï€}(39 rad/s)×(32.0sec)\right]+\frac{1}{2}\left[(-7.66 rad/s^2)×{(32.0 s)}^2\right]\\ & =& 3920rad\\ & =& 624 rev\\ & & \end{array}$

The number of revolutions rotated during the $32.0s$is$624 rev$

Only the mechanical energy that is converted to the thermal energy can still be computed without additional information.It is $4.59×{10}^{4}J$no matter howvaries with time, as longs as the system comes to rest.