Attached to each end of a thin steel rod of length$\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{}\mathit{m}$ and mass$\mathbf{6}\mathbf{.}\mathbf{40}\mathbf{}\mathit{k}\mathit{g}$ is a small ball of mass$\mathbf{1}\mathbf{.}\mathbf{06}\mathbf{}\mathit{k}\mathit{g}$ . The rod is constrained to rotate in a horizontal plane about a vertical axis through its midpoint. At a certain instant, it is rotating at $\mathbf{}\mathbf{39}\mathbf{.}\mathbf{0}\mathbf{}\mathit{r}\mathit{e}\mathit{v}\mathbf{/}\mathit{s}$. Because of friction, it slows to a stop in$\mathbf{32}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$ .Assuming a constant retarding torque due to friction, compute

(a) the angular acceleration,

(b) the retarding torque,

(c) the total energy transferred from mechanical energy to thermal energy by friction, and

(d) the number of revolutions rotated during the$\mathbf{32}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$ .

(e) Now suppose that the retarding torque is known not to be constant. If any of the quantities (a), (b), (c), and (d) can still be computed without additional information, give its value.

The lengthof road$1.20m$

Mass of the rod $M=6.40kg$

Mass of the ball$m=1.06kg$

The initial angular velocity${\omega }_0=39.0\text{rev}/\text{s}$

Time to come to rest $t=32.0s$

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

Formula:

$\omega ={\omega }_0+\alpha t$

(a)

$\omega ={\omega }_0+\alpha t$Using this equation,$\omega =0$

$\begin{array}{rcl}\alpha & =& \frac{{\omega }_0}{t}\\ & =& \frac{39.0rev/s}{32.0s}\\ & =& -1.22rev/s^2\\ & =& -7.66rad/s^2\\ & & \end{array}$

(b)

$\tau =l\alpha$,$\tau$ is the torque and Iis the rotational inertia. The contribution of the rod to

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

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

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

The retarding torque is $-11.7N.m$

(c)

$\begin{array}{rcl}Ki& =& \frac{1}{2}I (\omega 0{)}^2\\ & =& \frac{1}{2}\left[1.53kg.m^2\right]{\left[2\pi \times (39rad/s)\right]}^2\\ & =& 4.59\times {10}^{4}J\\ & & \end{array}$

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

(d)

$\begin{array}{rcl}\theta & =& {\omega }_{0}t+\frac{1}{2}\alpha t^2\\ & & \text{=}\left[2\pi (39 rad/s)\times (32.0sec)\right]+\frac{1}{2}\left[(-7.66 rad/s^2)\times {(32.0 s)}^2\right]\\ & =& 3920rad\\ & =& 624 rev\\ & & \end{array}$

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 $4.59\times {10}^{4}J$no matter howvaries with time, as longs as the system comes to rest.