Question

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.

Short answer

1. The angular acceleration is$-7.66rad/s^2$
2. The retarding torque is $-11.7N.m$
3. Thetotal energy transferred from mechanical energy to thermal energy by frictionis $4.59\times {10}^{4}J$
4. The number of revolutions rotated during the$32.0s$ is$624 rev$
5. Yes, Thermal energy can still be computed without additional information

Step-by-step solution

Listing the given quantities

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 $t=32.0s$

Understanding the concept of angular velocity and displacement

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

Formula:

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

Calculation of theangular acceleration

(a)

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

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

Calculation of theretarding torque

(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 $-11.7N.m$

 Calculation of thetotal energy transferred from mechanical energy to thermal energy by friction

(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 $4.59\times {10}^{4}J$

Calculation of the number of revolutions rotated during the  32.0 s.

(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$624 rev$

(e) Explanation

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.