Question

Racing disks. Figure $\mathbf{10}\mathbf{-}\mathbf{51}\mathbf{}$shows two disks that can rotate about their centers like a merry-go-round. At time$\mathit{t}\mathbf{=}\mathbf{0}$, the reference lines of the two disks have the same orientation. Disk A is already rotating, with a constant angular velocity of$\mathbf{9}\mathbf{.5}\mathbf{}\mathbf{\text{\hspace{0.17em}rad}}\mathbf{/}\mathbf{\text{s}}$. Disk B has been stationary but now begins to rotate at a constant angular acceleration of$\mathbf{2}\mathbf{.2}\mathbf{}\mathbf{\text{rad}}\mathbf{/}{\mathbf{\text{s}}}^{\mathbf{2}}\mathbf{}$. (a) At what timewill the reference lines of the two disks momentarily have the same angular displacement$\mathit{\theta }$? (b) Will that timebe the first time since$\mathit{t}\mathbf{=}\mathbf{0}$that the reference lines are momentarily aligned?

Short answer

a) The time at which reference lines of two disks momentarily have the same angular displacement$\theta$is$8.6\text{\hspace{0.17em}s}$

b) There are two times for which the reference lines are momentarily aligned.

Step-by-step solution

The given data

a) Angular velocity of disk A,${\omega }_A=9.5\text{rad/s}$

b) Angular acceleration of disk B,${\alpha }_B=2.2{\text{rad/s}}^2$${\alpha }_B=2.2{\text{rad/s}}^2$

Understanding the concept of angular form of kinematics

The study of the rotational motion of the body is given as the angular form of kinematics. The angular entities like displacement, velocity, and acceleration are related to the linear kinematics by a radial value. Further, using this radial theorem, we can observe the relatable kinematic equations in angular form.

Formulae:

The angular displacement of a body in rotational motion analogy to 2^nd law of kinematic equations, $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$ (i)

where,${\omega }_0$is the initial angular velocity of the body,$t$is the angular acceleration of the body,is the time of motion.

The roots of a quadratic equation$a{x}^2+bx+c=0$ , $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ (ii)

a) Calculation of the reference time at which both the disks will have same angular displacement

Using equation (i), the angular displacement of the disk A can be given as:

(As the angular velocity is constant, that implies no angular acceleration)

${\theta }_A={\omega }_{A}t$………………………… (I)

And, for disc B using second kinetic equation in rotational motion,

Using equation (i), the angular displacement of the disk B can be given as:

(As the angular acceleration is constant, that implies no angular velocity)

${\theta }_B=\frac{1}{2}{\alpha }_B{t}^2$………………………… (II)

AS per our requirement for both the angular displacements to be equal, we get the reference time using the above equations as follows:

$\begin{array}{c}{\theta }_A={\theta }_B\\ {\omega }_{A}t=\frac{1}{2}{\alpha }_B{t}^2\\ (9.5\text{rad/s})t=\frac{1}{2}\left(2.2{\text{rad/s}}^2\right)t^2\\ (9.5\text{rad/s})=\left(1.1{\text{rad/s}}^2\right)t\\ t=\frac{(9.5)}{\left(1.1\right)}\text{s}\\ =8.6\text{\hspace{0.17em}s}\end{array}$

Therefore, the time at which reference lines of two disks momentarily have the same angular displacement$\theta$ is.$8.6\text{\hspace{0.17em}s}$

b) Calculation of the number of times at which the reference lines are aligned 

Now, the change in the angular displacement of the two disks can be given using equations (I) and (II) as follows:

${\theta }_A-{\theta }_B={\omega }_{A}t-\frac{1}{2}{\alpha }_B{t}^2$

And, the reference line of both the disks will be aligned if the following condition is satisfied:

${\theta }_A-{\theta }_B=2\pi \text{n}$

Thus, equating above two equations, we get the reference time as follows:

$\begin{array}{c}{\omega }_{A}t-\frac{1}{2}{\alpha }_B{t}^2=2\pi n\\ (9.5\text{rad/s})t-\frac{1}{2}\left(2.2{\text{rad/s}}^2\right)t^2=2\pi \text{n}\\ \left(1.1{\text{rad/s}}^2\right)t^2-(9.5\text{rad/s})t+2\pi \text{n}=0\end{array}$

Thus, the root values can be given using equation (ii) as follows:

(for the coefficients values as:$a=1.1,b=-9.5,c=2\pi n$)

$t_n=\frac{\left(9.5\right)\pm \sqrt{{\left(9.5\right)}^2-8\pi \left(1.1\right)\left(n\right)}}{2\left(1.1\right)}$

Thus at,$n=0$ we get the reference time for the positive root value as:

$\begin{array}{c}{t}_0=\frac{\left(9.5\right)+\sqrt{{\left(9.5\right)}^2-8\pi \left(1.1\right)\left(0\right)}}{2\left(1.1\right)}\\ =\left[\frac{\left(9.5\right)+\left(9.5\right)}{2.2}\right]\text{s}\\ =\left(\frac{9.5}{1.1}\right)\text{s}\\ =8.6\text{s}\end{array}$

Similarly, the reference time for the negative root value as:

$\begin{array}{c}{t}_0=\frac{\left(9.5\right)-\sqrt{{\left(9.5\right)}^2-8\pi \left(1.1\right)\left(0\right)}}{2\left(1.1\right)}\\ =\left[\frac{\left(9.5\right)-\left(9.5\right)}{2.2}\right]\text{s}\\ =0\text{s}\end{array}$

Again, on further calculations, we can see that the reference time values remain same for$n=1,2,3$as$t_1=8.6\text{s}\&t_2=0\text{\hspace{0.17em}s}$:

Therefore, for the two reference time values, the reference lines of both disks are momentarily aligned.