Answer to Question #162266 in Mechanics | Relativity for Alexis

Question

The tape in a videotape cassette has a total length 264 m and can play for 1.8 h. As the tape starts to play, the full reel has an outer radius of 46 mm and an inner radius of 12 mm. At some point during the play, both reels will have the same angular speed. What is this common angular speed? Answer in units of rad/s.

Answer 1

Explanations & Calculations

• Both the reels have the same tangential speed (at which the tape normally runs throughout for 1.8hours) during the playing process, otherwise, the released length from the feed reel will not be wound in the catchup reel.
• Then at a given instance from both reels, it could be written,

$\qquad\qquad \begin{aligned} \small V_{common}&= \small r_{catchup}\omega_{catchup}=r_{feed}\omega_{feed} \end{aligned}$

• As the angular speeds become equal to each other the outer radii of both the reels also become equal.
• At that moment due to the symmetry of the system outer radius of each is $r_{common}=\frac{(46+12)mm}{2}=\small29\,mm=0.029\,m$
• What actually happens is the initial tape width ($\small 46mm-12mm=34mm$) is reduced to its half ($\small 17mm$), which we got altogether as the diameter ($\small 17mm+12mm=29mm$).
• Then,

$\qquad\qquad \begin{aligned} \small V_{common}&= \small r_{common}\times\omega_{omega}\\ \small \frac{264\,m}{1.8\times3600\,s}&= \small 0.029m\times\omega_{common}\\ \small \omega_{common}&= \small \bold{1.405\,rad/s} \end{aligned}$