Answer:
.94 rad/s
Explanation:
Total length of videotape = 240 m
Total time of play = 2.1 h -> 7560s
- Linear speed of the video tape is given as:
[tex]v = \frac{d}{t}[/tex]
[tex]v = \frac{240}{7560} = .032 m/s[/tex]
- we know that:
[tex]v = rw[/tex]
Since we know that both reels have the same speed,
now, we incorporate v = rw to find w
[tex].032 = .03413w\\w = \frac{.032}{.03413} = .94 rad/s[/tex]
Both reels will have the same angular speed of 0.930 rad/s.
Given data:
The total length of videotape cassette is, L = 240 m.
The time interval for the play is, [tex]t = 2.1 \;\rm h = 2.1 \times 3600 =7560 \;\rm s[/tex]
The outer radius of tape is, [tex]r_{o}=47 \;\rm mm[/tex].
The inner radius of tape is, [tex]r_{i}=11 \;\rm mm[/tex].
The videotape will undergo rotational motion and the value of angular velocity is given as,
[tex]\omega =\dfrac{v}{r}[/tex] ....................................................................(1)
Here, v is the linear speed of the cassette and r is the effective radius of the cassette, and its value is,
[tex]r = \sqrt{\dfrac{r^{2}_{o}+r^{2}_{i}}{2}} \\\\r = \sqrt{\dfrac{47^{2}+11^{2}}{2}} \\\\r = 34.13 \;\rm mm\\\\r \approx 0.03413 \;\rm m[/tex]
And, the linear speed of the cassette is given as,
[tex]v=\dfrac{L}{t}\\\\v=\dfrac{240}{7560}\\\\v=0.0317 \;\rm m/s[/tex]
Substitute the values in equation (1) as,
[tex]\omega =\dfrac{0.0317}{0.03413}\\\\ \omega = 0.930 \;\rm rad/s[/tex]
Thus, we can conclude that at some point during the play, both reels will have the same angular speed of 0.930 rad/s.
Learn more about the angular speed here:
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