In a tape recorder, the magnetic tape moves at a constant linear speed of 5.6 cm/s. To maintain this constant linear speed, the angular speed of the driving spool (the take-up spool) has to change accordingly.
(a) What is the angular speed of the take-up spool when it is empty, with radius r1 = 0.79 cm?
(b) What is the angular speed when the spool is full, with radius r2 = 2.32 cm?
(c) If the total length of the tape is 100.8 m, what is the average angular acceleration of the take-up spool while the tape is being played?
a. C=pi*D = 3.14 * (2*0.79) = 4.96 cm.
V=5.6cm/s * 6.28rad/4.96cm = 7.1rad/s.
b. C = 3.14 * (2*2.32) = 14.57 cm.
V = 5.6cm * 6.28rad/14.57cm = 2.4rad/s.
C. 0.00294 rad/s^2
C. 100.8 1m*100cm/m /5.6 cm/s = 1800s
(7.1-2.4)rad/s/1800s =.00261 rad/s^2
To answer these questions, we need to understand the relationship between linear speed, angular speed, and radius.
The linear speed of the tape, v, is related to the angular speed of the spool, ω, and the radius of the spool, r, by the equation:
v = ω * r
(a) To find the angular speed when the take-up spool is empty, with radius r1 = 0.79 cm, we can rearrange the equation:
ω = v / r
Plugging in the values, we get:
ω = 5.6 cm/s / 0.79 cm = 7.09 rad/s
So the angular speed of the spool when it is empty is 7.09 rad/s.
(b) To find the angular speed when the spool is full, with radius r2 = 2.32 cm, we can use the same equation:
ω = v / r
Plugging in the values, we get:
ω = 5.6 cm/s / 2.32 cm = 2.41 rad/s
So the angular speed of the spool when it is full is 2.41 rad/s.
(c) To find the average angular acceleration of the take-up spool while the tape is being played, we need to use the equation:
ωf - ωi = α * t
where ωf is the final angular speed, ωi is the initial angular speed, α is the angular acceleration, and t is the time.
Since the initial angular speed is the speed when the spool is empty and the final angular speed is the speed when the spool is full, we can use the values we already calculated:
ωf = 2.41 rad/s (final angular speed)
ωi = 7.09 rad/s (initial angular speed)
We also know that the total length of the tape, s, is related to the angular displacement, θ, by the equation:
s = r2 * θ
Plugging in the values, we get:
100.8 m = 2.32 cm * θ
Converting cm to meters, we have:
100.8 m = 0.0232 m * θ
Simplifying, we get:
θ = 100.8 m / 0.0232 m ≈ 4358.62 rad
Using the equation for angular acceleration, we can solve for α:
2.41 rad/s - 7.09 rad/s = α * t
Simplifying, we find:
-4.68 rad/s = α * t
We don't have a specific value for t, but we can use the relationship between angular displacement and time:
θ = ωi * t + (1/2)αt²
Plugging in the values, we get:
4358.62 rad = 7.09 rad/s * t + (1/2)αt²
Unfortunately, we don't have enough information to solve for α and t with the available equations and values.