Respuesta :
Answer:
a)
[tex]1.53[/tex] rad s⁻¹
b)
[tex] 3[/tex] rev
Explanation:
[tex]d[/tex] = diameter of the merry-go-round = 5.1 m
[tex]r[/tex] = radius of the merry-go-round = [tex]\frac{d}{2}[/tex] = [tex]\frac{5.1}{2}[/tex] = 2.55 m
[tex]T[/tex] = time period of the merry-go-round = 4.1 s
[tex]w_{i}[/tex] = initial angular velocity of merry-go-round
Initial angular velocity is given as
[tex]w_{i}[/tex] = [tex]\frac{2\pi }{T}[/tex]
[tex]w_{i}[/tex] = [tex]\frac{2\pi }{4.1}[/tex]
[tex]w_{i}[/tex] = 1.53 rad s⁻¹
b)
[tex]w_{f}[/tex] = final angular velocity of merry-go-round = 0 rad s⁻¹
[tex]t[/tex] = time taken to stop = 25 s
[tex]\theta[/tex] = angular displacement = ?
[tex]n[/tex] = Number of revolutions made before stopping
Angular displacement is given as
[tex]\theta =\frac{(w_{f}+ w_{i})t}{2}[/tex]
[tex]\theta =\frac{(0 + 1.53)(25)}{2}[/tex]
[tex]\theta =19.125[/tex] rad
Number of revolutions are given as
[tex]n = \frac{\theta}{2\pi }[/tex]
[tex]n = \frac{19.125}{2\pi }[/tex]
[tex]n = 3[/tex] rev
For the merry-go-round that is initially turning with a 4.1 s period and stops in 25 seconds, we have:
A. The speed of the child on the rim before slowing is 1.5 rad/s.
B. The merry-go-round makes 3.1 revolutions as it stops.
Part A. The speed of a child on the rim
We can find the speed with the following equation:
[tex] \omega = \frac{2\pi}{T} [/tex] (1)
Where:
- T: is the period = 4.1 s
After entering the value of T into equation (1), we get:
[tex]\omega = \frac{2\pi}{T} = \frac{2\pi}{4.1 s} = 1.5 \:rad/s[/tex]
Hence, the speed of the child on the rim is 1.5 rad/s.
Part B. Number of revolutions
We can calculate the number of revolutions of the merry-go-round as follows:
[tex] \theta_{f} = \theta_{i} + \omega_{i}t + \frac{1}{2}\alpha t^{2} [/tex] (2)
Where:
- [tex]\theta_{i}[/tex]: is the initial number of revolutions = 0
- [tex]\theta_{i}[/tex]: is the final number of revolutions =?
- [tex]\omega_{i}[/tex]: is the angular speed = 1.5 rad/s
- [tex]\alpha[/tex]: is the angular acceleration
- t: is the time at which the merry-go-round stops = 25 s
The angular acceleration is given by:
[tex] \omega_{f} = \omega_{i} + \alpha t [/tex]
Where:
[tex] \omega_{f}[/tex]: is the final angular speed of the merry-go-round = 0 (it stops)
Hence, the angular acceleration is:
[tex]\alpha = \frac{\omega_{f} - \omega_{i}}{t} = \frac{0 - 1.53 \:rad/s}{25 s} = -0.06 \:rad/s^{2}[/tex]
By entering the acceleration value into equation (2), we have:
[tex]\theta_{f} = 1.5 \:rad/s*25 s + \frac{1}{2}*(-0.06 \:rad/s^{2})*(25 s)^{2} = 19.5 \:rad*\frac{1 \:rev}{2\pi rad} = 3.1 \:rev[/tex]
Therefore, the merry-go-round makes 3.1 revolutions as it stops.
Find more about angular acceleration here:
brainly.com/question/26133378
I hope it helps you!
