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3 years ago

14

A certain CD has a playing time of 74.0 minutes. When the music starts, the CD is rotating at an angular speed of 480 revolution

s per minute (rpm). At the end of the music, the CD is rotating at 210 rpm. Find the magnitude of the average angular acceleration of the CD. Express your answer in rad/s2.

1 answer:

Rashid [163]3 years ago

3 0

Answer: $0.00636\ rad/s^2$

Explanation:

Given

CD has a playing time of $t=74\ min\ or\ 74\times 60\ s$

Initial angular speed of CD is $480\ rpm$

Final angular speed of DC is $210\ rpm$

Angular speed, when rpm is given

$\omega =\dfrac{2\pi N}{60}$

$\omega_i=\dfrac{2\pi \times 480}{60}\\\\\Rightarrow \omega_i=16\pi \ rad/s$

Final speed

$\Rightarrow \omega_f=\dfrac{2\pi \times 210}{60}\\\\\Rightarrow \omega_f=7\pi \ rad/s$

Using equation of angular motion

$\Rightarrow \omega_f=\omega_i+\alpha t$

Insert the values

$\Rightarrow 7\pi =16\pi +\alpha \times 74\times 60\\\Rightarrow -9\pi =\alpha \cdot (4440)\\\\\Rightarrow \alpha=-\dfrac{9\pi}{4440}\\\\\Rightarrow \alpha=-0.00636\ rad/s^2$

Magnitude of angular acceleration is $0.00636\ rad/s^2$

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The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

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$s=u_y t + \frac{1}{2}gt^2$

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$g=-9.8 m/s^2$

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

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Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

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So, the ball lands 123.3 m far from the initial point.

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