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2 years ago

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A playground ride consists of a disk of mass M = 49 kg and radius R = 1.7 m mounted on a low-friction axle. A child of mass m =

29 kg runs at speed v = 2.6 m/s on a line tangential to the disk and jumps onto the outer edge of the disk.

1 answer:

HACTEHA [7]2 years ago

4 0

Answer:

The angular speed is 0.83 rad/s.

Explanation:

Given that,

Mass of disk M=49 kg

Radius = 1.7 m

Mass of child m= 29 kg

Speed = 2.6 m/s

Suppose if the disk was initially at rest , now how fast is it rotating

We need to calculate the angular speed

Using conservation of momentum

$m\omega_{i}=(mr^2+\dfrac{Mr^2}{2})\omega_{f}$

$mvR=(mr^2+\dfrac{Mr^2}{2})\omega$

Put the value into the formula

$29\times2.6\times1.7=(29\times1.7^2+\dfrac{49\times1.7^2}{2})\omega_{f}$

$\omega_{f}=\dfrac{29\times2.6\times1.7}{(29\times1.7^2+\dfrac{49\times1.7^2}{2})}$

$\omega_{f}=0.83\ rad/s$

Hence, The angular speed is 0.83 rad/s.

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3 years ago

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$f'=(\frac{v+v_r}{v+v_s})f$

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$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

$v_s$ is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

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In the reflector frame (= on surface B), we have also

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$\lambda=\frac{v}{f}$

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(d) 0.225 m

The wavelength of the wave is given by

$\lambda=\frac{v}{f}$

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f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

$\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m$

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