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.

Question

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Respuesta :

CarliReifsteck CarliReifsteck · Answer 1 · 2019-12-04T12:24:01+01:00

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

[tex]m\omega_{i}=(mr^2+\dfrac{Mr^2}{2})\omega_{f}[/tex]

[tex]mvR=(mr^2+\dfrac{Mr^2}{2})\omega[/tex]

Put the value into the formula

[tex]29\times2.6\times1.7=(29\times1.7^2+\dfrac{49\times1.7^2}{2})\omega_{f}[/tex]

[tex]\omega_{f}=\dfrac{29\times2.6\times1.7}{(29\times1.7^2+\dfrac{49\times1.7^2}{2})}[/tex]

[tex]\omega_{f}=0.83\ rad/s[/tex]

Hence, The angular speed is 0.83 rad/s.

okpalawalter8 okpalawalter8 · Answer 2 · 2022-04-12T11:37:50+02:00

The angular speed is mathematically given as

wf=0.83rad/s

Question Parameter(s):

Generally, the equation for the conservation of momentum is mathematically given as

[tex]m\omega_{i}=(mr^2+\frac{Mr^2}{2})\omega_{f}[/tex]

Therefore

[tex]29\*2.6*1.7=(29\times1.7^2+\frac{49*1.7^2}{2})\omega_{f}[/tex]

wf=0.83rad/s

In conclusion, the angular speed is

wf=0.83rad/s