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Darya [45]
3 years ago
10

A disk is spinning about its center with a constant angular speed at first. Let the turntable spin faster and faster, with const

ant angular acceleration ?. Which sketch qualitatively shows the direction of the acceleration vector a of the bug?

Physics
1 answer:
hoa [83]3 years ago
7 0

Answer:

4 (please see the attached file)

Explanation:

While the angular speed (counterclockwise) remained constant, the angular acceleration was just zero.

So, the only force acting on the bug (parallel to the surface) was the centripetal force, producing a centripetal acceleration directed towards the center of the disk.

When the turntable started to spin faster and faster, this caused a change in the angular speed, represented by the appearance of an angular acceleration α.

This acceleration is related with the tangential acceleration, by this expression:

at = α*r

This acceleration, tangent to the disk (aiming in the same direction of the movement, which is counterclockwise, as showed in the pictures) adds vectorially with the centripetal force, giving a resultant like the one showed in the sketch Nº 4.

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v: velocity of the person just when he touches the fire net = ?

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v^2=v_o^2+2gy

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y: height from the person jumps = 20.0m

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}

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mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

W=F_e\\\\mg=kx

you solve the last expression for x:

x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m

When the person is lying on the fire net the elongation of the fire net is 0.098m

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v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}

Next, you calculate x from the equation (1):

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m

The net fire is stretched 2.72 m

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