Sanjay and Ting, each with a ma of 25 kg, are riding oppoite each other on the edge of a 150 kg, 3. 0-m-diameter playground merry-go-round that' rotating at 21 rpm. Each walk traight inward and top 35 cm from the center

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20-12-2022
Physics

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Sanjay and Ting, each with a ma of 25 kg, are riding oppoite each other on the edge of a 150 kg, 3. 0-m-diameter playground merry-go-round that' rotating at 21 rpm. Each walk traight inward and top 35 cm from the center

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tutorconsortium125 tutorconsortium125 · Answer 1 · 2022-12-26T05:48:03-05:00

The linear velocity of Sanjay and Ting is 0.98 m/s. Linear velocity is a measure of how fast an object is moving in a straight line.

How to find linear velocity ?

To find the linear velocity of Sanjay and Ting, we can use the formula:

Linear velocity = Angular velocity * Radius

The angular velocity of the merry-go-round is 21 rpm, or 21 * 2 * pi / 60 = 0.7 rad/s.

The radius of the merry-go-round is 1.75 m,

which is the radius of the playground merry-go-round minus the distance that Sanjay and Ting walk inwards (1.75 m - 0.35 m = 1.4 m).

So the linear velocity of Sanjay and Ting is 0.7 rad/s * 1.4 m = 0.98 m/s.

It's important to note that this linear velocity is relative to the merry-go-round itself, not the ground. To find the linear velocity relative to the ground, we would need to take into account the rotation of the merry-go-round.

To learn more about angular velocity refer :

https://brainly.com/question/20432894

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