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12345 [234]
3 years ago
9

A rotating merry go-round makes one complete revolution in 4.2 s. (a) what is the linear speed of a child seated 1.3 m from the

center?
Physics
2 answers:
gregori [183]3 years ago
6 0
The radius, r, of the child from the center of the wheel is
r = 1.3 m

The wheel makes one revolution in 4.2 s. Its angular velocity is
ω = (2π rad)/(4.2 s) = 1.496 rad/s

The linear speed of the child is the tangential velocity, given by
v = rω
   = (1.3 m)*(1.496 rad/s)
   = 1.945 m/s

Answer: 1.95 m/s  (nearest hundredth)

Lera25 [3.4K]3 years ago
3 0
Remember in solving any Physics problem it is important to write all the given parameter. Given are 1 complete revolution of merry go-round or can be written as 1rev/s, this should be converted to radians, since it makes one complete revolution. Radius given equal to 1.3 m from the center. Simply follow the process,  ω=1rev/s x 2πrads/rev  ω=answer will in "rad/s" or cancel the unit "rev". The formula to follow is Speed=rω. just multiply radius and the angular speed. 
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<em><u>The </u></em><em><u>new </u></em><em><u>sp</u></em><em><u>e</u></em><em><u>ed </u></em><em><u>of </u></em><em><u>the </u></em><em><u>car </u></em><em><u>is </u></em><em><u>3</u></em><em><u>0</u></em><em><u> </u></em><em><u>m/</u></em><em><u>s.</u></em>

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Doubt clarification - use comment section.

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