A playground merry-go-round of radius 2.00 m has a moment of inertia I = 265 kg·m2 and is rotating about a frictionless vertical axle. As a child of mass 25.0 kg stands at a distance of 1.00 m from the axle, the system (merry-go-round and child) rotates at the rate of 13.5 rev/min. The child then proceeds to walk toward the edge of the merry-go-round.
What is the angular speed of the system when the child reaches the edge?
rad/s
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Verified answer
The angular speed is the same everywhere.
To find angular speed we divide the angle covered by the time it is covered in.
We convert rev/min to rad /sec.
1 rev. = 2*pi(full circle)
13.5 rev/min = ((13.5*2*pi)/(60)) = 1.4 rad/sec. (FINAL ANSWER)
If you wish to find linear speed, just multiply the angular speed by the radius of the object.
I hope this helps.
A l. a. campagne, le rythme de l. a. vie est plus lent et j'adore marcher sur les chemins de terre ou sur le bord de l'eau, je ressens un regain d'énergie entourée de cette belle nature autour de moi.