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Question

A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of 3 m/s and a centripetal acceleration →a of magnitude 2 m/s2. Position vector →r locates him relative to the rotation axis. What is the magnitude of →r ? What is the direction of →r when →a is directed due east?

A

1.5 m, east

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B

4.5 m, east

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C

1.5 m, west

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D

4.5 m, west

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Solution

The correct option is D 4.5 m, west v=3m/s,a=2m/s2 a=v2r⇒r=v2a=322=92=4.5m ∴|→r|=4.5m When →a is due east →r will be due west as r points to the particle's position W.r.t. centre As →a always points toward the centre of motion Please watch the video if you are still doubtful about why a:v2r

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