A carnival merry-go-round rotates about a vertical axis at a constant rate. A

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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

\to a

of magnitude 2 m/s². Position vector

\to r

locates him relative to the rotation axis.
What is the magnitude of

\to r

? What is the direction of

\to r

when

\to a

is directed due east?

Options:

A . 1.5 m, east

B . 4.5 m, east

C . 1.5 m, west

D . 4.5 m, west

Answer: Option D
:
D

v = 3 m / s, a = 2 m / s^{2}

a = \frac{v^{2}}{r} \Rightarrow r = \frac{v^{2}}{a} = \frac{3^{2}}{2} = \frac{9}{2} = 4.5 m

∴ | \to r | = 4.5 m

When

\to a

is due east

A Carnival Merry-go-round Rotates About A Vertical Axis At...

\to r

will be due west as r points to the particle's position W.r.t. centre
As

\to a

always points toward the centre of motion
Please watch the video if you are still doubtful about why a:

\frac{v^{2}}{r}

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