Question

A particle is moving with 10m/s parallel to x axis start from the (0,2m) find angular speed of particle about origin at t=2sec

Answer 1

Shortest distance between origin and direction of speed is 2 m

So,
Angular momentum = mvr
= M × 10 m/s × 2 m
= 20 M

Mass of particle (M) is unknown.

∴ Momentum of particle about origin is 20 M (where M is mass of particle)

Answer 2

Answer:

2.5 rad/s

Distance travelled after 0.2 sec = 0.2 * 10 = 2             [as a = 0]

position of particle (2,2)

Radius after 0.2 sec = distance between origin and particle

                                   =  $\sqrt{2^{2} +2^{2} }$ = $2\sqrt{2}$

By conservation of angular momentum

$mv_{1} r_{1} =mv_{2} r_{2}$

$(10)(2)=(v_{2} )(2\sqrt{2} )$

$v_{2} =\frac{10}{\sqrt{2} }$

​ω = $\frac{v_{2} }{r_{2} }$

   = $\frac{10}{(\sqrt{2})(2\sqrt{2}) }$

   = $\frac{10}{4}$

   = 2.5 rad/s

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