A particle of mass $5 g$ is moving in a circle of radius $0.5 m$ with an angular velocity of $6 r a d / s$ . Find (i) the change in linear momentum in half a revolution (ii) the magnitude of the acceleration of the particle.

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Solution

Mass $m = 5 \times 10^{- 3} k g$

Radius $r = 0.5 m$

Angular velocity $w = 6 r a d / s$

$(i)$ After half a revolution, the velocity changes its direction

Change in linear momentum

$= m v - (- m v) = 2 m v = 2 m \times r w = 2 \times 5 \times 10^{- 3} \times 0.5 \times 6 = 0.03 k g m / s$

$(i i)$ Change in linear momentum

$= \frac{d p}{d t} = F = m a ∴ a = \frac{1}{m} \frac{d p}{d t} = \frac{1}{5 \times 10^{- 3}} \times 0.03 = 6 m / s^{2}$

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A particle of mass 5g is moving in a circle of radius 0.5m with an angular velocity of 6rad/s. Find (i) the change in linear momentum in half a revolution (ii) the magnitude of the acceleration of the particle.

Mass m=5×10−3kg Radius r=0.5m Angular velocity w=6rad/s (i) After half a revolution, the velocity changes its direction Change in linear momentum =mv−(−mv)=2mv=2m×rw=2×5×10−3×0.5×6=0.03kgm/s (ii) Change in linear momentum =dpdt=F=ma∴a=1mdpdt=15×10−3×0.03=6m/s2

A particle of mass $5 g$ is moving in a circle of radius $0.5 m$ with an angular velocity of $6 r a d / s$ . Find (i) the change in linear momentum in half a revolution (ii) the magnitude of the acceleration of the particle.