Question

A particle of mass 5 g is moving in a circle of radius 0.5 metre with an angular velocity of 6 Radian per second find the change in linear velocity in half for resolution and the magnitude of acceleration of the particle​

Answer 1

Explanation:

$\sf\large\underline\purple{Given:-}$

$\displaystyle \sf \star \: mass \: = 5 \times {10}^{ - 3} \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \star \: \: r = 0.5m \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \star \: angular \: velocity \: = 6 \frac{rad}{s}$

$\sf\large\underline\purple{Solution:-}$

$\small\boxed{ \sf{ \underline{ \bold{1) : - after \: half \: a \: revolution \: the \: velocity \: changes its \: direction }}}}$

$\displaystyle \sf \bigstar \: change \: in \: momentum \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow \: mv - ( - mv) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow 2mv = 2m \times rw \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \longrightarrow 2mv = 2 \times 5 \times {10}^{ - 3} \times 0.5 \times 6 \\ \\ \displaystyle \sf \longrightarrow 0.3kg {m}^{ - s} \:$

__________________________________________________

$\sf\large\underline\purple{Answer\:\:\:2:-}$

$\small \boxed{ \sf{ \underline{ \bold{2.) \: change \: in \: linear \: momentum}}}}$

$\displaystyle \sf \longrightarrow \frac{dp}{dt} = f = ma \\ \\ \displaystyle \sf \longrightarrow a \: = \frac{1}{m} \frac{dp}{dt} \\ \\ \displaystyle \sf \longrightarrow a = \frac{1}{5 \times {10}^{ - 3} } \times 0.03 = 6 \ {m}^{ { - s}^{2} }$