A particle moves in a circle of radius 0.5 m at a speed that uniformly increases. Find the angular acceleration of particle if its speed changes from 2.0 m/s to 4.0 m/s in 4.0 s.

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Solution

The correct option is A
The tangential acceleration of the particle is
$a_{t} = \frac{d v}{d t} = \frac{4.0 - 2.0}{4.0} = 0.5 m / s^{2} α = \frac{a_{t}}{r} = \frac{0.5}{0.5} = 1 r a d / s^{2}$

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A particle moves in a circle of radius 0.5 m at a speed that uniformly increases. Find the angular acceleration of particle if its speed changes from 2.0 m/s to 4.0 m/s in 4.0 s.

The correct option is A The tangential acceleration of the particle is at=dvdt=4.0−2.04.0=0.5m/s2α=atr=0.50.5=1 rad/s2

The correct option is A
The tangential acceleration of the particle is
$a_{t} = \frac{d v}{d t} = \frac{4.0 - 2.0}{4.0} = 0.5 m / s^{2} α = \frac{a_{t}}{r} = \frac{0.5}{0.5} = 1 r a d / s^{2}$