Question
The total acceleration of a particle in circular motion is $5 \mathrm{~m} / \mathrm{s}^{2}$. If the speed of the particle decreases at a rate of $3 \mathrm{~m} / \mathrm{s}^{2}$, assuming $r=2 \mathrm{~m}$, the angular speed $\omega$ of the particle relative to the centre of the circle is(a) $1 \mathrm{rad} / \mathrm{s}$(b) $\sqrt{2} \mathrm{rad} / \mathrm{s}$(c) $2 \mathrm{rad} / \mathrm{s}$(d) $2 \sqrt{2} \mathrm{rad} / \mathrm{s}$
Step 1
Given that the total acceleration $a = 5 \, m/s^2$ and the tangential acceleration $a_t = 3 \, m/s^2$, we can substitute these values into the formula to find the centripetal acceleration $a_c$. Show more…
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Key Concepts
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Motions in Two and Three Dimensions
Section A
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