A particle moves along a circle of radius left( {frac{{20}}{π }} right),m with constant

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3-2.Motion in Plane

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A particle moves along a circle of radius $\left( {\frac{{20}}{\pi }} \right)\,m$ with constant tangential acceleration. If the velocity of the particle is $80 \,m/s$ at the end of the second revolution after motion has begin, the tangential acceleration is

A$40$

B$640$

C$160\,\pi$

D$40\,\pi$

(AIPMT-2003)

Solution

$a_{t}=r \alpha=$ const. $\Rightarrow \alpha=$ const.
${\omega _i} = 0$ $\theta=2 \pi$
$\omega_{f}=\frac{V}{r}=\frac{80}{(20 / \pi)}=4 \pi$
$\alpha=\frac{\omega_{f}^{2}-\omega_{i}^{2}}{2 \theta}=2 \pi \operatorname{rad} / \sec ^{2}$
$\mathrm{a}_{\mathrm{t}}=\frac{20}{\pi} \times 2 \pi=40 \mathrm{\,m} / \mathrm{s}^{2}$

Standard 11

Physics

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