# A particle moves along a circle of radius 20/π 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 began the tangential acceleration is ??

Jan 13, 2018

The tangential acceleration is $= 40 m {s}^{-} 2$

#### Explanation:

Assuming that the particle starts from rest

The initial angular velocity is ${\omega}_{0} = 0 r a {\mathrm{ds}}^{-} 1$

The radius of the circle is $r = \frac{20}{\pi} m$

The final velocity of the particle is $v = 80 m {s}^{-} 1$

The final angular velocity is

$\omega = \frac{v}{r} = \frac{80}{\frac{20}{\pi}} = 4 \pi r a {\mathrm{ds}}^{-} 1$

The angle is $\theta = 2 \cdot 2 \pi = 4 \pi$

Applying the equation

${\omega}^{2} = {\omega}_{0}^{2} + 2 \alpha \theta$

Where $\alpha$ is the angular acceleration

$\alpha = \frac{{\omega}^{2} - {\omega}_{0}^{2}}{2 \theta}$

$= \frac{{\left(4 \pi\right)}^{2} - 0}{2 \cdot 4 \pi} = \frac{16 {\pi}^{2}}{8 \pi} = \left(2 \pi\right) r a {\mathrm{ds}}^{-} 2$

The tangential acceleration is

${a}_{T} = \alpha \cdot r = 2 \pi \cdot \frac{20}{\pi} = 40 m {s}^{-} 2$