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A disc of radius R rolls on a horizontal ground with linear acceleration $'a'$ and angular acceleration $' \alpha '$ as shown. The magnitude of the acceleration $'\alpha'$ of point P at an instant , when its linear velocity is V and angular velocity is $w$ will be,

 

\[\begin {array} {1 1} (a)\;\sqrt{(a+r\alpha)^2+(rw^2)^2} & \quad (b)\;\frac{ar}{R} \\ (c)\;\sqrt {r^2a^2+r^2w^4} & \quad  (d)\;r \alpha \end {array}\]

1 Answer

Total tangential acceleration $a_t =a+ r\alpha$
Total radial axes $a_r=\large\frac{V^2}{r}$
$\qquad= rw^2$
Hence $a= \sqrt {{a_t}^2+{a_r}^2}$
answered Nov 29, 2013 by meena.p
edited Jun 22, 2014 by lmohan717
 

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