A force of 49 N acts tangentially at the hig | vedclass

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(A) Let $F = 49 \ N$ be the applied force,$m = 20 \ kg$ be the mass of the solid sphere,and $r$ be its radius.For a solid sphere,the moment of inertia

Vedclass · Accepted Answer

$3.5$

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(A) Let $F = 49 \ N$ be the applied force,$m = 20 \ kg$ be the mass of the solid sphere,and $r$ be its radius.For a solid sphere,the moment of inertia about its center of mass is $I_{cm} = \frac{2}{5} mr^2$.The moment of inertia about the point of contact (bottom point) is $I = I_{cm} + mr^2 = \frac{2}{5} mr^2 + mr^2 = \frac{7}{5} mr^2$.The torque about the point of contact is $	au = F 	imes (2r)$.Using $	au = I \alpha$,we have $F 	imes 2r = (\frac{7}{5} mr^2) \alpha$.$49 	imes 2r = \frac{7}{5} 	imes 20 	imes r^2 	imes \alpha$.$98r = 28r^2 \alpha$.Since the sphere rolls without slipping,the linear acceleration $a$ of the center is $a = r \alpha$,so $\alpha = \frac{a}{r}$.Substituting this into the equation: $98r = 28r^2 (\frac{a}{r}) = 28ra$.$98 = 28a$.$a = \frac{98}{28} = 3.5 \ m/s^2$.

$A$ force of $49 \ N$ acts tangentially at the highest point of a solid sphere of mass $20 \ kg$,kept on a rough horizontal plane. If the sphere rolls without slipping,then the acceleration of the center of the sphere is (in $m/s^2$)

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