On a solid sphere lying on a horizontal surfa | vedclass

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(A) Let $M$ be the mass and $R$ be the radius of the solid sphere. The moment of inertia about the center of mass is $I = \frac{2}{5}MR^2$.Applying Ne

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$\frac{15F}{7M}$

Vedclass · Answer

(A) Let $M$ be the mass and $R$ be the radius of the solid sphere. The moment of inertia about the center of mass is $I = \frac{2}{5}MR^2$.Applying Newton's second law for linear motion: $F + f = Ma$,where $f$ is the friction force.Applying torque equation about the center of mass: $F(R/2) - fR = I\alpha = (\frac{2}{5}MR^2)\alpha$.Since there is no slipping at the point of contact,the acceleration of the center of mass $a = R\alpha$,so $\alpha = a/R$.Substituting $\alpha$: $F/2 - f = \frac{2}{5}Ma \implies f = F/2 - \frac{2}{5}Ma$.Substituting $f$ into the linear equation: $F + F/2 - \frac{2}{5}Ma = Ma \implies \frac{3F}{2} = \frac{7}{5}Ma \implies a = \frac{15F}{14M}$.The angular acceleration is $\alpha = a/R = \frac{15F}{14MR}$.The acceleration of the top point is $a_{top} = a + R\alpha = a + a = 2a$.Therefore,$a_{top} = 2 	imes \frac{15F}{14M} = \frac{15F}{7M}$.

On a solid sphere lying on a horizontal surface,a force $F$ is applied at a height of $R/2$ from the centre of mass. The initial acceleration of a point at the top of the sphere is (there is no slipping at any point).

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