A uniform spherical object of mass M and radi | vedclass

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Question

(A) For an object rolling down an inclined plane without slipping,the forces acting along the plane are the component of gravity $Mg \sin 	heta$ and

Vedclass · Accepted Answer

$\frac{g \sin 	heta}{1 + I/MR^2}$

Vedclass · Answer

(A) For an object rolling down an inclined plane without slipping,the forces acting along the plane are the component of gravity $Mg \sin 	heta$ and the static friction $f$. The equation of motion is $Mg \sin 	heta - f = Ma$,where $a$ is the linear acceleration.For rotational motion about the center of mass,the torque equation is $	au = I \alpha = fR$,where $\alpha$ is the angular acceleration. Since it rolls without slipping,$a = R \alpha$,so $f = I \alpha / R = I(a/R)/R = Ia/R^2$.Substituting $f$ into the force equation: $Mg \sin 	heta - Ia/R^2 = Ma$.Rearranging for $a$: $Mg \sin 	heta = Ma + Ia/R^2 = Ma(1 + I/MR^2)$.Thus,$a = \frac{g \sin 	heta}{1 + I/MR^2}$.

$A$ uniform spherical object of mass $M$ and radius $R$ has a moment of inertia $I$. It rolls down an inclined plane of angle $\theta$ without slipping. What is its acceleration?

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