A hollow cone of radius R and height H = 2R i | vedclass

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(B) For a hollow cone,the center of mass is located at a height $h_{cm} = H/3$ from the base along the axis of symmetry.Given the height $H = 2R$,the

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$tan^{-1}(3/2)$

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(B) For a hollow cone,the center of mass is located at a height $h_{cm} = H/3$ from the base along the axis of symmetry.Given the height $H = 2R$,the center of mass is at $h_{cm} = (2R)/3 = 2R/3$ from the base.The cone will topple when the line of action of the gravitational force (passing through the center of mass) falls outside the base of the cone.At the point of toppling,the line of action of the weight passes through the edge of the base.In this configuration,the angle of inclination $	heta$ of the plane is equal to the angle between the vertical and the line connecting the center of mass to the edge of the base.Considering the right-angled triangle formed by the height of the center of mass $(2R/3)$ and the radius of the base $(R)$,we have $	an 	heta = \frac{	ext{radius}}{	ext{height of CM}} = \frac{R}{2R/3} = \frac{3}{2}$.Therefore,$	heta = 	an^{-1}(3/2)$.

$A$ hollow cone of radius $R$ and height $H = 2R$ is placed on an inclined plane of inclination $\theta$. If $\theta$ is increased gradually,at what value of $\theta$ will the cone topple? Assume sufficient friction is present to prevent slipping.

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