Angle made by the plane of hemisphere with inclined plane

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A uniform thin hemispherical shell is kept static on an inclined plane of angle
$\theta = 30$ as shown. If the surface of the inclined plane is sufficiently rough to prevent
sliding then what is the angle $\alpha$ made by the plane of hemisphere with inclined plane .

asked Nov 4, 2018
retagged Nov 5, 2018
There is no acceleration. The hemispherical shell is *kept static*.

Take moments about the point of contact P. Friction $F$ and normal reaction $N$ act through P so weight $W$ must also act through P. ie COM must lie vertically above P.

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Centroid C is at the midpoint of the axis OA of the hemisphere, ie distance OC=R/2. In equilibrium position C must lie vertically above point of contact P.

Apply Sine Rule in triangle OPC : $$\frac{R/2}{\sin\theta}=\frac{R}{\sin\beta}$$ $$\sin\beta=2\sin\theta=2\sin30^{\circ}=1$$ $$\beta=180^{\circ}-(\theta+\alpha)=90^{\circ}$$ $$\alpha=90^{\circ}-\theta=60^{\circ}$$

answered Nov 5, 2018 by (28,448 points)
selected Mar 12 by koolman
Is torque zero because all force vectors pass through P?
Yes, that is correct.