Question 16: Toppling of Cylinder with Spheres
1. Geometry:
Let \( \alpha \) be the angle of the line joining the centers with the horizontal. The horizontal distance between the centers is \( 2(R-r) \). The distance between centers is \( 2r \). $$ \cos\alpha = \frac{2(R-r)}{2r} = \frac{R-r}{r} $$
Let \( \alpha \) be the angle of the line joining the centers with the horizontal. The horizontal distance between the centers is \( 2(R-r) \). The distance between centers is \( 2r \). $$ \cos\alpha = \frac{2(R-r)}{2r} = \frac{R-r}{r} $$
2. Normal Force (N):
Considering the equilibrium of the top sphere, the horizontal reaction from the wall \( N \) balances the horizontal component of the inter-sphere normal force. $$ N = mg \cot\alpha $$
Considering the equilibrium of the top sphere, the horizontal reaction from the wall \( N \) balances the horizontal component of the inter-sphere normal force. $$ N = mg \cot\alpha $$
3. Toppling Condition:
The couple generated by the wall reactions on the two spheres (\( \tau_{couple} \)) tries to topple the cylinder. The weight of the cylinder (\( Mg \)) provides the restoring torque about the pivot (bottom edge). $$ \tau_{couple} = N \times (2r \sin\alpha) $$ $$ \tau_{restoring} = Mg \times R $$ For critical stability: $$ (mg \cot\alpha)(2r \sin\alpha) = Mg R $$ $$ 2 mg r \cos\alpha = Mg R $$
The couple generated by the wall reactions on the two spheres (\( \tau_{couple} \)) tries to topple the cylinder. The weight of the cylinder (\( Mg \)) provides the restoring torque about the pivot (bottom edge). $$ \tau_{couple} = N \times (2r \sin\alpha) $$ $$ \tau_{restoring} = Mg \times R $$ For critical stability: $$ (mg \cot\alpha)(2r \sin\alpha) = Mg R $$ $$ 2 mg r \cos\alpha = Mg R $$
4. Final Result:
Substitute \( \cos\alpha = \frac{R-r}{r} \): $$ 2 m r \left( \frac{R-r}{r} \right) = M R $$ $$ \frac{m}{M} = \frac{R}{2(R-r)} $$
Substitute \( \cos\alpha = \frac{R-r}{r} \): $$ 2 m r \left( \frac{R-r}{r} \right) = M R $$ $$ \frac{m}{M} = \frac{R}{2(R-r)} $$