Auniform cylinder rests on a cart as shown. The coefficient of static friction between the cylinder and the…
Auniform cylinder rests on a cart as shown. The coefficient of static friction between the cylinder and the cart is \(0.5\). If the cylinder is \(4 \mathrm{~cm}\) in diameter and \(10 \mathrm{~cm}\) in height, then what is the minimum acceleration (in \(\mathrm{ms}^{-2}\)) of the cart 'needed to cause the cylinder to tip over?
Solution
Maximum acceleration of cart so that cylinder does not slip: \(\mathrm{a}_{\mathrm{m}}=\mu \mathrm{g}=\) \(0.5 \times 10=5 \mathrm{~ms}^{-2}\)
For tipping over: Let acceleration of cart is a. Considering torque about A:
\(\begin{array}{l}
\mathrm{Ma} \times 5 \geq \mathrm{mg} \times 2 \\
\Rightarrow \mathrm{a} \geq 2 \mathrm{~g} / 5=4 \mathrm{~ms}^{-2}
\end{array}\)
Asked in: JEE Mains - Rotational Motion - Chapter Test
