A solid wooden cube of side l and mass M is resting on a horizontal table, as shown in the figure. The cube is constrained to rotate about an axis through D and perpendicular to the face ABCD. A bullet of mass m moving with speed v strikes the block at a height $$2l/3$$ as shown. Let the line along which the bullet moves be in the plane passing through the centre of mass of the block and parallel to the face ABCD. If the minimum value of v that topples the block is Mmxgℓ$$2$$−$$121/2$$ then find x. Assume bullet comes to rest after collision.

Question

A solid wooden cube of side l and mass M is resting on a horizontal table, as shown in the figure. The cube is constrained to rotate about an axis through D and perpendicular to the face ABCD. A bullet of mass m moving with speed v strikes the block at a height $$2l/3$$ as shown. Let the line along which the bullet moves be in the plane passing through the centre of mass of the block and parallel to the face ABCD. If the minimum value of v that topples the block is  Mmxgℓ$$2$$−$$121/2$$ then find x. Assume bullet comes to rest after collision.

Infinity Learn · Accepted Answer

Let ω be the angular velocity of the cube (just$$ after the bullet $$strikes) about an axis passing through D. Conservation of angular momentum about this axis gives $$mv2$$ℓ$$3=I$$ω  where I is the moment of inertia about the axis through D. I $$=$$  Mℓ$$26+M$$ℓ$$22=2M$$ℓ$$23$$  ⇒ω$$=mvMl$$  The cube will topple if the center of mass is just able to rise from  $$l2$$ to $$l2$$. In such a case, the rotational energy must be equated to the change of potential energy. Thus, $$12I$$ω$$2=Mg$$ℓ$$2$$−ℓ$$2$$  Using the values of I and ω, we get the expression for v that will just topple the cube:  $$v=Mm3gl2$$−$$1212$$.

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