*April*

Came across a question recently that many students answered incorrectly.

Close to the surface of the Earth the gravitational field strength is uniform. A pair of unequal masses are joined by a light, rigid horizontal bar and suspended by a string from their centre of gravity as shown. The mass *M* of the ball on the left is larger than the mass *m* of the ball on the right.

The supporting string is now cut and the system begins to fall. Air resistance is negligible.

Which statement is correct?

A | The bar will remain horizontal as it falls. |

B | The bar will rotate clockwise as it falls. |

C | The bar will rotate anti-clockwise as it falls. |

D | The bar will first rotate clockwise and then rotate anticlockwise as it falls. |

**Without air resistance**

This question supposes that air resistance is negligible and so the only forces initially acting on the object is weight. The answer that many students gave incorrectly as B because they assume that the larger weight acting on the larger mass will bring about a larger acceleration.

Since the object begins in equilibrium, and the acceleration of both objects is just gravitational acceleration, the bar will remain horizontal.

**With air resistance**

This then invites a question: What if there is air resistance?

To consider the vertical acceleration on both balls, we need to consider the net force $F_{net}$, which is the vector sum of weight $W$ and air resistance $F_R$, ignoring the tension exerted by the rod at the initial stage of the fall.

$$F_{net} = W – F_R = V \rho_{ball}g – \dfrac{1}{2} \rho_{air}v^2C_DA$$

The volume V of a sphere is proportional to $r^3$ and its cross-sectional area A is proportional to $r^2$,

A larger radius will imply a larger increase in V than A, and hence, a large $W$ than $F_R$. This will then allow the larger mass to experience a larger acceleration than the smaller mass in the initial stage.