Question

W, the turntable stops rotating, and the ladybug and the gentleman bug move to opposite edges of the outer rim. the gentleman bug has not watched his weight and is much more massive. the turntable then begins rotating ever more rapidly. who will fall off first? (assume non-zero frictional force and the same coefficients of friction for both bugs.)

Answer 1

The gentleman bug will fall off the turntable last due to its greater mass and resulting greater frictional force.

Step-by-step explanation:

When the turntable stops rotating and the ladybug and the gentleman bug move to opposite edges of the outer rim, the turntable then begins rotating more rapidly. In this scenario, the gentleman bug is more massive than the ladybug. Since both bugs experience the same coefficient of friction and the same non-zero frictional force, the bug with the greater mass will be more resistant to being pushed off the turntable.

The frictional force can be calculated using the formula: force of friction = coefficient of friction x normal force. Since the normal force is the weight of the bug, the gentleman bug will have a greater normal force due to its greater mass. Therefore, the frictional force acting on the gentleman bug will be greater, allowing it to resist being pushed off the turntable for a longer time compared to the ladybug.

In conclusion, the gentleman bug will fall off the turntable last because its greater mass and the resulting greater normal force will result in a stronger frictional force, allowing it to resist being pushed off the turntable for a longer period of time.

Answer 2

The ladybug will fall off first because she is lighter than the gentleman bug.

This is because the inertia (tendency to remain at rest) is affected by weight. The more a body has weight the more its inertia force is high. The opposite is also true.