A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turntable (initially at rest) begins to rotate with its rate of rotation constantly increasing.

What is the first event that will occur?(Assume non-zero frictional force and the same coefficients of friction for both bugs.)

a. The ladybug begins to slide

b. The gentleman bug begins to slide

c. Both bugs begin to slide at the same time

d. Nothing ever happens

e. Not enough information to determine

The centripetal force acting on a rotating body or bugs can be written as,

[tex]F=mrw^2[/tex]

[tex]m=[/tex] mass of the corresponding bugs

[tex]r=[/tex] corresponding radial distance of each bug

[tex]w=[/tex] angular speed of the turntable

The centripetal force tries to slide the bugs in an outward direction and it is directly proportional to the products of its mass and radial distance from the axis of rotation of the turntable

[tex]F[/tex] ∝ [tex]mr[/tex]

Since the radial distance from the axis of rotation of the turntable for each bug is given, but the mass is not given, the given information is therefore not enough to determine which bugs will slide first.

Option "e" is correct.

oeerivona oeerivona · Answer 2 · 2020-03-03T08:46:55+01:00

Answer:

What is the first event that will occur? Assume non zero Nonzero frictional force and the same coefficient of friction for both bugs

a. The lady bug begins to slide

b. The gentleman bug begins to slide

c. Both bugs begin to slide at the same time

d. Nothing ever happens

e. Not enough information to determine

The correct answer to the question is

a. The lady bug begins to slide

Explanation:

To solve the question we have to write out the forces acting on the bugs

The frictional force is given by

f = μN = m×g×μ and the centripetal force is given by [tex]F_c=\frac{mv^2}{r}[/tex]

For equilibrium the two forces are equal and the bugs remain, that is

m×g×μ = [tex]\frac{mv^2}{r}[/tex] canceling like terms gives g×μ = [tex]\frac{v^2}{r}[/tex]= ω²r

Therefore as the acceleration increases the centripetal force will eventually exceed the frictional force

therefore for the ladybug, if the distance from the center = r then

g×μ = ω²r

while for the gentleman bug we have the radius = r/2, therefore

g×μ = ω²r/2 or the centripetal force is half of that of the ladybug hence the effect of the frictional force is grater on the gentleman bug than on the ladybug for the same angular velocity

which means the ladybug begins to slide first as it has two times the centripetal force of the gentleman bug to overcome the force of friction