Question

An ant rotates its mandible, of length 1.30 mm and mass 130 μg (which we can model as a uniform rod rotated about its end), at a high angular speed. As the tip strikes the ground, it undergoes an angular acceleration of 3.5×108rad/s2.

Answer 1

The force on the trip is
`1.97*10^(-2)\ N`.

Step-by-step explanation:

Given that,

Length = 1.30 mm

Mass = 130 μg

Angular acceleration
`\alpha=3.5*10^(8)\ rad/s^2`

If we assume that the tip of the mandible hits perpendicular to the ground, what is the force on the tip

We need to calculate the torque

Using formula of torque

`\tau=I*\alpha`

Put the value of torque and moment of inertia into the formula

`F* l=(ML^2)/(3)*\alpha`

`F=(ML)/(3)*\alpha`

Put the value into the formula

`F=(130*10^(-9)*1.30*10^(-3))/(3)*3.5*10^(8)`

`F=0.0197\ N`

`F=1.97*10^(-2)\ N`

Hence, The force on the trip is
`1.97*10^(-2)\ N`.

Answer 2

`1.97*10^(-2) N`

Step-by-step explanation:

The expression for torque

`\tau=I\alpha`

I= moment of inertia and α = angular acceleration

now,

`L* F= (ML^2)/(3) \alpha`

⇒
`F= (ML)/(3) \alpha`

plugging values we get

`F= (130*10^(-9)*1.3*10^(-3))/(3) \3.5*10^8`

=
`1.97*10^(-2) N`

Hence, the required force is
`1.97*10^(-2) N`