Answer:
[tex]1.97\times10^{-2} N[/tex]
Explanation:
The expression for torque
[tex]\tau=I\alpha[/tex]
I= moment of inertia and α = angular acceleration
now,
[tex]L\times F= \frac{ML^2}{3} \alpha[/tex]
⇒[tex]F= \frac{ML}{3} \alpha[/tex]
plugging values we get
[tex]F= \frac{130\times10^{-9}\times1.3\times10^{-3}}{3} \3.5\times10^8[/tex]
= [tex]1.97\times10^{-2} N[/tex]
Hence, the required force is [tex]1.97\times10^{-2} N[/tex]
Answer:
The force on the trip is [tex]1.97\times10^{-2}\ N[/tex].
Explanation:
Given that,
Length = 1.30 mm
Mass = 130 μg
Angular acceleration [tex]\alpha=3.5\times10^{8}\ rad/s^2[/tex]
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
[tex]\tau=I\times\alpha[/tex]
Put the value of torque and moment of inertia into the formula
[tex]F\times l=\dfrac{ML^2}{3}\times\alpha[/tex]
[tex]F=\dfrac{ML}{3}\times\alpha[/tex]
Put the value into the formula
[tex]F=\dfrac{130\times10^{-9}\times1.30\times10^{-3}}{3}\times3.5\times10^{8}[/tex]
[tex]F=0.0197\ N[/tex]
[tex]F=1.97\times10^{-2}\ N[/tex]
Hence, The force on the trip is [tex]1.97\times10^{-2}\ N[/tex].
