Torque and inertia of a wooden rod

[ac7597]

Homework Statement

A wooden rod sits on the ice of a hockey rink. The rod has a length L=1 meters, but it is made of a peculiar sort of wood. The density changes from the left end to the right, in the following way.

If we measure position x from the left end of the rod, the linear mass density is:
λ(x)=0.2 kg/m + 0.061(x/L)^2 kg/m

What is the mass of the rod?

Joe sticks a nail through the left-hand end of the rod, as shown, and down into the ice. The rod can now rotate around the nail.
What is the moment of inertia of the rod around this nail?

Joe kicks the rod at its middle, applying a force F=53 N perpendicular to the rod as shown in the figure.
What is the magnitude of the torque around the nail due to this force?

The rod now starts to rotate around the nail. What is the magnitude of angular acceleration of the rod?

Relevant Equations

I=mass(radius)^2

∫ λ(x)=0.2 kg/m + 0.061(x/L)^2 kg/m = 0.2(x) + (0.061/3) (x^3) /(1/L^2)
mass of rod = 0.2+ (0.061/3) =0.22 kg

inertia of rod through nail = (1/3) (mass) (L)^2
inertia of rod through nail = (1/3) (0.22kg) (1m)^2 = 0.073 kg*m^2

torque magnitude = (53N) (0.5m) = 26.5N*m

angular acceleration of the rod= (26.5N*m) / (0.073 kg*m^2)=363 rad/s^2

 

[ac7597]

inertia of rod through the nail is not 0.073 kg*m^2. I don't know why

 

[TSny]

inertia of rod through nail = (1/3) (mass) (L)^2

This formula assumes that the mass is uniformly distributed along the rod.

 

[ac7597]

dI= (0.2+(0.061/L^2)*x^2 ) dx *x^2
∫dI= I = (0.2x^3)/3 + (0.061x^5)/ (5L^2) |x=1, x=0
I=78.8E-3 kg*m^2

 

[ac7597]

angular acceleration = 26.5N*m / (78.8E-3 kg*m^2) = 336 rad/s^2