Angular acceleration of plank hanging off edge with a weight

[pbnj]

Homework Statement

A heavy, 6m long uniform plank has a mass of 30 kg. It is positioned so that 4m is supported on the deck of a ship and 2m sticks out over the water. It is held in place only by its own weight. You have a mass of 70kg and walk the plank past the edge of the ship.
a) How far past the edge do you get before the plank starts to tip, in m?
b) If you go 10cm past the point determined above, what is the angular acceleration of the board in rad/s^2 ?

Relevant Equations

$$\tau_{net} = I\alpha$$
$$I = \sum_j m_j r_j^2$$
$$\tau = rF\sin\theta$$

Supposing L = 6m, m = 30kg, M = 70kg, h = 2m. Also set the origin at the leftmost edge of the plank.

Free body diagram description:
The center of mass of the plank is at $\frac{L}{2}$, so considering it as a point mass, it feels the force of gravity $F_m = mg$.
The person is at some point $x$, and feels the force of gravity $F_M = Mg$.
The pivot point is where the plank meets the edge of the ship, this is at $L - h$.
The lever arm of the plank is the distance from the pivot point to the plank's center of mass, $r_m = |L - h - \frac{L}{2}| = |\frac{L}{2} - h|$.
The lever arm of the person is $r_M = x$.
As the person walks towards the rightmost point possible while not falling down, the normal force of the ship on the plank becomes localized at the pivot point.

For a), since the normal force is at the pivot point, its torque is 0.
The torque of the plank is $\tau_m = r_m F_m$, positive since it would induce a counterclockwise rotation about the pivot.
The torque of the person is $\tau_M = -r_M F_M$, negative since it would induce a clockwise rotation.
Hence $\tau_m + \tau_M = 0$ implies $r_M = \frac{\tau_m}{F_M} \approx 0.43m$.
This is apparently correct.

For b), I define a new lever arm and torque to take into account the extra 10cm; $r_M' = r_M + 0.1m$ and $\tau_M' = -r_M'F_M$.
Using the formula for moment of inertia, I calculate $I = mr_m^2 + M(r_M')^2$, and the net torque is $\tau_{net} = \tau_m + \tau_M'$.
So the angular acceleration should be $\alpha = \frac{\tau_{net}}{I} \approx -1.38$, in units of radians/s^2. However, this is apparently wrong.

 

[haruspex]

What is the moment of inertia of a uniform plank about its centre?
(I would choose the pivot point as origin.)

 

[pbnj]

What is the moment of inertia of a uniform plank about its centre?
(I would choose the pivot point as origin.)

I figure since we're only given the length and mass of the plank, it should be $\frac{1}{12}mL^2$. Then I should use the parallel-axis theorem to get the moment of inertia of the plank about the pivot, which is $I_m = \frac{1}{12}mL^2 + mr_m^2$. So moment of inertia of the plank and human should be $I = I_m + I_M$ where $I_M = M(r_M')^2$, then using $\alpha = \frac{\tau_{net}}{I}$ I get a final answer of -0.49 rad/s^2, which is apparently still wrong.

 

[haruspex]

I figure since we're only given the length and mass of the plank, it should be $\frac{1}{12}mL^2$. Then I should use the parallel-axis theorem to get the moment of inertia of the plank about the pivot, which is $I_m = \frac{1}{12}mL^2 + mr_m^2$. So moment of inertia of the plank and human should be $I = I_m + I_M$ where $I_M = M(r_M')^2$, then using $\alpha = \frac{\tau_{net}}{I}$ I get a final answer of -0.49 rad/s^2, which is apparently still wrong.

Your method looks ok now, but I do get different second significant digit. Please post your detailed working.

 

[pbnj]

Your method looks ok now, but I do get different second significant digit. Please post your detailed working.

I've been using lean4 as my calculator, here's my source code:

def g := 9.81
def m := 30.0
def M := 70.0
def L := 6.0
def h := 2.0

def rm := (0.5*L - h).abs
def Fm := m*g
def FM := M*g
def τm := rm*Fm
def rM := τm/FM
def τM := -rM*FM

#eval rM

def rM' := rM + 0.1
def τM' := -rM'*FM

def τnet := τM' + τm

def Im := (1/12)*m*L^2 + m*rm^2
def IM := M*(rM')^2
def I := Im + IM

def α := τnet / I

#eval α // -0.492057

and it can be pasted here, then place the cursor on the last line to see the answer I got. I double-checked that the numeric literals are getting inferred as Floats, so there's no weird integer division going on.

 

[haruspex]

I've been using lean4 as my calculator, here's my source code:

def g := 9.81
def m := 30.0
def M := 70.0
def L := 6.0
def h := 2.0

def rm := (0.5*L - h).abs
def Fm := m*g
def FM := M*g
def τm := rm*Fm
def rM := τm/FM
def τM := -rM*FM

#eval rM

def rM' := rM + 0.1
def τM' := -rM'*FM

def τnet := τM' + τm

def Im := (1/12)*m*L^2 + m*rm^2
def IM := M*(rM')^2
def I := Im + IM

def α := τnet / I

#eval α // -0.492057

and it can be pasted here, then place the cursor on the last line to see the answer I got. I double-checked that the numeric literals are getting inferred as Floats, so there's no weird integer division going on.

Apologies, it was my error. I now get the same as you.
A possibility is that you should answer -.5 on the basis that some given data only have one significant figure, but that seems unlikely.

 

[erobz]

Why would we include the mass moment of inertia of the human? The human doesn’t rotate with the plank. They are an external force on the plank, and it seems likely (to me) they just fall straight down when the plank flips?

 

[haruspex]

Why would we include the mass moment of inertia of the human? The human doesn’t rotate with the plank. They are an external force on the plank, and it seems likely (to me) they just fall straight down when the plank flips?

The person's weight causes the plank to rotate, therefore, at least to begin with, they remain in contact. Whether the person is merely standing on the plank or is glued to it cannot make a difference at that stage.
If you prefer, you can treat them separately, but you have to consider that since the person has a downward acceleration of $\alpha x$ the force exerted on the plank is $M(g-x\alpha)$. As a torque, that is $Mgx-Mx^2\alpha$. This produces the same equation as before.