- #1

SilverSoldier

- 26

- 3

- Homework Statement
- A uniform rectangular rod of length ##L##, having mass ##M## is placed at the edge of a frictionless table so that some portion of the rod rests on the table, and the other portion extends out of the table, and its center of gravity ##G## is a distance ##x## away from the edge of the table.

A block of mass ##m## is initially placed at the far end of the part of rod resting on the table, and given a velocity ##v##. If the coefficient of kinetic friction between the rod and the object is ##mu##, what must be the minimum value of ##v##, so that the rod can be made to topple over the table?

- Relevant Equations
- Equations of motion

Kinetic friction = ##\mu_k\cdot N##

Momentum = ##mv##

Torque = ##F\times d##

Here is a diagram of the situation.

I first marked the forces acting on the system.

I understand that for this rod to topple over, there must be a net moment about point ##A##. But how exactly should I approach this problem?

If by any chance that were not the case, I tried solving it as follows.

First I calculate the distance to the center of gravity, say ##G'##, of the system from ##A##. We have

$$

GG'=\dfrac{m}{m+M}\cdot\dfrac{L}{2},

$$

and so we have

$$

\begin{align*}

AG'&=x+GG'\\

\\

&=\dfrac{2Mx+2mx+mL}{2(m+M)}.

\end{align*}

$$

Then, I observe that momentum in the system is conserved horizontally as there is no external horizontal force. Therefore,

$$

mv=(m+M)w\\

\text{ }\\

w=\dfrac{mv}{m+M},

$$

where ##w## is the velocity at which the center of gravity of the system will move after the block is given a velocity ##v##.

Next, I calculate the acceleration ##a## of the little block and get ##a=\dfrac{\mu mg}{m}=\mu g##.

If ##t## is the time that the block will take to come to rest, I have ##0=v-\mu gt##, so ##t=\dfrac{v}{\mu g}##.

If the distance that the center of gravity of the system moves in time ##t## is equal to ##AG'##, then the system will be at the verge of toppling over. Therefore,

$$

\begin{align*}

w\cdot t&=\dfrac{2Mx+2mx+mL}{2(m+M)}\\

\\

\dfrac{mv}{m+M}\cdot \dfrac{v}{\mu g}&=\dfrac{2Mx+2mx+mL}{2(m+M)}\\

\\

\dfrac{mv^2}{\mu g}&=\dfrac{2Mx+2mx+mL}{2}\\

\\

v^2&=\dfrac{\mu g}{2m}\cdot\left(2Mx+2mx+mL\right)\\

\\

v&=\sqrt{\mu g\left(x+\dfrac{L}{2}+\dfrac{Mx}{m}\right)}.

\end{align*}

$$

Is this correct?

Thank you very much for your assistance.

**Isn't it so, that irrespective of the velocity given to the block, at some point both the rod and block will acquire equal velocities as a result of their acceleration and deceleration, and then just continue moving forward at that velocity?**If by any chance that were not the case, I tried solving it as follows.

First I calculate the distance to the center of gravity, say ##G'##, of the system from ##A##. We have

$$

GG'=\dfrac{m}{m+M}\cdot\dfrac{L}{2},

$$

and so we have

$$

\begin{align*}

AG'&=x+GG'\\

\\

&=\dfrac{2Mx+2mx+mL}{2(m+M)}.

\end{align*}

$$

Then, I observe that momentum in the system is conserved horizontally as there is no external horizontal force. Therefore,

$$

mv=(m+M)w\\

\text{ }\\

w=\dfrac{mv}{m+M},

$$

where ##w## is the velocity at which the center of gravity of the system will move after the block is given a velocity ##v##.

Next, I calculate the acceleration ##a## of the little block and get ##a=\dfrac{\mu mg}{m}=\mu g##.

If ##t## is the time that the block will take to come to rest, I have ##0=v-\mu gt##, so ##t=\dfrac{v}{\mu g}##.

If the distance that the center of gravity of the system moves in time ##t## is equal to ##AG'##, then the system will be at the verge of toppling over. Therefore,

$$

\begin{align*}

w\cdot t&=\dfrac{2Mx+2mx+mL}{2(m+M)}\\

\\

\dfrac{mv}{m+M}\cdot \dfrac{v}{\mu g}&=\dfrac{2Mx+2mx+mL}{2(m+M)}\\

\\

\dfrac{mv^2}{\mu g}&=\dfrac{2Mx+2mx+mL}{2}\\

\\

v^2&=\dfrac{\mu g}{2m}\cdot\left(2Mx+2mx+mL\right)\\

\\

v&=\sqrt{\mu g\left(x+\dfrac{L}{2}+\dfrac{Mx}{m}\right)}.

\end{align*}

$$

Is this correct?

Thank you very much for your assistance.