Question

An ice cube of mass $M$ and with sides of length a is sliding without friction with speed $v_0$ when it hits a ridge $O$ at the edge of counter (see figure). This collision causes the cube to tilt as shown. The minimum value of $v_0$ needed for the cube to fall off the table is given by $\sqrt{kag}.$ Find value of $k.$ (Approximate the answer to nearest integer.)

Answer 1

Answer: 1

$M{v}_0\left(\frac{a}{2}\right)=I_0\omega$ (apply COAM)
$=\left(\frac{M{a}^2}{6}+M{\left(\frac{a}{\sqrt{2}}\right)}^2\right)\omega \Rightarrow \omega =\frac{3v_0}{4a}$
Now, from conservation of energy.
$\frac{1}{2}I{\left(\omega \right)}^2=mg(\Delta h)$
$\Rightarrow \frac{1}{2}\left(\frac{2}{3}{\left(Ma\right)}^2\right]{\left(\left(\frac{3v_0}{4a}\right]\right)}^2=Mg\left(\frac{a}{\sqrt{2}}-\frac{a}{2}\right)$
$\Rightarrow v_0=\sqrt{1.1ag}$