Question
A cubical block of side a is moving with velocity $$V$$ on a horizontal smooth plane as shown in Figure. It hits a ridge at point $$O.$$ The angular speed of the block after it hits $$O$$ is
A cubical block of side a is moving with velocity $$V$$ on a horizontal smooth plane as shown in Figure. It hits a ridge at point $$O.$$ The angular speed of the block after it hits $$O$$ is
A.
$$\frac{{3V}}{{\left( {4a} \right)}}$$
B.
$$\frac{{3V}}{{\left( {2a} \right)}}$$
C.
$$\frac{{\sqrt {3V} }}{{\left( {\sqrt {2a} } \right)}}$$
D.
$$Zero$$
Answer :
$$\frac{{3V}}{{\left( {4a} \right)}}$$
Solution :
$$r = \sqrt 2 \frac{a}{2}\,\,\,\,\,\,or,\,{r^2} = \frac{{{a^2}}}{2}$$
Net torque about $$O$$ is zero.
Therefore, angular momentum $$\left( L \right)$$ about $$O$$ will be conserved, or $${L_i} = {L_f}$$
$$\eqalign{ & MV\left( {\frac{a}{2}} \right) = {I_0}\,\omega = \left( {{I_{cm}} + M{r^2}} \right)\omega \cr & \omega = \left\{ {\frac{{M{a^2}}}{6} + M\left( {\frac{{{a^2}}}{2}} \right)} \right\}\omega = \frac{2}{3}M{a^2}\omega = \frac{{3v}}{{4a}} \cr} $$
$$r = \sqrt 2 \frac{a}{2}\,\,\,\,\,\,or,\,{r^2} = \frac{{{a^2}}}{2}$$
Net torque about $$O$$ is zero.
Therefore, angular momentum $$\left( L \right)$$ about $$O$$ will be conserved, or $${L_i} = {L_f}$$
$$\eqalign{ & MV\left( {\frac{a}{2}} \right) = {I_0}\,\omega = \left( {{I_{cm}} + M{r^2}} \right)\omega \cr & \omega = \left\{ {\frac{{M{a^2}}}{6} + M\left( {\frac{{{a^2}}}{2}} \right)} \right\}\omega = \frac{2}{3}M{a^2}\omega = \frac{{3v}}{{4a}} \cr} $$
