A cubical block of side moving with a velocity v on a horizontal smooth plane as shown. It hits a ridge at point O and sticks to it (collision is perfectly inelastic). The angular speed of the block after it hits O is

Question

A cubical block of side moving with a velocity v on a horizontal smooth plane as shown. It hits a ridge at point O and sticks to it (collision is perfectly inelastic). The angular speed of the block after it hits O is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-su0l9dp9tyxw8vgo.jpg" /> (A) (3 v/4 a) (B) (3 v/2 a) (C) (√3 textitv/√2 textita) (D) Zero

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-su0l9dp9tyxwkb9x.jpg" /> Net torque about O is zero. Therefore, angular momentum (L) about point O will be conserved or, textitL texti= textitL textf operatornameMv((a/2))=I0 ω=(I text com + Mr 2) ω = (( Ma 2/6))+M((a2/2)) ω=(2/3) Ma 2 ω ∴ ω=3 v / 4 a

Q. A cubical block of side moving with a velocity $v$ on a horizontal smooth plane as shown. It hits a ridge at point $O$ and sticks to it (collision is perfectly inelastic). The angular speed of the block after it hits $O$ is

$\frac{3 v}{4 a}$

$\frac{3 v}{2 a}$

$\frac{3 v}{2 a}$

Zero

Net torque about $O$ is zero.
Therefore, angular momentum $(L)$ about point $O$ will be conserved
or, $L_{i} = L_{f}$
$Mv (\frac{a}{2}) = I_{0} ω = (I_{com} + M r^{2}) ω$
$= {(\frac{M a ^{2}}{6}) + M (\frac{a ^{2}}{2})} ω = \frac{2}{3} M a^{2} ω$
$∴ ω = 3 v /4 a$

Q. A cubical block of side moving with a velocity v on a horizontal smooth plane as shown. It hits a ridge at point O and sticks to it (collision is perfectly inelastic). The angular speed of the block after it hits O is

4a3v​

2a3v​

2​a3​v​

Zero

Net torque about O is zero. Therefore, angular momentum (L) about point O will be conserved or, Li​=Lf​ Mv(2a​)=I0​ω=(Icom ​+Mr2)ω ={(6Ma2​)+M(2a2​)}ω=32​Ma2ω ∴ω=3v/4a

Q. A cubical block of side moving with a velocity $v$ on a horizontal smooth plane as shown. It hits a ridge at point $O$ and sticks to it (collision is perfectly inelastic). The angular speed of the block after it hits $O$ is

$\frac{3 v}{4 a}$

$\frac{3 v}{2 a}$

$\frac{3 v}{2 a}$

Net torque about $O$ is zero.
Therefore, angular momentum $(L)$ about point $O$ will be conserved
or, $L_{i} = L_{f}$
$Mv (\frac{a}{2}) = I_{0} ω = (I_{com} + M r^{2}) ω$
$= {(\frac{M a ^{2}}{6}) + M (\frac{a ^{2}}{2})} ω = \frac{2}{3} M a^{2} ω$
$∴ ω = 3 v /4 a$