Rotation 5 Question 25
37. Two heavy metallic plates are joined together at $90^{\circ}$ to each other. A laminar sheet of mass $30 kg$ is hinged at the line $A B$ joining the two heavy metallic plates. The hinges are frictionless. The moment of inertia of the
laminar sheet about an axis parallel to $A B$ and passing through its centre of mass is $1.2 kg-m^{2}$. Two rubber obstacles $P$ and $Q$ are fixed, one on each metallic plate at a distance $0.5 m$ from the line $A B$. This distance is chosen, so that the reaction due to the hinges on the laminar sheet is zero during the impact. Initially the laminar sheet hits one of the obstacles with an angular velocity $1 rad / s$ and turns back. If the impulse on the sheet due to each obstacle is $6 N$-s.
(2001, 10M)
(a) Find the location of the centre of mass of the laminar sheet from $A B$.
(b) At what angular velocity does the laminar sheet come back after the first impact?
(c) After how many impacts, does the laminar sheet come to rest?
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Answer:
Correct Answer: 37. (a) $0.1 m$
(b) $1 rad / s$
(c) sheet will never come to rest
Solution:
- Let $r$ be the perpendicular distance of $CM$ from the line $A B$ and $\omega$ the angular velocity of the sheet just after colliding with rubber obstacle for the first time.
Obviously the linear velocity of $CM$ before and after collision will be $v _i=(r)(1 rad / s)=r$ and $v _f=r \omega$.
$\mathbf{v} _i$ and $\mathbf{v} _f$ will be in opposite directions.
Now, linear impulse on $CM$
$=$ change in linear momentum of $CM$
or
$$ \begin{gathered} 6=m\left(v _f+v _i\right)=30(r+r \omega) \\ r(1+\omega)=\frac{1}{5} \end{gathered} $$
$$ \text { or } $$
Similarly, angular impulse about $A B=$ change in angular momentum about $A B$
Angular impulse $=$ Linear impulse
$\times$ perpendicular distance of impulse from $A B$
Hence, $\quad 6(0.5 m)=I _{A B}(\omega+1)$
(Initial angular velocity $=1 rad / s$ )
$$ \begin{array}{ll} \text { or } & 3=\leftI _{CM}+M r^{2}\right \\ \text { or } & 3=\left1.2+30 r^{2}\right \end{array} $$
Solving Eqs. (i) and (ii) for $r$, we get
$$ r=0.4 m \text { and } r=0.1 m $$
But at $r=0.4 m, \omega$ comes out to be negative $(-0.5 rad / s)$ which is not acceptable. Therefore,
(a) $r=$ distance of $CM$ from $A B=0.1 m$
(b) Substituting $r=0.1 m$ in Eq. (i), we get $\omega=1 rad / s i e$, the angular velocity with which sheet comes back after the first impact is $1 rad / s$.
(c) Since, the sheet returns with same angular velocity of $1 rad / s$, the sheet will never come to rest.
