Centre of Mass, Momentum and Collision Concepts Page - 9

Example

Motion of rigid bodies in vertical circle involving collision

Example: A uniform rod OA$OA$ of length l$l$ resting on smooth surface is slightly distributed from its vertical position. P$P$ is a point on the rod whose locus is a circle during the subsequent motion of the rod. Then what is the distance OP$OP$ ?
Solution:Two forces normal reaction and weight are the only forces acting on the rod during motion. Both forces are vertical. So centre of mass will fall downwards in a vertical line.
XP=x(say)=l2cosθ−dcosθ${\displaystyle X_P=x(say)=\frac{l}{2}\cos \theta -d\cos \theta }$
or, cosθ=x(l2−d)....(i)${\displaystyle \cos \theta ={\displaystyle \frac{x}{({\displaystyle \frac{l}{2}}-d)}}....(i)}$
Similarly, yp=y=dsinθ$y_p=y=d\sin \theta$ ∴sinθ=yd.....(ii)${\displaystyle \therefore \sin \theta =\frac{y}{d}.....(ii)}$
Squaring and adding Eqs. (i) and (ii), we get
x2(l2−d)2+y2d2=1${\displaystyle {\displaystyle \frac{x^2}{{({\displaystyle \frac{l}{2}}-d)}^2}}+{\displaystyle \frac{y^2}{d^2}}=1}$
This is an equation of a circle for d=l4${\displaystyle d={\displaystyle \frac{l}{4}}}$ For any other value of d$d$ it is equation of ellipse.

Example

Equation of conservation of angular momentum for elastic collision

Example: A small disc and a thin uniform rod of length l$l$, whose mass is
η$\eta$ times greater than the mass of the disc, lie on a smooth horizontal plane. The disc is set in motion, in horizontal direction and perpendicular to the rod, with velocity v$v$, after which it elastically collides with the end of the rod. Find the velocity of the disc after the collision.

Solution:
Let mass of the disc be one unit.
By conservation of linear momentum,
v=v′+ηvo$v=v^{{}^{\prime }}+\eta v_o$
v′$v^{{}^{\prime }}$ is the velocity of disc after collision
vo=v−v′η$v_o={\displaystyle \frac{v-v^{{}^{\prime }}}{\eta }}$
Conservation of angular momentum,
vl2=v′l2+ηl212w$v{\displaystyle \frac{l}{2}}=v^{{}^{\prime }}{\displaystyle \frac{l}{2}}+{\displaystyle \frac{\eta l^2}{12}}w$
w$w$ is the angular velocity of rod.
w=6(v−v′)ηl$w={\displaystyle \frac{6(v-v^{{}^{\prime }})}{\eta l}}$
Conservation of energy,
12v2=12v′2+12ηv2o+12ηl212w2${\displaystyle \frac{1}{2}}{v}^2={\displaystyle \frac{1}{2}}{v}^{{}^{\prime }2}+{\displaystyle \frac{1}{2}}\eta v_o^2+{\displaystyle \frac{1}{2}}{\displaystyle \frac{\eta l^2}{12}}{w}^2$
Substituting vo$v_o$ and w$w$ in energy equation
(v2−v′2)=η((v−v′)2η2+3(v−v′)2η2)$(v^2-v^{{}^{\prime }2})=\eta ({\displaystyle \frac{(v-v^{{}^{\prime }}{)}^2}{{\eta }^2}}+3{\displaystyle \frac{(v-v^{{}^{\prime }}{)}^2}{{\eta }^2}})$
(v2−v′2)=4(v−v′)2η$(v^2-v^{{}^{\prime }2})={\displaystyle \frac{4(v-v^{{}^{\prime }}{)}^2}{\eta }}$Since v≠v′$v\ne v^{{}^{\prime }}$
v+v′=4(v−v′)η$v+v^{{}^{\prime }}={\displaystyle \frac{4(v-v^{{}^{\prime }})}{\eta }}$
v′=4−η4+ηv$v^{{}^{\prime }}={\displaystyle \frac{4-\eta }{4+\eta }}v$

Example

Energy equation in collision involving rigid bodies in case of elastic collision

Example: A solid smooth uniform sphere A of mass m$m$ rolls without sliding on a smooth horizontal surface. It collides elastically and head-on with another stationary smooth hollow sphere B of the same mass m and same radius. What is the ratio of kinetic energy of B to that of A just after the collision?

Solution:
Since the two bodies have same mass and collide head-on elastically, the linear momentum gets interchanged.
Hence just after the collision 'B' will move with velocity v′0$v_0^{\prime }$ and 'A'
becomes stationary but continues to rotate at the same initial angular velocity (v0R).$({\displaystyle \frac{v_0}{R}}).$
Hence, after collision. (K.E.)B=12mv20$(K.E.{)}_B={\displaystyle \frac{1}{2}}m{v}_0^2$
and
(K.E.)A=12Iω2=12(25mR2).(v0R)2${\displaystyle (K.E.{)}_A={\displaystyle \frac{1}{2}}I{\omega }^2={\displaystyle \frac{1}{2}}\left({\displaystyle \frac{2}{5}}m{R}^2\right).{\left({\displaystyle \frac{v_0}{R}}\right)}^2}$
⇒(K.E.)B(K.E.)A=52$\Rightarrow {\displaystyle \frac{(K.E.{)}_B}{(K.E.{)}_A}}={\displaystyle \frac{5}{2}}$
Note : Sphere 'B' will not rotate, because there is no torque on
'B' during the collision as the collision is head-on.

Formula

Impulse

When a force acts on an object for a short (infinitesimally small) duration of time, it's called impulse. Impulse is the measure of how much the force changes the momentum of an object.

F×△t=m△v=m(v−u)$F\times \mathrm{△}t=m\mathrm{△}v=m(v-u)$

Definition

Impulsive and Non-impulsive Force

The process of minimizing an impact force can be approached from the definition of the impulse of force:
If an impact stops a moving object, then the change in momentum is a fixed quantity, and extending the time of the collision will decrease the impact force by the same factor.

Non-impulsive Force:  A constant force acting on a body is an example of non-impulsive force.

Example

Find impulsive force using impulse momentum relation

Example: A particle of mass 2$2$ kg moving with a velocity of 3$3$ m/s is acted upon by a force which changes its direction of motion by an angle of 90$90$o${}^o$ without changing its speed. What is the magnitude of impulse experienced by the particle?

Solution:
Initial velocity of the particle    
v1→=3i^$\overrightarrow{v_1}=3\hat{i}$ m/s$m/s$
Final velocity of the particle   v2→=3j^$\overrightarrow{v_2}=3\hat{j}$   m/s$m/s$
Change in momentum  ΔP⃗ =m(v2→−v1→)=2×3(j^−i^)$\mathrm{\Delta }\overrightarrow{P}=m(\overrightarrow{v_2}-\overrightarrow{v_1})=2\times 3(\hat{j}-\hat{i})$                   
|ΔP⃗ |=2×32√=62√$|\mathrm{\Delta }\overrightarrow{P}|=2\times 3\sqrt{2}=6\sqrt{2}$ N s
Magnitude of impulse         I=|ΔP⃗ |=62√$I=|\mathrm{\Delta }\overrightarrow{P}|=6\sqrt{2}$   N-s

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