A rod collides elastically with smooth horizontal surface after falling from a height. For maximum angular
speed of the rod just after impact, the rod should be released in such a way that it makes an angle ‘α’ with horizontal, the value of ‘α’ will be
Answers
Answer:
ANSWER
Given: A rod AB of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface hits the end A of the rod with a velocity v
0
in a direction perpendicular to AB. The collision is elastic. After the collision the particle comes to rest
Solution:
Let C be centre of mass of the rod of mass M and length L. Consider the rod and the particle together as a system. Let v be velocity of C and ω be angular velocity of the rod just after collision. The linear momentum of the system just before and just after the collision is,
p
i
=mv
0
,p
f
=Mv
where p
i
,
f
are the initial and final linear momentum of the system.
There is no external force on the system in x direction. Hence, linear momentum in x-direction is conserved i.e., p
i
=p
f
, which gives,
Mv=mv
0
⟹v
0
=
m
Mv
........(i).
The angular momentum of the system about C just before and just after the collision is,
L
i
=mv
0
2
L
,L
f
=ωI
c
=Mω
12
L
2
.
There is no external torque on the system about C. Hence, angular momentum of the system about C is conserved i.e., L
i
=L
f
, which gives,
mv
0
=Mω
6
L
........(ii).
The kinetic energy before and after the collision is,
K
i
=
2
1
mv
0
2
,K
f
=
2
1
Mv
2
+
2
1
I
c
ω
2
.
Since kinetic energy is conserved in elastic collision, K
i
=K
f
, i.e.,
mv
0
2
=Mv
2
+(
12
ML
2
)ω
2
.......(iii).
Solve above equations we get
M
m
=
4
1
,v=
4
v
0
,ω=
2L
3v
0
.
The velocity of a point P with position vector
r
PC
from C is given by
v
P
=
v
C
+
ω
×
r
PC
. Just after the collision
r
PC
=y
j
^
. Thus, P is at rest if,
v
P
=
4
v
0
i
^
+(
2L
3v
0
k
^
)×(y
j
^
)=
4
v
0
−
2L
3yv
0
=0,
which gives y=
6
L
.
The distance AP=AC+CP=
2
L
+
6
L
=
3
2L
.
After the collision, C keeps moving with
v
C
=
4
v
0
i
^
and angular velocity of the rod remains
ω
=
2L
3v
0
k
^
.
The angular displacement of the rod in time t=
3v
0
πL
is ωt=
2
π
and hence after time t position vector of P w.r.t. C is
r
PC
=−
6
L
i
^
+0
j
^
.
The velocity of P is,
v
P
=
v
C
+
ω
×
r
PC
=
4
v
0
i
^
+(
2L
3v
0
k
^
)×(−
6
L
i
^
)=
4
v
0
i
^
−
4
v
0
j
^
,
and its magnitude is ∣
v
P
∣=
2
2
v
0