- (b) : The CM of the rod has fallen
by
l
2
(1 – cos q).
? 'U = – Mg l
2
(c 1 os )
;
'K =^1
2
IZ^2 =
1
23
2
Ml 2
.
'U + 'K = 0 or 'K = –'U
1
23
Ml^22
. = Mg
l
2
.(c 1 os )
Z^2 =
31 g
l
(cos )
Z^2 =
6
2
g^2
l
sin
Z =^6
2
g
l
.sin
- (c) : Torque due to F must
be more than torque due
to mg about P.
Fmin(r – h) = mgx
Fmin(r – h)
= mg rr^22 ()h
Fmin =
mg rrhrh
rh
(^22) (^2) 2
()
()
=
mg hrh
rh
()
()
2
.
- (c) : For string to be wound around the nail,
vgt 5 r
vg 51 () 0 L
Now
1
2
mv^2 = mg(10) v^2 = 2g (10)
(^21) gg. 05 () 10 L
20 g > 5g(10 – L) 20 > 50 – 5L 5 L > 30
L > 6
Lmin = 6 cm.
- (a) : Mass of square plate M 1 = l^2 V
Mass of circular disc M 2 = πσ
l^2
4
l/2 l/
l/2 l/2 cos q
q
F
r
P
x h
mg
x
M ll
l
CM=
×+
+
1
2
2
0
4
1
4
πσ
σ
π
xCM= ll
+
=
+
π
π
π
4 4 π
4
() 4
xCM<l
2
, i.e., COM lies in the square
- (a) : Acceleration of system
a
mg mg
m
=
sins 60 °− in 30 °
2
Here m = mass of each block
or ag=
−
31
4
Now
a
ma ma
CM m
=
12 +
2
Here
a
1 and
a 2 are^31
4
−
g at the right angles.
Hence, aaCM== g
−
2
2
31
42
- (a) : If v is the velocity by which balls strike the
bases, then their velocities aer collision are 0.6 v
and 0.4 v as shown in the gure ‘u’ is the velocity
of COM just aer collision and Z is the angular
velocity acquired by the rod.
A B
0.6 v u 0.4 v
Z
For end A : 0.6v = u
+l
2
.ω ...(i)
For end B : 0.4v = u
l
−
2
.ω ...(ii)
Subtracting (i) and (ii) we get
0.2v = Zl ω=
v
5 l
Now using [v^2 = u^2 + 2as]
v^2 = 0 + 2 × 10 × 0.2 = 4 v = 2 m s–
ω=
×
= −
2
51
2
5
rad s^1
- (d) : We can observe each
and every element of rod in
rotating with dierent radius
about the axis of rotation.
Take an elementary mass dm
of the rod.
l
r
dl
q
dm