- (b) : The CM of the rod has fallen
byl
2(1 – cos q).? 'U = – Mg l
2(c 1 os );
'K =^1
2IZ^2 =
1
23
2
Ml 2
.
'U + 'K = 0 or 'K = –'U
1
23
Ml^22
. = Mgl
2 .(c 1 os )
Z^2 =
31 g
l(cos )Z^2 =
6
2
g^2lsin 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 ()hFmin =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 LNow1
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 = πσl^2
4l/2 l/l/2 l/2 cos q
qF
r
P
x hmgxM lllCM=
×+
+
1220
4
1
4
πσσπ xCM= ll
+=
+
π
ππ4 4 π
4() 4
xCM<l
2, i.e., COM lies in the square- (a) : Acceleration of system
amg mg
m=
sins 60 °− in 30 °
2
Here m = mass of each blockor ag= −
31
4
Now
ama 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 B0.6 v u 0.4 v
ZFor end A : 0.6v = u
+l
2.ω ...(i)For end B : 0.4v = ul
−
2.ω ...(ii)
Subtracting (i) and (ii) we get0.2v = 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.
lrdlqdm