SOLUTION SET-62- (c) : According to question
Mg sinq – f = ma ...(i)
Now, Torque acting about centre of mass is due to
friction force
τ
==fR Iα
R
⇒=fMR a =
R(^2) Ma
5
2
5
2
2 ...(ii)
From eqn (i) and (ii)a Mg
Mg
==sin
θ sinθ
7
55
7
qqMgsinq
Mg Mgcosqa fIf slides without friction. Let the acceleration be ac∴′ag==sinθ a7
5
- (a) : e perpendicular distance of the weight
acting through P from OR =l
2cosT.
Again the perpendicular distance of the weight
acting through Q from O is,OQ = BQ – BO =l
l
2 cos
At equilibrium of two weightsmg
×=l mg×−l l
22 coscθθos ⇒=^3
21
2
cosθ⇒=cosθ^1
3
⇒=θ cos−^1 ^1
3- (b) : Conserving angular momentum
mv RmvRv
v.c 12 os ..;^2
1() 60 ° 42 ==
Conserving energy of the system−+GMm =− +
Rmv GMm
Rmv
41
2
1
(^12)
2
2
2
1
2
1
2
3
4
1
(^22)
2
1
2
1
vvGM^2
R
v GM
R
−= or =
v 1 1 61
2
64 10^8
2
=×= kms−AB q
R O
mgmgCQ- (b) : Let mass of the ball is ‘m’.
(^1) =−()
2
mv^2 mV V
AB
1
2
3
2
mv^2 m GM
RGM
R
=− −−
1
22
mv^2 GMm
RvGM
R
Velocity of ball just a er collision,ve′==vGM
R
1
5
Let r be the distance from the centre upto where
the ball reaches a er collision. en,
1
2mv^2 mVrV
′ =−[()]()centre1
503
2
3
(^322)
GMm^22
R
m
GM
R
GM
R
Rr
1
50
3
2
3
(^22)
2
2
r
R
⇒=⇒=
1
25 5
2
2r
Rr RThe desired distance,SRRRR
55
7
5
- (c) : F = mg sinq | mgtan q, (∵ q is small)
x^2 = 40y
24 x 0dy
dx=
dy
dx=^2 x
40Fmgdy
dxmgx
ax
==−×⇒=−2
40 2
⇒=ω^1
2(^) ()∵ax=−ω^2
q mgcosq
mgsinqmg
Focus NEET/JEE (XI) : Gravitation
Focus NEET/JEE (XII) : Electromagnetic Waves
and Optics
Monthly Tune Up (XI) : Gravitation
Monthly Tune Up (XII) : Electromagnetic Waves
and Optics
Brain Map
: Work, Energy and Power
Focus NEET/JEE (XI)
:^ Gravitation
Focus NEET/JEE (XII)
:Electromagnetic Waves
and Optics
Monthly Tune Up (XI)
: Gravitation
Monthly Tune Up (XII)
:Electromagnetic Waves
and Optics
Brain Map^
:Work, Energy and Power
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