A uniform thin cylindrical disk of mass M and radius R is attached to two identical massless springs of spring constant k which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance d from its centre. The axle is massless and both the springs and the axle are in a horizontal plane. The unstretched length of each spring is L. The disk is initially at its equilibrium position with its centre of mass (CM) at a distance L from the wall. The disk rolls without slipping with velocity 
. The coefficient of friction is µ.
(i) The net external force acting on the disk when its centre of mass is at displacement x with respect to its equilibrium position is
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i) Applying equation of torque about lowest point
(2Kx) R = 
αR = 
as there is no slipping
a = α R = 
Net force = Ma = 
Which is directed opposite to displacement
F net = 
(ii) F net = 
= –M(ꞷ 2 x)
ꞷ = 
(iii) 
MV 0 2 + 
= 
(2K)x 2 max
= 
MV 0 2 = 2kx 2 max
⇒ x max = 
At extreme position, friction will have maximum value.
2kx max – f max = 
⇒ f max = 
kx max
µMg = 
⇒ µMg = 
V 0 = 
(iv) From graph
T = 8 second., A = 1 cm, x = A sinꞷt. = 1sin 
t.
a = – ꞷ 2 x = – 
sin 
t cm/s 2
At, t = 
second
a = 
– 
sin = – 
cm/s
(v)
Torque about P = (kx) 
+ (kx) 
= kxL (θ= 
)
τ = Iα
⇒ – 
⇒ 
= α
⇒ α = – 
θ = -ꞷ 2 θ
⇒ ꞷ= 
and f = 
= 
(vi)
Extensions in springs are x 1 and x 2 then
k 1 x 1 = k 2 x 2
and x 1 + x 2 = A
⇒ x 1 + 
= A
⇒ x 1 = 
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