The correct option is
D Friction force between cylinder and ground is
157 NLet the acceleration of the centre of mass
(acom) of the cylinder be
a.
Since the tangential acceleration
(at=rα) at the top-most point and acceleration of the centre of mass of the cylinder
(acom) are in the same direction,
Acceleration of topmost point is,
a′=acom+at ...(i)
From condition of no slipping on ground,
acom=rα=a
From Eq.
(i),
a′=rα+rα=2rα
⇒a′=2a ...(ii)
Since, slipping is not taking place at any contact, therefore acceleration of plank
(aP) will be equal to the acceleration of the point of contact with the cylinder i.e
aP=2a.
Friction
(f) will act in forward direction on cylinder and in backward direction on the plank to prevent slipping (at top-most position).
For plank, on applying Newton's second law along its direction of acceleration,
15−f=1×2a=2a ...(1)
For cylinder, along its direction of acceleration of
COM
f+f′=2a ...(2)
[For cylinder, the point of contact with the ground will tend to slip backwards on application of
15 N, hence friction
f′ will act along forward direction.]
On taking instantaneous axis of rotation (
I.A.R) at the point of contact of cylinder and ground (because point of contact for cylinder with ground is always at rest)
Applying equation of torque for cylinder about
I.A.R,
∑τ=Iα
∵ Normal reaction from ground,
f′, mg will pass through
I.A.R, their torque will vanish.
⇒τf=Iα
⇒2rf=3mr22.ar
[
I=I0+md2,I0=mr22, d=r and
a=αr for pure rolling]
⇒f=3ma4
⇒f=3×2×a4=3a2 ...(3)
On putting Eq.
(3) in Eq.
(1),
15−3a2=2a
⇒a=307 m/s2 is the acceleration of COM of cylinder.
Acceleration of plank is,
aP=2a=607 m/s2
On putting the value of
a in Eq.
(3), we get friction between plank and cylinder
⇒f=457 N
On putting
f in Eq.
(2), we get friction between cylinder and ground
⇒f′=157 N
Hence options
(a), (b), (c), (d) are correct.