Given, mass, radius of rigid body be
m and
R respectively,
g is acceleration due to gravity and
θ is inclination.
Let
I be the moment of inertia, then
I=nmR2 Taking force along inclined
mgsinθ−f=ma. . . (i)
where,
f is friction force and
a is acceleration.
τ=Iα=fR. . . (ii)
where,
τ= torque,
I= moment of inertia and α= angular acceleration =. Substituting in Eq. (ii), we get
I=fR nmR2⋅=fR nma=f . . . (iii)
Substituting in Eq. (i), we get
⇒mgsinθ−nma=ma ⇒gsinθ=a(n+1) ⇒a= Substituting in Eq. (iii), we get
f=nma =nm . . . (iv)
(A) For ring,
I=mR2 ∴n=1 and fring == (B) For solid sphere,
I=mR2 ∴n= fsolid sphere = = = ∴n= fsolid sphere = = = (C) For solid cylinder, I= ∴n= ∴fsolid cylinder = = ∴ n= fsolid cylinder = = (D) For hollow cylinder
I=MR2 ∴ n=1 fhollow cylinder = = ∴ Hence,
A→4,B→3,C→2,D→4 is the correct.
