A rigid body of mass M and radius R rolls without slipping on an inclined plane of inclination θ, under gravity. Match the type of body Column I with magnitude of the force of friction Column II.
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 ‌=‌
a
R
‌. ‌ Substituting in Eq. (ii), we get I‌
a
R
=fR nmR2⋅‌
a
R
=fR nma=f . . . (iii) Substituting in Eq. (i), we get ⇒‌‌mg‌sin‌θ−nma‌‌=ma ⇒‌‌g‌sin‌θ‌‌=a(n+1) ⇒‌‌a‌‌=‌
g‌sin‌θ
n+1
Substituting in Eq. (iii), we get f‌‌=nma ‌‌=nm‌
g‌sin‌θ
n+1
. . . (iv) (A) For ring, I=mR2 ‌∴‌n‌=1 ‌‌ and ‌‌f‌ring ‌‌=‌
mg‌sin‌θ
1+1
‌=‌
mg‌sin‌θ
2
(B) For solid sphere, I=‌
2
5
mR2 ∴‌‌n‌‌=‌
2
5
f‌solid sphere ‌‌‌=‌
‌
2
5
mg‌sin‌θ
2
5
+1
‌‌=‌
2mg‌sin‌θ
7
‌‌=‌
mg‌sin‌θ
3.5
∴‌‌n‌‌=‌
2
5
f‌solid sphere ‌‌‌=‌
‌
2
5
mg‌sin‌θ
2
5
+1
‌‌=‌
2mg‌sin‌θ
7
‌‌=‌
mg‌sin‌θ
3.5
‌ (C) For solid cylinder, ‌I‌‌=‌
mR2
2
∴‌‌n‌‌=‌
1
2
∴‌‌f‌solid cylinder ‌‌‌=‌
‌
1
2
mg‌sin‌θ
1
2
+1
‌‌=‌
mg‌sin‌θ
3
∴n‌‌=‌
1
2
f‌solid cylinder ‌‌‌=‌
‌
1
2
mg‌sin‌θ
1
2
+1
‌‌=‌
mg‌sin‌θ
3
(D) For hollow cylinder I=MR2 ∴n=1 f‌hollow cylinder ‌‌‌=‌
mg‌sin‌θ
1+1
‌‌=‌
mg‌sin‌θ
2
∴ Hence, A→4,B→3,C→2,D→4 is the correct.