AP EAPCET 20-Aug-2021 Shift 1 Paper

Moment of inertia of sphere and hollow cylinder are, Isphcre=

2

5

MR2 and Icylinder=MR2
Angle of inclination, θ=30° Let, acceleration of body =a
Friction force =f
Mass of body =m
Radius of body =R
mg‌sin‌θ−f=ma ...(i)
Torque, τ=Rf=Iα
⇒f=

Iα

R

...(ii)
From Eqs. (i) and (ii), we get
mg‌sin‌θ−

Iα

R

=ma
Angular acceleration, α=

a

R

⇒mg‌sin‌θ−

Ia

R2

=ma
⇒mg‌sin‌θ=a(m+

I

R2

)
a=

mg‌sin‌θ

m+ I R2

=

g‌sin‌θ

1+ I mR

∵ Distance covered by body, s=

1

2

at2
∵Ssphere=Scylinder

1

2

asphere×t2=

1

2

×Scylinder×t2
⇒asphcre=acylinder
⇒

g‌sin‌30°

1+ 2 5 mR2 mR2

=

g‌sin‌θ

1+ mR2 mR2

⇒

1∕2

7∕5

=

sin‌θ

2

⇒sin‌θ=

5

7

⇒θ=sin−1(

5

7

)=45.6°≈45°