Given that, θ=30°
Speed of C.M. of cylinder at the bottom,v=5ms–1
(a) As cylinder goes up, it attains potential energy at the expense of its kinetic energy of translational and rotational motion. Suppose that the cylinder goes up to the height h on the inclined plane.
According to the principle of conservation of energy

1

2

mv2+

1

2

Iω2=mgh
∴

1

2

mv2+

1

4

mv2=mgh
⇒

3

4

mv2=mgh
h=

3v2

4g

=

3×52

4×9.8

=1.913m
Suppose that the cylinder covers a distance S along the inclined plane, when it goes up to the height h on the plane. If θ is the inclination of the plane, then
sin‌θ=

h

S

⇒S=

h

sin‌θ

=

1.913

sin‌30°

⇒h=

1.913

1∕2

=3.826 m
(b) Time taken by the cylinder to return to the bottom, T=2t,
where t = Time of ascending or descending.
Acceleration of a body rolling down an inclined plane is given by
a=

g‌sin‌θ

1+ k2 r2

where k is radius of gyration of rolling body
a=

2

3

g‌sin‌θ
Now,S=ut+

1

2

at2
Since u=0, we get
t=√

2S

a

=√

2×3.826

2 3 g‌sin‌θ

;
t=√

2×3.826×3

2×9.8×sin‌30°

=1.53s
T=2t=2×1.53=3.065≈3.0s