Question

If a plane is inclined at an angle 50 with the horizontal and a cylindrical metal is allowed to roll between two fixed height 92cm apart on the plane. Given that the acceleration of the cylinder to be (2gSinθ/3) and g as 9.8m/s2, the time of descent i

Answer 1

The distance of 10 m on the hypotenuse on an incline of 30

0

corresponds to a height difference of 5. Thus change in potential energy is given as 5mg.

The change in the kinetic energy is given as

2

1

mv

2

+

2

1

Iω

2

moment of inertia of a hollow cylinder about centroidal axis is mr

2

Thus we get the kinetic energy as

2

1

mv

2

+

2

1

mr

2

r

2

v

2

or

mv

2

Equating both the energies we get

5mg=mv

2

or

v=

49

=7m/s

Answer 2

Given:

Inclination of plane, θ = 50°

Height, h = s = 92 cm = 0.92 m

Acceleration of cylinder, a = (2/3) × g sinθ

Acceleration due to gravity, g = 9.8 m/s²

To Find:

Time of descent of the cylinder.

Calculation:

- Using 2nd equation of motion, we get:

s = (1/2) at²

⇒ 0.92 = (9.8 × sin 50°/ 3) × t²

⇒ t² = 0.92 × 3/ 9.8 × sin 50°

⇒ t² = 2.76 / (9.8 × 0.766)

⇒ t² = 0.37

⇒ t = 0.61 second

- Hence, the time period of descent of the cylinder is 0.61 seconds.