Question

A solid cylinder of mass 2 kg and radius 50 cm rolls up an inclined plane of angle inclination 30°. The centre of mass of cylinder has speed of 4 m/s. The distance travelled by the cylinder on the incline surface will be :
(Take g = 10 m/s\(^2\))

• A. 2.2 m
• B. 1.6 m
• C. 1.2 m
• D. 2.4 m

Answer 1

The Correct Option is D

To solve this problem, we need to determine the distance travelled by a solid cylinder rolling up an inclined plane. We'll take the following steps:

1. Identify the type of energy involved: The cylinder has both translational and rotational kinetic energy as it rolls up.
2. Apply the conservation of energy principle: As the cylinder rolls up the incline, its initial kinetic energy will be converted into potential energy. At the highest point in its path, all the kinetic energy will be converted into potential energy.

Step 1: Calculate Initial Kinetic Energy
The total initial kinetic energy for the rolling cylinder is the sum of its translational and rotational kinetic energy.

Translational kinetic energy, T_{KE} = \frac{1}{2} m v^2
Rotational kinetic energy, R_{KE} = \frac{1}{2} I \omega^2
For a solid cylinder, the moment of inertia, I = \frac{1}{2} m r^2 and the angular speed v = r \omega gives us \omega = \frac{v}{r}

Substituting these in, the total initial kinetic energy becomes:

E_{\text{initial}} = \frac{1}{2} m v^2 + \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v}{r}\right)^2 = \frac{1}{2} m v^2 + \frac{1}{4} m v^2 = \frac{3}{4} m v^2

Given, m = 2 \text{ kg}, v = 4 \text{ m/s}, thus:

E_{\text{initial}} = \frac{3}{4} \times 2 \times 4^2 = \frac{3}{4} \times 2 \times 16 = 24 \text{ J}

Step 2: Equate Initial Energy to Potential Energy at the Highest Point
The potential energy at the highest point is U = mgh. Here, h is the height the cylinder reaches, and can be related to the distance d travelled along the incline by h = d \sin(\theta).

Equating gives us:

24 = 2 \times 10 \times d \sin(30^\circ)

The value of \sin(30^\circ) = \frac{1}{2}, so:

24 = 2 \times 10 \times d \times \frac{1}{2}

24 = 10d

Solve for d:

d = \frac{24}{10} = 2.4 \text{ m}

Thus, the distance travelled by the cylinder on the incline surface is 2.4 m.