A uniform disk of mass m and radius R rolls, without slipping, down a fixed plane inclined at an angle 30 to the horizontal. The linear acceleration of the disk (in m/sec2) is _______________.Correct answer is '3.266'. Can you explain this answer?

Question

Answer

Solution:

Acceleration of the rolling body can be calculated by considering the rolling motion and linear motion separately.

Rolling Motion:

- For pure rolling, the velocity of the center of mass of the disk is equal to the velocity of the point of contact with the plane, i.e., v = Rω.
- The angular velocity of the disk can be calculated using the conservation of energy principle. The initial potential energy is mgh, and the final energy is a combination of kinetic energy of the center of mass and rotational energy about the center of mass. So, mgh = (1/2)mv^2 + (1/2)Iω^2, where I is the moment of inertia of the disk, which is (1/2)mr^2.
- Substituting v = Rω and I = (1/2)mr^2 in the above equation and solving for ω, we get ω = (2gh/R)^0.5.

Linear Motion:

- The gravitational force acting on the disk is mg sin(30), where 30 is the angle of inclination of the plane.
- The frictional force acting on the disk is mg cos(30) μ, where μ is the coefficient of static friction between the disk and the plane.
- The net force acting on the disk is F = mg sin(30) - mg cos(30) μ.
- Using Newton's second law, F = ma, where a is the linear acceleration of the disk.
- Substituting the value of F and solving for a, we get a = g (sin(30) - cos(30) μ).

Final Answer:

- Substituting the value of μ = 0.5 and g = 9.8 m/s^2 in the above equation, we get a = 3.266 m/s^2 (approx).

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