A body is rolling down an inclined plane without slipping. How does th...
The acceleration of a rolling body is independent of its mass and radius.
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A body is rolling down an inclined plane without slipping. How does th...
**Explanation:**
When a body rolls down an inclined plane without slipping, there are two types of forces acting on it: gravitational force and frictional force. The acceleration of the rolling body depends on the balance between these two forces.
**Gravitational Force:**
The gravitational force acting on the rolling body is given by:
Fg = m * g * sinθ
where m is the mass of the body, g is the acceleration due to gravity, and θ is the angle of inclination of the plane.
**Frictional Force:**
The frictional force acting on the rolling body opposes its motion. This force is responsible for the rolling motion of the body without slipping. The frictional force depends on the coefficient of friction (μ) and the normal force (N) acting on the body.
Frictional force, Ff = μ * N
The normal force can be decomposed into two components: the gravitational force component perpendicular to the plane (N⊥) and the gravitational force component parallel to the plane (N‖).
N⊥ = m * g * cosθ
N‖ = m * g * sinθ
Therefore, the normal force can be written as:
N = N⊥ + N‖ = m * g * (cosθ + sinθ)
**Acceleration of the Rolling Body:**
The net force acting on the rolling body is the difference between the gravitational force and the frictional force:
Fnet = Fg - Ff
Substituting the expressions for Fg and Ff, we get:
Fnet = m * g * sinθ - μ * m * g * (cosθ + sinθ)
Fnet = m * g * (sinθ - μ * (cosθ + sinθ))
The acceleration of the rolling body is given by:
a = Fnet / m
a = g * (sinθ - μ * (cosθ + sinθ))
We can observe that the radius of the rolling body does not appear in the equation for acceleration. Therefore, the acceleration of the rolling body does not depend on its radius. Hence, the correct answer is option C: It does not depend upon radius.
A body is rolling down an inclined plane without slipping. How does th...
C) It is directly proportional to the square root of radius
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