A uniform solid sphere rolls down an incline. A) What must be the incline angle (deg) if the linear acceleration of the center of the sphere is to have a magnitude of 0.21g? B) If a frictionless block were to slide down the incline at that angle, would its acceleration magnitude be more than, less than, or equal to 0.21g?

a) θ = 12.12°

b) equal to 0.21g

Explanation:

Solution:-

Declare variables:

- The mass of solid sphere, m

- The inclination angle, θ

- The linear acceleration a down the slope of the solid sphere is a = 0.21g

Where, g: The gravitational acceleration constant.

- The component of weight of solid sphere directed down the slope is given by:

                          Ws = m*g*sin ( θ )

- Apply Newton's second law of motion down the slope, state:

                        F_net = m*a

- The only net force acting on the solid sphere is the Weight. So, the equation of motion in the coordinate axis ( down the slope ).

                        Ws = m*a

                        m*g*sin ( θ ) = m*0.21*g

- Solve for inclination angle ( θ ):

                        sin ( θ ) = 0.21

                        θ = arcsin ( 0.21 )

                        θ = 12.12°

- If a friction-less block of mass ( m ) moves down the same slope then block has weight component down the slope as:

                         Wb = m*g*sin ( θ )

- Apply Newton's second law of motion down the slope, state:

                        F_net = m*a

- The only net force acting on the solid sphere is the Weight. So, the equation of motion in the coordinate axis ( down the slope ).

                        Ws = m*a

                        m*g*sin ( θ ) = m*a

- Solve for linear acceleration ( a ):

                        g*sin ( θ ) = a

                        a = sin ( 12.12 ) * g

                        a = 0.21g

The acceleration is independent of mass and only depends on the inclination angle θ.

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