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24 December, 13:33

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?

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  1. 24 December, 14:47
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    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

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