A uniform solid sphere rolls down an incline. What must be the incline angle (deg) if the linear acceleration of the center of the sphere is to have a magnitude of 0.16g?
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To determine the incline angle, we need to analyze the forces acting on the uniform solid sphere while it rolls down the incline.
When the sphere rolls without slipping, both translational motion (center of mass) and rotational motion are present. Let's denote the following variables:
- θ: incline angle (in degrees)
- m: mass of the sphere
- R: radius of the sphere
- I: moment of inertia of the sphere
- g: acceleration due to gravity
First, we need to calculate the moment of inertia of the solid sphere. For a solid sphere, the moment of inertia is given by the formula:
I = (2/5) * m * R^2
Next, we need to analyze the forces acting on the sphere. There are two main forces involved: the gravitational force (mg) acting vertically downwards, and the component of the gravitational force parallel to the incline (mg * sinθ).
The net force along the incline acting on the sphere will be given by the difference between the parallel component of the gravitational force and the force of friction.
Let's denote the acceleration of the center of the sphere as "a". For a solid sphere rolling without slipping, the relationship between linear acceleration (a) and angular acceleration (α) is given by:
a = R * α
Since we know that a = 0.16g, we can rewrite the above equation as:
0.16g = R * α
Using the relationship between torque (τ) and angular acceleration (α) for a solid sphere rolling without slipping:
τ = I * α
The torque acting on the sphere is caused by the force of friction, which can be calculated as the product of the coefficient of rolling friction (μ) and the normal force (mg * cosθ). Therefore:
τ = (R * μ * m * g * cosθ)
Setting the torque equation equal to the angular acceleration equation, we can solve for the coefficient of rolling friction:
(R * μ * m * g * cosθ) = (2/5) * m * R^2 * α
Simplifying and substituting α = (0.16g) / R, we get:
μ * g * cosθ = (2/5) * (0.16g)
Now, let's solve for the incline angle θ:
cosθ = [(2/5) * (0.16g)] / (μ * g)
θ = arccos{[(2/5) * (0.16g)] / (μ * g)}
Simply substitute the value of the coefficient of rolling friction (μ) to obtain the final incline angle (θ) in degrees.