A uniform solid sphere rolls down an incline. (a) What must be the incline angle if the linear acceleration of the center of the sphere is to have a magnitude of 0.10g? (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 0.10g? Why?

Question

Physics

Colton

17 October, 09:09

A uniform solid sphere rolls down an incline. (a) What must be the incline angle if the linear acceleration of the center of the sphere is to have a magnitude of 0.10g? (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 0.10g? Why?

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Answers (1)

Know the Answer?

Kaitlynn Ward · Answer 1

a. 87.7° b. It is less than 0.10g

Explanation:

a. Angle of incline

To find the angle of incline, we resolve the weight of the sphere into its horizontal and vertical component. mgcosθ and mgsinθ respectively. The horizontal component provides a torque along the incline which produces the linear acceleration, a of the center of mass.

The torque, τ = mgRcosθ = Iα

where m = mass of sphere, R = radius of sphere, θ = angle of incline, I = moment of inertia of sphere = 2/5MR² and α = angular acceleration of sphere = a/R where a = linear acceleration of center of mass of sphere

So, MgRcosθ = Iα = Ia/R

MgRcosθ = 2/5MR²a/R

MgRcosθ = 2aMR/5

gcosθ = 2a/5

a = 5gcosθ/2 if a = 0.10g

0.10g = 5gcosθ/2

2 * 0.10g/5 = gcosθ

2 * 0.10/5 = cosθ

cosθ = 0.04

θ = cos⁻¹0.04 = 87.7°

So, the angle of incline must be 87.7°

b

If a frictionless block were to slide down the incline at that angle, the magnitude of its acceleration would less than 0.10g. This is because the component of acceleration along the incline is gcosθ = gcos87.7° = 0.04g.

So, it is obviously less.