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 0.13g? (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 0.13g?

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aristocles aristocles · Answer 1 · 2019-08-18T06:50:55+02:00

Answer:

Part a)

[tex]\theta = 10.5 degree[/tex]

Part b)

acceleration would be

a = g sin10.5 = 0.18 g

so it is more than the above acceleration

Explanation:

As the sphere rolls down on the inclined plane then in that case the force equation on the sphere is given as

[tex]mg sin\theta - f = ma[/tex]

torque equation about the center of the sphere is given as

[tex]\tau = I\alpha[/tex]

[tex]f R = \frac{2}{5}mR^2 (\frac{a}{R})[/tex]

[tex]f = \frac{2}{5}ma[/tex]

now we have

[tex]mgsin\theta = ma + \frac{2}{5}ma[/tex]

[tex]mgsin\theta = \frac{7}{5}ma[/tex]

[tex]a = \frac{5}{7} gsin\theta[/tex]

now we have

[tex]a = 0.13 g = \frac{5}{7}g sin\theta[/tex]

[tex]0.13 = \frac{5}{7} sin\theta[/tex]

[tex]0.182 = sin\theta[/tex]

[tex]\theta = 10.5 degree[/tex]

Part b)

acceleration of the object sliding on smooth inclined plane is given as

[tex]a = gsin\theta[/tex]

[tex]a = 9.8 sin10.5[/tex]

[tex]a = 0.18 g[/tex]

so this acceleration is more than the acceleration of sphere on same inclined plane