Question

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?

Answer 1

Answer:

Part a)

$\theta = 10.5 degree$

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

$mg sin\theta - f = ma$

torque equation about the center of the sphere is given as

$\tau = I\alpha$

$f R = \frac{2}{5}mR^2 (\frac{a}{R})$

$f = \frac{2}{5}ma$

now we have

$mgsin\theta = ma + \frac{2}{5}ma$

$mgsin\theta = \frac{7}{5}ma$

$a = \frac{5}{7} gsin\theta$

now we have

$a = 0.13 g = \frac{5}{7}g sin\theta$

$0.13 = \frac{5}{7} sin\theta$

$0.182 = sin\theta$

$\theta = 10.5 degree$

Part b)

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

$a = gsin\theta$

$a = 9.8 sin10.5$

$a = 0.18 g$

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