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

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

Step-by-step 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 = (2)/(5)mR^2 ((a)/(R))`

`f = (2)/(5)ma`

now we have

`mgsin\theta = ma + (2)/(5)ma`

`mgsin\theta = (7)/(5)ma`

`a = (5)/(7) gsin\theta`

now we have

`a = 0.13 g = (5)/(7)g sin\theta`

`0.13 = (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