A solid sphere of 1.31 kg and radius 0.17 m, is rolling down a rough plane that

Question

A solid sphere of 1.31 kg and radius 0.17 m, is rolling down a rough plane that is inclined at an angle 300 to the horizontal. The figure shows the three forces acting on the sphere (its weight W, friction force F and the normal reaction N), together with the origin O, the unit vectors i and j in the x- and y- directions respectively. The sphere starts rolling down when its point of contact with the plane is at the origin O.

You may take the moment of inertia of a solid sphere about an axis through its centre of rotation as 2/5 MR2 and the angular acceleration is to be 12.67 rad s-2.

By considering Newton

Explanation / Answer

a = alpha * R = 12.67 * .17 = 2.154 m/s^2      linear acceleration of sphere
av = a * sin 30 = 1.077     vertical acceleration of sphere
m g - N cos 30 - Ff sin 30 = m av     vertical acceleration of sphere
N = m (g - av) / cos 30 - Ff tan theta     solving for N where Ff is the frictional force
N = 1.31 * (9.80 - 1.077) / cos 30 - Ff * tan 30
N = 13.95 - .577 Ff
2 M R^2 * alpha / 5 = Ff * R       since Ff provides the angular acceleration
Ff = 2 M R * alpha / 5 = 2 * 1.31 * .17 * 12.67 / 5 = 1.129 N
N = 13.95 - .577 * 1.129 = 13.30 Newtons

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