A solid sphere of 7.02 kg and radius 0.32 m, is rolling down a rough plane that

Question

A solid sphere of 7.02 kg and radius 0.32 m, is rolling down a rough plane that is inclined at an angle 30 degree 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- direction 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 MR^2 and the angular acceleration is to be 12.61 rad s^-2. By considering the torque for the motion about the centre of mass of the sphere, calculate the magnitude of the friction force F: giving your answer to 3 decimal places.

Explanation / Answer

F=2*m*g*sin thetha/7 =( 2*7.02*9.8*sin 30)/7=9.828 N

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