r = 2rCcosè using the arm OA. If the arm has an angular velocity ù = .4 rad/s an

Question

r = 2rCcosè using the arm OA. If the arm has an angular velocity ù = .4 rad/s and an angular

acceleration of á = .8 rad/s2 at the instant è = 30o, determine the force of the arm AO on the ball and the normal force of the semi-cylinder on the ball. Neglect friction and the size of the ball

The 0.2 kg ball is guided along a smooth semi-cylinder with rC = 0.4 m and path r = 2rCcos using the arm OA. If the arm has an angular velocity u = .4 rad/s and an angular acceleration of a = .8 rad/s2 at the instant e = 30o, determine the force of the arm AO on the ball and the normal force of the semi-cylinder on the ball. Neglect friction and the size of the ball

Explanation / Answer

basiclly, there are three forces acting on the ball, one is from the moving arm, one is the force of gravity, one is from the surface of the cylinder.

to calculate the total force :

F=ma, a=á*r

a=á.r=0.8rad/s2 * 2rCcosè=0.8*2*0.4*cos30=0.554m/s2
so F=0.2*0.554=0.111N

to calculate the force from the gravity:

G=mg=0.2*0.98=0.196N

(sorry , I am not able to draw the graph to better understand)

if we make the direction of total force as y-axis, and the F1(the force by the surface of the cylinder) direction as the x-axis.

then, Gcos60=F2( the force by the moving arm)*cos30- F

and Gsin60-F2*sin30=F1

so 0.196cos60=cos30F2-0.111

F2=0.241N

F1=0.196*sin60-0.241*sin30=0.0492N

so the force by the moving arm is 0.241N and the force by the surface of the cylinder is 0.0492N.

Hope my answer will help you.

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