The left end of a massless stick with length | vedclass

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(A) Let $N$ be the normal force exerted by the table on the stick at the left end. The stick is massless,and a point mass $m$ is at the center (distan

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(A) Let $N$ be the normal force exerted by the table on the stick at the left end. The stick is massless,and a point mass $m$ is at the center (distance $l/2$ from the pivot point).Immediately after release,the stick starts rotating about the corner of the table (the pivot).The torque $	au$ about the pivot is given by $	au = mg(l/2)$.The moment of inertia $I$ of the point mass $m$ about the pivot is $I = m(l/2)^2$.Using $	au = I\alpha$,we have $mg(l/2) = m(l/2)^2 \alpha$,which gives $\alpha = g/(l/2) = 2g/l$.The linear acceleration $a$ of the center of mass is $a = \alpha(l/2) = (2g/l)(l/2) = g$.Applying Newton's second law for the vertical motion of the center of mass: $mg - N = ma$.Substituting $a = g$,we get $mg - N = m(g)$,which implies $N = 0$.

The left end of a massless stick with length $l$ is placed on the corner of a table,as shown in the figure. $A$ point mass $m$ is attached to the center of the stick,which is initially held horizontal. It is then released. Immediately afterward,what normal force does the table exert on the stick?

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