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Solving Confusions on Pulling a Spool
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[QUOTE="kuruman, post: 6858145, member: 192687"] The accompanying image is useful as a free body diagram and can be used to examine the force of static friction that was brought up. We write the usual Newton's second law for the translational motion $$T-f_s=ma_{\text{cm}}.\tag{1}$$ For the rotational motion, we express the torque about the contact point P and note that ##\alpha = a_{\text{cm}}/R.## Then $$T(R+r)=mR^2[I/(mR^2)+1]\frac{a_{\text{cm}}}{R}.\tag{2}$$We solve for the unknown acceleration of the CM and the force of friction to find $$a_{\text{cm}}=\frac{T}{m}\frac{(1+r/R)}{[I/(mR^2)+1]}\tag{3}.$$At this point the kinematic equation ##~2a_{\text{cm}}L=v_{\text{cm}}^2~## yields directly the speed of the center of mass without need to worry about the work done by static friction or the distance over which the disembodied hand exerts a force. The force of static friction is interesting because its magnitude and direction depend on the radius of the axle. $$f_s=\frac{T[I/(mR^2)-r/R]}{[I/(mR^2)+1]}\tag{4}$$For example, if ##I/(mR^2)=1/2## then ##f_s = 0## when ##r =R/2.## When ##r >R/2##, the force of static friction is in the same direction as the motion. In this case, the accompanying picture can be forgotten but that's OK because it has already served its purpose. Another interesting case is ##r=-R##, i.e. the string is wrapped around the wheel and comes out from its bottom. Then ##f_s=T## which means that the net force is zero and the wheel will not accelerate. This explains why bottles placed on the supermarket belt with their axes perpendicular to the direction of the belt's motion, roll in place and don't move closer to the cashier. [/QUOTE]
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Solving Confusions on Pulling a Spool
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