A 4.00-kg block is attached to a vertical rod bymeans of two strings. When the s

Question

A 4.00-kg block is attached to a vertical rod bymeans of two strings. When the system rotates about the axis of therod, the strings are extended and the tension in the upper stringis 80.0 N. The space between the strings on therod is 2.00m and each string is 1.25m long. Values for sine, cosine, and r were found to be: sin = (4/5) cos= (3/5) r= (1.25m)/(3/5) How were these values determined? A 4.00-kg block is attached to a vertical rod bymeans of two strings. When the system rotates about the axis of therod, the strings are extended and the tension in the upper stringis 80.0 N. The space between the strings on therod is 2.00m and each string is 1.25m long. Values for sine, cosine, and r were found to be: sin = (4/5) cos= (3/5) r= (1.25m)/(3/5) How were these values determined?

Explanation / Answer

________________________ From mass 4kg drop perpendicular on vertical rod of 2m This perpendicular will now become BASE of triangle whosehypoteneous is upper cord of 1.25m The perpendicular of that triangle is upper half of vertical rod.This upper half is 1 m The perpendicular of that triangle is 1m The BASE = sq rt [ (1.25)^2 - 1^2]= sq rt [1.5625 - 1] =sq rt0.5625 =0.75 the upper cord makes angle O with BASE (horizontal line from 4 kgto vertical rod) sinO=perpendicular / hypoteneous=1/1.25=100/125=0.8 cosO=base/hypoteneous=0.75/1.25=0.6 Suppose tension in upper cord =T1 =80 N the tension in the lower cord = T2 Radius of circular path = r= sq rt [1.25^2 -1]=sq rt 0.5625=0.75m Perpendicular distance of circulating mass from verticalrod=radius=0.75 m mass =m=4.00 kg Let the number of revolutions made by the system in 1 second= f If v is speed in circular path and w is angular speed, then v=rw w=2(pi)f Centripetal force = F = mv^2/r = mrw^2= mr* [4(pi)^2]*f^2 Suppose upper cord makes angle O with the horizontal , the lowercord also makes angle O with horizontal because both the cords areof equal length. sinO=0.8 cosO=0.6 Resolving T1 and T2 into vertical and horizontal components, Horizontal components T1cosO and T2 cosO being in same direction,add up to provide the centripetal force F Vertical component T1sinO is vertically upwards but verticalcomponent T2sinO is vertically downwards . weight 'mg' is also downwards. T1sinO =T2sinO + mg [T1 - T2 ] sinO =mg [T1 - T2 ] =mg / sinO T2= T1 - (mg/sinO ) T2=80 -4*9.8 /0.8 T2= 31 N The tension in the lower cord is T2= 31 N ______________________________________… Part B Horizontal components T1cosO and T2 cosO being in same direction,add up to provide the centripetal force F F=T1cosO + T2 cosO mr* [4(pi)^2]*f^2 = [ T1+T2 ] cosO mr* [4(pi)^2]*f^2=[80 +31]0.6=111*0.6=66.6 N 4*0.75*4*9.8696*f^2=66.6 f^2=66.6/118.4353=0.5623 f =0.7498 revolutions per second revolutions per minute =f*60=44.99 rpm The system mskes 44.99 revolutions per minute ______________________________________… Part C When lower cord is about to slack, onlt T1sinO malances weightmg T1sinO=mg T1=mg/sinO=4*9.8/ 0.8=49 N Tension in upper cord changes to 49 N Horizontal component of tension in upper cord provides centripetalforce T1cosO = F = mr* [4(pi)^2]*f^2 T1cosO =4*0.75*4*9.8696*f^2 4*0.75*4*9.8696*f^2 = 49*0.6 f^2 =29.4/ 118.4353=0.248236 f =0.4982 revolutions per second=29.89 revolutions per minute the lower cord just goes slack when system makes 29.89 revolutionsper minute

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.