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

Question

A 4.00-kg block is attached to a vertical rod by means of two strings. When the system rotates about the axis of the rod, the strings are extended as shown in the figure below and the tension in the upper string is 70.0 N.

(a) What is the tension in the lower cord?

(b) How many revolutions per minute does the system make?

(c) Find the number of revolutions per minute at which the lower cord just goes slack.

(d) Explain what happens if the number of revolutions per minute is less than that in part (c).

Explanation / Answer

From mass 4kg drop perpendicular on vertical rod of 2m

This perpendicular will now become BASE of triangle whose hypoteneous 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 rt 0.5625 =0.75

the upper cord makes angle O with BASE (horizontal line from 4 kg to 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 =70 N

the tension in the lower cord = T2

Radius of circular path = r= sq rt [1.25^2 -1]=sq rt 0.5625=0.75 m

Perpendicular distance of circulating mass from vertical rod=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 lower cord also makes angle O with horizontal because both the cords are of 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 vertical component 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=70 -4*9.8 /0.8

T2= 21 N

The tension in the lower cord is T2= 21 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=[70 +21]0.6=91*0.6=54.6 N

4*0.75*4*9.8696*f^2=54.6

f^2=54.6/118.4353=0.461011201

f =0.678978057 revolutions per second

revolutions per minute =f*60=40.73868342 rpm

The system mskes 40.73868342 revolutions per minute

______________________________________...

Part C

When lower cord is about to slack, onlt T1sinO malances weight mg

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 centripetal force

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 revolutions per minute

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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