Problem 5.104 art A What is the tension in the lower cord? The 4.00-kg block in
ID: 1575410 • Letter: P
Question
Problem 5.104 art A What is the tension in the lower cord? The 4.00-kg block in the figure (Figure 1 )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 diagram and the tension in the upper string is 82.0 N Submit Request Answer Part B How many revolutions per minute does the system make? Figure of 1 rev/min Submit Request Answer 1.25 m Part C 2.00 m 4.00 kg Find the number of revolutions per minute at which the lower cord just goes slack. 1.25 m rev minExplanation / Answer
Part A:
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 = [(1.25)2 - 12]
= [1.5625 - 1]
= 0.5625 = 0.75m
the upper cord makes angle with BASE (horizontal line from 4 kg to vertical rod)
sin = perpendicular / hypotenuse = 1m/1.25m
= 100m/125m = 0.8
cos = base/hypotenuse
= 0 .75m/1.25m = 0.6
Suppose tension in upper cord = T1 = 82 N
That means the tension in the lower cord = T2
Radius of circular path = r = [1.252 -1]= 0.5625 = 0.75 m
Perpendicular distance of circulating mass from vertical rod = radius = 0.75 m
We know the 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 we have
v = r
Where we know:
= 2f......(1)
Centripetal force = F = mv2/r = (mr)2= mr* [42]*f2
Suppose upper cord makes angle with the horizontal, then the lower cord also makes angle with horizontal because both the cords are of equal length.
sin = 0.8
cos = 0.6
Which come from the previous solution in the beginning of the explanation!
Resolving T1 and T2 into vertical and horizontal components, we get:
Horizontal components:
T1cos and T2cos being in same direction, add up to provide the centripetal force F
Vertical Components:
Vertical component T1sin is vertically upwards but vertical component T2sin is vertically downwards, and additionally weight, 'mg', is also downwards.
T1sin = T2sin + mg
[T1 - T2 ] sin = mg
[T1 - T2 ] =mg / sin
T2 = T1 - (mg/sin )
T2 = 82N - [(4kg*9.8m/s2)/0.8)
T2 = 33 N
The tension in the lower cord is T2 = 33 N
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Part B
Horizontal components T1cos and T2cos being in same direction, add up to provide the centripetal force F
F = T1cos + T2cos
mr* [42]*f2 = [ T1 + T2 ]cos
mr* [42]*f2 = [82N +33N]0.6 = 115N*0.6 = 69 N
4kg * 0.75m * (4*9.8696) * f2 = 69 N
f2 = 69 N / 118.4353 = 0.583
f = 0.764 revolutions per second
However, we want rev/min:
f * 60s = 45.82 rev/min
The system makes 45.82 revolutions per minute
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Part C
When lower cord is about to slack, only T1sinO balances the weight 'mg':
T1sin = mg
T1 = mg/sin = (4kg*9.8m/s2)/ 0.8 = 49 N
Tension in upper cord changes to 49 N
Horizontal component of tension in upper cord provides centripetal force
T1cos = F = mr * [42]*f2
T1cos = 4kg * 0.75m * (4*9.8696) * f2
4kg * 0.75m * (4*9.8696) * f2 = (49N * 0.6)
f2 = (29.4N/ 118.4353) = 0.248236
f = 0.4982 revolutions per second =
f * 60s = 29.89 revolutions per minute
The lower cord just goes slack when system makes 29.89 revolutions per minute