A 3.05 kg object is attached to a vertical rod by two strings as shown ...
A 3.05 kg object is attached to a vertical rod by two strings as shown in Figure. The
object rotates in a horizontal circle at constant speed 6.75 m/s.
(a) Find the tension in the upper string.
(b) Find the tension in the lower string.
Let the tension in the upper string is T1 and that in the lower one is T2
The angle the strings make with the vertical is given by
Cos? = 1.5/2 = 0.75
Or
? = 41.40
Now the vertical components of the tensions in the strings balances the weight of the
object and hence writing the equation of motion in vertical direction we have
T1*cos? - T2* cos? - mg = 0
Or
T1 ¨C T2 = mg/cos? = 3.05*9.8/cos41.4 = 39.85 N ---------- (1)
Now the radius of circular path is given by
R = 2*sin?
= 2*0.6613 = 1.323 m
The horizontal components of the tensions provide the necessary centripetal
acceleration and hence
T1*sin? + T2* sin? = mv2/R
Or
(T1+T2)* sin? = mv2/R
Or
T1 + T2 = mv2/(R sin?? = 3.05*6.752/(1.323*0.6613) = 158.8 N
Adding the two equations
2T1 = 39.85 + 158.8
or
T1 = 99.34 N
And on subtracting the two equations (1) and (2) we get
2T2 = 158.8 - 39.85
or
T2 = 59.47 N
--(2)
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