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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