A heavy ball with a weight of 110 N is hung from the ceiling of a lecture hall on a 4.4-m-long rope. The ball is pulled to one side and released to swing as a pendulum, reaching a speed of 5.3 m/s as it passes through the lowest point. What is the tension in the rope at that point?

The forces acting on the ball is the tension [tex]T[/tex] on the string and the force of gravity [tex]mg[/tex]. Their sum contributes to the centripetal force:

[tex]T -mg = \dfrac{mv^2}{R}[/tex]

Solving for tension [tex]T[/tex] we get:

[tex]\boxed{T = \dfrac{mv^2}{R} +mg}[/tex]

Putting in numerical values

[tex]m = \dfrac{110N}{10ms^{-2}} =11kg[/tex]

[tex]v = 5.3m/s[/tex]

[tex]R = 4.4m[/tex]

[tex]T = \dfrac{(11kg)(5.3m/s)^2}{4.4m} +(11kg)(10ms^{-2})[/tex]

[tex]\boxed{T = 180.225N}[/tex]

which is the tension in the rope.

andromache andromache · Answer 2 · 2022-03-22T17:13:35+01:00

andromache andromache

22-03-2022

The tension in the rope at that point is 181.63 N

Calculation of the tension in the rope:

Since

The weight of the ball W = 110 N

The length of the rope l=4.4m

And, the velocity of the ball at the lowest point v=5.3m/s

Now the mass of the ball is

= 110 / 9.8

= 11.22 kg

Now the tension should be

= 110 + (11.22*5.3^2)/ 4.4

= 181.63 N

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