Question

A stone of mass 0.3 kg is tied to one end of a string 0.8 m long and rotated in a vertical circle. At what speed of the ball will the tension in the string be zero at the highest point of the circle? What would be the tension at the lowest point in this case ? Given g = 9.8 ms^(-2) ?

Answer 1

answer;

For tension to be zero at highest point

mv^2 / r = mg

v = sqrt(rg)

v = sqrt(0.8 * 9.8)

v = sqrt(7.84)

v = 2.8 m/s

Tension at lowest point

T = mv^2/r + mg

= m * (5gr) / r + mg

= 6mg

= 6 * 0.3 * 9.8

= 17.64 N

Tension at lowest point is 17.64N

Answer 2

Answer:

The highest point of the circle=2.8 m/s

Tension at the lowest point=17.64N

Explanation:

Concept

• Tension is described as a pair of opposing forces acting on each end of the specified parts. Apart from the endpoints, every portion of a rope feels the tension force in both directions. The ends are subjected to tension on one side and the force of the attached weight on the other. In other cases, the strain on the string varies.
• A force over the length of a medium, particularly a force carried by a flexible medium like as a rope or cable, is known as tension.

Given

m=0.3 kg

g=9.8$ms^{-2}$

Find

Speed of the highest point of the circle

Tension at the lowest point

Solution

Hеге, $m=0.3 k g, r=0.8 m, v=?, T= ?$

When tension in the string is zero at the highest point speed of stone at the highest point is

$v_{H}=\sqrt{g r}=\sqrt{9.8 \times 0.8}=2.8 \mathrm{~m} / \mathrm{s}$

$\text { As } T_{L}-T_{H}=6 m g, \text { and } T_{H}=0$

$T_{L}=6 m g=6 \times 0.3 \times 9.8=17.64 N$

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