Physics

The formula for Tension in a wire under circular motion

An object under circular motion undergoes centripetal acceleration due to continuous changes in direction. In cases when the object is tied to a rope, the centripetal force is provided by the tension in the rope. The circular motion can be vertical or horizontal against the force of gravity. In this article, we will look at examples, formulas, and problems for tension in a wire under circular motion.

Examples of tension in a wire under circular motion

A Gravitron gives a flying-like feel due to the centrifugal force. But, the tension of the wires hold the people safe. — A Gravitron gives a flying-like feel due to the centrifugal force. But, the tension of the wires holds the people safe.

Hammer throw
Gravitron
Circus

The formula for tension In a horizontal circular motion

It is very difficult to maintain the object exactly at 90 degrees. But, in such an ideal scenario, the tension in the string provides the centripetal force, which is balanced by an equal centrifugal force acting in the outward direction. The weight of the body equal to mass*acceleration due to gravity will be acting in the downward direction.

In the x-direction, only two forces are on the body-centrifugal and tension force. Hence, in such a case the tension will be equal to the centrifugal force.

Formula for tension = centrifugal force = mv2/r

The formula for tension In a vertical circular motion

When we try to maintain an object under vertical circular motion we can experience the changes in tension in our hands at different positions. So that means the formula for tension in a vertical circular motion is dependent on the position. This is basically due to the fact that we are acting against gravity.

The formula for tension when the object is at top

Formula for tension in a vertical circular motion

When the object is at the top of the circle, the centrifugal force is acting against the force of gravity.

So the formula of tension will be = centripetal force – force of gravity = mv2/r – mg = m(v2/r-g)

The formula for tension when the object is at bottom

When the object is at the bottom the centrifugal is acting in the direction of the gravitational force.

The formula of tension will be = centripetal force + force of gravity = mv2/r + mg = m(v2/r+g)

The formula for tension when the object is at an angle in a circular motion

Here, in this case, the centripetal force, tension, and the weight of the object act in a different direction. So, we take the vector product to derive the formula for tension.

Therefore, formula for tension = sqrt((mg)2 + (mv2/r)2)

Numerical problems on the calculation of tension in a string under circular motion

1)A ball of weight m is attached to a rope and rotated at 5m/s in such a fashion that the ball is always at 90 degrees to the axis. The length of the rope is 100 cm with a tension of 100 N. Find the weight of the ball.

First, let’s convert all units into SI units.

Length(r) = 1 m
speed(v) = 5
Tension = 100 N

Now let us apply the formula for tension

Centripetal force = tension = mv2/r
100 = m*5*5/1
m = 4
The weight of the object = m*g = 4*9.8 = 39.2 N

2) In the case of exactly horizontally uniform circular motion, if the length is double find the effect on the tension.

Let us first start with the formula for tension.

Tension = mv2/r
Now let us substitute r with 2r
Tension = mv2/2r
So, the tension becomes halved by doubling the length of the rope.

3) A object of mass 10 Kg is tied to a rope and rotated in a uniform circular motion. The length of the rope is 0.5 m. The rope can bear a maximum tension of 50 N. Find the maximum speed at which the object can be rotated in a uniform circular motion without breaking the rope.

We have mass = 10 Kg, radius = 0.5 m, tension = 50 N.

Let us apply the formula of tension to this.

T = mv2/r

50 = 10 * v2/0.5

v = 1.58 m/s

So any speed exceeding 1.58 m/s2 will result in the fracture of the rope.

4) In a Gravitron, suppose the maximum possible weight is 100 Kg, with the length of the cords being 10 meters. The operating speed of the Gravitron is 10 m/s. Find the maximum tension on the cords of the Gravitron.

Now here the circular motion is not perfectly vertical, so we have to apply the general formula of tension for an object in a circular motion.

Tension = sqrt((mg)2 + (mv2/r)2)

T = sqrt((100*9.8)2 + (100*10*10/10)2)

Tension = sqrt(19,60,400) =1,400.14 N

Maximum tension of the cords = 1,400.14 for a weight of 100 Kg at a speed of 10 m/s with a cord length of 10 m.

Why does not the wire become slack when the object is at the top of a vertical circle?

When the object is at the top in a vertical circular motion, ideally the wire/rope should go slack. But the centrifugal force is enough to keep the object in motion in a circular motion. Since centrifugal force = mv2/r, the force is dependent on the speed of motion. So, under slow speed, the wire may go slack till the required amount of centrifugal force is reached.