a small mass of 0.2kg is whirled round a horizontal circle at the end of the string of 0.5m at a constant angular speed of 4 radian per second,the tension in the string is?

Answer 1

It would be extremely helpful for you to draw a picture of the situation as seen from the side.

Let A be the point where the string is anchored,

let B be the point where the rotating object is located.

The object is rotating in a horizontal plane, which in your picture is

just a line, somewhat below A.

Draw the point B on that line of rotation at the rightmost point.

From the point A drop a line perpendicular to the plane of rotation;

in your picture it will intersect the line of rotation in a point C.

So we have a right angle triangle ABC, with right angle at point C.

The line AB is the string.

The line CB is the radius of rotation.

Let

m = 0.2 kg be the mass of the object,

L = 0.5 m be the length of the string,

ω = 4s^-1 be the angular velocity,

g = 9.81 m/s² be gravitational acceleration,

α (unknown) be the angle in the triangle ABC at the point B,

r = Lcos(α) be the radius of rotation.

The tension in the string is there because the string needs to provide

two forces:

(Please draw the forces as acting on the object B.)

- upward force W (weight) needed to keep the object from falling down,

- horizontal (leftward) centripetal force F needed to keep the object rotating.

The tension in the string is the vector sum of W and F.

The magnitudes of the forces are

W = mg

F = mrω² = mLcos(α)ω²

(If you are not familiar with these formulas let me know.)

By Pythagorean Theorem the tension in the string is

T = √(W² + F²).

To calculate T we just need the angle α.

We can calculate it from the triangle ABC:

tan(α) = W/F

To solve for α we just need some trigonometric manipulation

sin(α)/cos(α) = W/F = mg/mLcos(α)ω² = g/Lcos(α)ω²

sin(α) = g/Lω²

cos(α) = √(1-sin²(α)) = √(1 - g²/L²ω⁴)

Putting everything together

T = √(m²g² + m²L²(1 - g²/L²ω⁴)ω⁴)

= m√(g² + L²ω⁴ - g²)

= mLω²

Notice that the result is the same as you would get in the absence of gravity,

in which case the point A would be the center of rotation and the radius would be L.

Substituting actual numbers

T = 0.2×0.5×4² = 1.6 N