A 3.8 kg ball is swung in a horizontal circle at the end of a 1.5 meter rope. If the tension in the rope is kept constant at 79 N, what is the period of rotation, T?

Well, ignoring gravity (gravity makes the ball swing downward, the rope makes an angle, which changes the radius of rotation), then

tension= massball*v^2/r=m(2PIr/T)^2 /r

do the algebra, and solve for Period T.

To find the period of rotation, T, of the ball, we need to consider the forces acting on it and use the centripetal force equation.

The only force acting on the ball is the tension in the rope. According to Newton's second law, the net force acting on an object moving in a circular path is equal to the mass times acceleration towards the center of the circle.

The centripetal force (Fc) is calculated using the formula: Fc = m * (v^2 / r), where m is the mass, v is the velocity, and r is the radius of the circular path.

In this case, the centripetal force is equal to the tension in the rope, so we have Fc = T. Thus, T = m * (v^2 / r).

We also know that the velocity (v) is related to the period of rotation (T) and the circumference of the circle (C) by the formula v = C / T.

The circumference of the circle (C) can be calculated using the formula C = 2 * π * r, where π is a constant.

Combining these equations, we have T = m * (v^2 / r) = m * ([C / T]^2 / r). Simplifying further, we get T^2 = (m * C^2) / (r^3), and finally, T = √((m * C^2) / (r^3)).

Now we can substitute the given values into the equation:

m = 3.8 kg (mass of the ball)
T = unknown (period of rotation)
r = 1.5 m (radius of the circular path)
C = 2 * π * r (circumference of the circle)
T = √((m * C^2) / (r^3)) = √((3.8 kg * (2 * π * 1.5 m)^2) / (1.5 m)^3)

Evaluating this expression, we can find the period of rotation, T.