A 0.40-kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 2.0-m on a frictionless horizontal surface.
If the cord will break when the tension in it exceeds 75-N, what is the maximum speed the ball can have?
T(max) = mv²/R.
v=sqrt(R•T(max)/m) =
=sqrt(2•75/0.4)=19.4 m/s
To find the maximum speed the ball can have without the cord breaking, we need to determine the tension in the cord at maximum speed.
To start solving this problem, we can look at the forces acting on the ball when it is spinning in a circle:
1. The tension force in the cord, T, which provides the centripetal force to keep the ball in a circular path.
2. The gravitational force, mg, acting vertically downwards.
The centripetal force is given by the formula Fc = (mv^2)/r, where m is the mass of the ball, v is its speed, and r is the radius of the circle. Since the ball is spinning horizontally, the centripetal force is provided by the tension in the cord.
So, at the maximum speed, the tension in the cord, T, should not exceed 75 N. We can equate the centripetal force and the maximum tension force:
Fc = T
(mv^2)/r = T
Rearranging the equation, we get:
v^2 = (Tr) / m
Now we can substitute the given values:
T = 75 N
r = 2.0 m
m = 0.40 kg
Plugging these values into the equation, we get:
v^2 = (75 N * 2.0 m) / 0.40 kg
Simplifying further:
v^2 = 375 N m / 0.40 kg
v^2 = 937.5 N m/kg
Finally, taking the square root of both sides of the equation:
v = √(937.5 N m/kg)
Calculating this, we get:
v ≈ 30.64 m/s
Therefore, the maximum speed the ball can have without breaking the cord is approximately 30.64 m/s.