A 0.4kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.3 m on a frictionless horizontal surface. If the cord will break when the tension in it exceeds 65 N, what is the maximum rpm(revolution per minute) the ball can have?
HELP?
T=F=mv²/R=mω²R
ω=sqrt(T/mR)
f=ω/2π
To find the maximum RPM (revolutions per minute) that the ball can have without breaking the cord, we need to consider the tension in the cord as well as the maximum allowable tension.
First, let's find the centripetal force exerted on the ball. The centripetal force is given by the equation:
F = m * v^2 / r
where F is the centripetal force, m is the mass of the ball, v is the velocity of the ball, and r is the radius of the circle.
In this case, since the ball is rotating in a horizontal circle on a frictionless surface, there is no other force acting on the ball except the tension in the cord. So, the tension in the cord is equal to the centripetal force.
We can rearrange the centripetal force equation to solve for the velocity of the ball:
v = sqrt(F * r / m)
Now, let's substitute the given values into the equation:
m = 0.4 kg (mass of the ball)
r = 1.3 m (radius of the circle)
F = 65 N (maximum allowable tension)
v = sqrt(65 * 1.3 / 0.4)
v = sqrt(84.5)
v ≈ 9.2 m/s
Since RPM is a measure of the number of complete revolutions made per minute, we need to convert the velocity to the circumference of the circle traveled per minute.
The circumference of the circle is given by:
C = 2 * π * r
Converting the RPM to radians per second, we have:
1 revolution = 2 * π radians
1 minute = 60 seconds
Therefore, the maximum RPM can be calculated as:
RPM = v / (2 * π * r) * (1 minute / 60 seconds) * (1 revolution / 2 * π radians)
RPM = v / (2 * r) * (1 minute / 60 seconds)
RPM = v / (2 * r / 60)
Substituting the given values:
RPM = 9.2 / (2 * 1.3 / 60)
RPM = 9.2 / (2 * 1.3 / 60)
RPM = 9.2 * 60 / (2 * 1.3)
RPM ≈ 210
Therefore, the maximum RPM the ball can have without breaking the cord is approximately 210.