A 0.55- ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.5 on a frictionless horizontal surface.If the cord will break when the tension in it exceeds 95 , what is the maximum speed the ball can have?
T= m(v^2/r)
95 N = 0.55(v^2/1.5)
95 N = (0.55v^2)/1.5
(95 N)(1.5) = 0.55v^2
142.5 = 0.55v^2
divide both sides by 0.55
answer is : 259.09 = v^2
take the square root of 259.09
final answer is : v = 16.1 m/s = the maximum speed the ball can have
Well, that's a delicate situation! I guess you could say this ball is walking a tightrope, or should I say a tightcord?
To find the maximum speed of the ball, we need to find the point where the tension in the cord reaches its breaking point. In this case, the tension is maximum when the centripetal force is at its maximum.
The centripetal force acting on the ball is given by the equation F = mv²/r, where F is the force, m is the mass, v is the velocity, and r is the radius of the circle. In this case, the force is equal to the tension in the cord.
To solve for the maximum speed, we can rearrange the equation to v² = Fr/m. Plugging in the values you provided, F = 95 N, r = 1.5 m, and m = 0.55 kg, we can solve for v.
v² = (95 N)(1.5 m)/(0.55 kg)
v² = 258.18 N·m/kg
Taking the square root of both sides, we find:
v ≈ 16.06 m/s
So the maximum speed the ball can have without the cord breaking is approximately 16.06 m/s. Just remember, this is no time for the ball to be clowning around!
To find the maximum speed the ball can have without breaking the cord, we need to consider the tension in the cord when the ball is at its maximum speed.
The tension in the cord can be calculated using the centripetal force formula:
Tension = Mass * Centripetal Acceleration
The centripetal acceleration is given by the formula:
Centripetal Acceleration = (Velocity^2) / Radius
Simplifying the formulas, we have:
Tension = (Mass * Velocity^2) / Radius
In this case, the tension should not exceed 95 N. We are given the mass of the ball, 0.55 kg, and the radius of the circular path, 1.5 m.
Substituting these values into the formula, we can solve for the maximum velocity (speed) of the ball:
95 = (0.55 * Velocity^2) / 1.5
Multiply both sides of the equation by 1.5 to isolate the velocity:
142.5 = 0.55 * Velocity^2
Divide both sides of the equation by 0.55:
Velocity^2 = 142.5 / 0.55
Velocity^2 = 259.09
Taking the square root of both sides, we find:
Velocity = √(259.09)
Velocity ≈ 16.10 m/s
Therefore, the maximum speed the ball can have without breaking the cord is approximately 16.10 m/s.
To determine the maximum speed the ball can have without breaking the cord, we can use the centripetal force equation.
The centripetal force (F_c) is the force that keeps an object moving in a circular path. It can be calculated by multiplying the mass (m) of the object by the square of its velocity (v), divided by the radius (r) of the circular path.
In this case, the tension in the cord provides the centripetal force, so we equate it to the centripetal force:
Tension = Centripetal force
T = m * v^2 / r
We need to rearrange the equation to solve for the maximum speed (v):
v^2 = (T * r) / m
Now, we can plug in the given values:
T = 95 N (maximum tension before the cord breaks)
r = 1.5 m (radius of the circular path)
m = 0.55 kg (mass of the ball)
v^2 = (95 * 1.5) / 0.55
v^2 = 259.09
Taking the square root of both sides, we find:
v ≈ 16.1 m/s
Therefore, the maximum speed the ball can have without breaking the cord is approximately 16.1 m/s.