A 0.80-kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.7 m on a frictionless horizontal surface.If the cord will break when the tension in it exceeds 90 N, what is the maximum speed the ball can have?
Express your answer using two significant figures.
Well, it seems like the ball is living life on the edge, hanging by a thread! Let's find out its maximum speed before things get messy.
First, we'll calculate the centripetal force acting on the ball. The centripetal force is given by the equation F = m * v^2 / r, where F is the force, m is the mass, v is the speed, and r is the radius.
Rearranging the equation to solve for v, we have v = √(Fr / m), where √ represents the square root.
Plugging in the given values, F = 90 N, r = 1.7 m, and m = 0.80 kg, we have v = √(90 N * 1.7 m / 0.80 kg).
Calculating that out, we find v ≈ 6.0 m/s.
So, the ball can have a maximum speed of around 6.0 m/s before the cord reaches its breaking point. I hope that brings a bit of "circus" level excitement to your physics problem!
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 reaches its maximum speed.
The centripetal force acting on the ball is provided by the tension in the cord. The centripetal force can be calculated using the following formula:
F = (m * v^2) / r
Where:
F is the centripetal force,
m is the mass of the ball (0.80 kg),
v is the velocity of the ball (maximum speed),
and r is the radius of the circular path (1.7 m).
We know that the tension in the cord must not exceed 90 N. Therefore, we can set the centripetal force equal to the maximum tension:
90 N = (0.80 kg * v^2) / 1.7 m
To find the maximum speed, we can rearrange the equation:
v^2 = (90 N * 1.7 m) / 0.80 kg
v^2 = 191.25 Nm / kg
v = sqrt(191.25 Nm / kg)
Calculating this using a calculator gives us:
v ≈ 13 m/s
Therefore, the maximum speed the ball can have without breaking the cord is approximately 13 m/s.
To determine the maximum speed the ball can have without breaking the cord, we need to consider the centripetal force acting on the ball.
The centripetal force required to keep an object moving in a circle is given by the equation:
F = m * v^2 / r
Where:
F is the centripetal force
m is the mass of the object
v is the velocity of the object
r is the radius of the circular path
In this case, the tension in the cord provides the centripetal force to keep the ball in circular motion.
If the tension in the cord exceeds 90 N, the cord will break. Therefore, we can set up the equation:
Tension in the cord ≤ 90 N
Substituting the centripetal force equation, we have:
m * v^2 / r ≤ 90 N
Plugging in the given values:
m = 0.80 kg (mass of the ball)
r = 1.7 m (radius of the circular path)
Tension in the cord = 90 N
0.80 kg * v^2 / 1.7 m ≤ 90 N
Simplifying the equation, we can solve for v:
v^2 ≤ (90 N * 1.7 m) / 0.80 kg
v^2 ≤ 191.25 N*m / kg
v^2 ≤ 191.25 m^2 / s^2
Taking the square root of both sides:
v ≤ sqrt(191.25) m/s
Calculating the square root:
v ≤ 13.82 m/s
Therefore, the maximum speed the ball can have without breaking the cord is 13.82 m/s (rounded to two significant figures).
M V^2/R = Tmax = 90 N
Solve for V
M = 0.80 kg
R = 1.7 m