A small lead ball, attached to a 1.50-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled at a constant rate of three revolutions per second and is released on the upward part of the circular motion when it is 0.75 m above the ground. The ball travels straight upward. In the absence of air resistance, to what maximum height above the ground does the ball rise?
Well, let me put on my magical circus hat and juggle some numbers for you. We know that the ball is being whirled in a circular motion, and it's released on the upward part. So, we need to find the maximum height above the ground that the ball reaches.
To start, we can find the time it takes for the ball to complete one revolution. Since it's whirling at a constant rate of three revolutions per second, each revolution will take 1/3 second.
Next, we can find the velocity of the ball when it's released. We know that the ball is traveling in a circle, so it has a centripetal acceleration. Using the formula for centripetal acceleration, we have a = (v^2) / r, where v is the velocity and r is the radius. Since there's no air resistance, the centripetal acceleration is equal to the acceleration due to gravity, which is 9.8 m/s^2.
We can rearrange the formula to find the velocity: v = sqrt(a * r). Plugging in the values, we have v = sqrt(9.8 m/s^2 * 0.75 m) ≈ 4.189 m/s.
Now, we can find the maximum height above the ground that the ball reaches. Since the ball is traveling vertically, we can use the formula for maximum height reached in projectile motion: h = (v^2) / (2g), where h is the maximum height, v is the initial velocity, and g is the acceleration due to gravity.
Plugging in the values, we have h = (4.189 m/s)^2 / (2 * 9.8 m/s^2) ≈ 0.901 meters.
So, the ball reaches a maximum height of approximately 0.901 meters above the ground. Enjoy the circus of physics!
To find the maximum height above the ground that the ball rises, we can analyze the motion of the ball using the principles of circular motion and energy conservation.
Step 1: Determine the initial speed of the ball
The ball is whirled with a constant rate of three revolutions per second, which corresponds to an angular velocity of 3(2π) radians per second. Since the rope is 1.50 m long, the initial speed of the ball can be calculated using the formula:
v = ω * r
where v is the tangential velocity, ω is the angular velocity, and r is the radius of the circular path. In this case, r is the length of the rope, which is 1.50 m. Substituting the values, we can find the initial speed:
v = (3(2π)) * 1.50 m/s = 9π m/s
Step 2: Determine the initial kinetic energy of the ball
The initial kinetic energy of the ball is given by the formula:
K = (1/2) * m * v^2
where K is the kinetic energy, m is the mass of the ball (which we'll assume is constant and cancel out in later steps), and v is the initial speed. Since we don't have the mass of the ball, we can ignore it for now and calculate the kinetic energy:
K = (1/2) * (9π)^2 J ≈ 127.23 J
Step 3: Determine the maximum potential energy of the ball
At its maximum height, all of the initial kinetic energy is transformed into potential energy according to the conservation of energy principle. The potential energy can be calculated using the formula:
PE = m * g * h
where PE is the potential energy, m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height above the ground.
Since the mass of the ball cancels out when calculating the change in potential energy, we can write:
ΔPE = K
Substituting the calculated kinetic energy, we have:
m * g * h = 127.23 J
Step 4: Determine the maximum height
Solving the equation for h, we get:
h = K / (m * g) ≈ 127.23 J / (m * 9.8 m/s^2)
Since we don't have the mass of the ball, we cannot determine the exact maximum height. However, we can look at the ratio between the potential energy and the mass times gravity to observe that the maximum height is proportional to the initial kinetic energy.
In this case, the maximum height above the ground that the ball rises would be approximately 12.97 meters, assuming the mass of the ball is non-zero.
To find the maximum height above the ground that the ball rises to, we can analyze the energy of the system.
Let's break it down step by step:
Step 1: Calculate the initial velocity
The initial velocity of the ball can be calculated from the given information. Since the ball is whirled at a constant rate of three revolutions per second, and each revolution corresponds to the circumference of the circle, we can use the formula V = 2πr/T, where V is the velocity, r is the radius of the circular motion, and T is the time for one revolution.
In this case, T is 1/3 seconds per revolution, and the radius is 1.5 m (given). So the initial velocity of the ball is V = 2π(1.5 m)/(1/3 s) = 9π m/s.
Step 2: Calculate the initial kinetic energy
The initial kinetic energy (KE) of the ball is given by the formula KE = (1/2)mv^2, where m is the mass and v is the velocity.
Since the mass of the ball is not mentioned, we can assume that it is small enough to consider negligible.
So the initial kinetic energy is KE = (1/2)(0)m(9π m/s)^2 = 0 J.
Step 3: Calculate the potential energy at maximum height
At maximum height, the ball has no kinetic energy because its velocity is zero. Therefore, all the initial kinetic energy is converted into potential energy (PE).
The potential energy is given by the formula PE = mgh, where m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the height.
Since the mass is negligible, the potential energy at maximum height is PE = (0)(9.8 m/s^2)(h) = 0 J.
Step 4: Calculate the maximum height
We can set the initial kinetic energy equal to the potential energy at maximum height to find the height (h).
(1/2)(0)(9π m/s)^2 = (0)(9.8 m/s^2)(h)
Simplifying, we get:
0 = 0
Since the equation is satisfied for any value of h, the ball can theoretically rise to any height. However, in practical terms, the ball reaches a maximum height of approximately 0.75 m above the ground (as given in the question) before falling back down due to the force of gravity.
Therefore, in this case, the maximum height above the ground that the ball rises to is approximately 0.75 m.