A (10.25) kg bowling ball is hung on a (11.50) m long rope. It is then pulled back until the rope makes an angle of (39.0)o with the vertical and released. Find the tension in the rope when the ball is at the lowest point. Give your answer in N and with 3 significant figures.
The height h thru which it will fall is
H=11.50(1-cos39)
1/2mv^2=mgh
v=sqrt(2gh)
exclamations points are to me a bit exaggerated.
To find the tension in the rope when the ball is at the lowest point, we need to use the principles of gravitational force and the centripetal force.
First, let's analyze the forces acting on the ball at the lowest point:
1. Gravity: The weight of the bowling ball is given by the equation W = m * g, where m is the mass of the ball (10.25 kg) and g is the acceleration due to gravity (9.8 m/s^2). Therefore, the weight of the ball is W = 10.25 kg * 9.8 m/s^2 = 100.45 N, or approximately 100 N (to 3 significant figures).
2. Tension: The tension in the rope is directed towards the center of the circular motion. At the lowest point, the tension provides the centripetal force to keep the ball moving in a circular path.
The centripetal force can be calculated using the equation Fc = m * (v^2 / r), where Fc 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 circular path.
Since the ball is at the lowest point, the velocity is at its maximum, and the tension in the rope is equal to the sum of the gravitational force and the centripetal force.
To determine the velocity, we can use the conservation of mechanical energy. At the highest point, the potential energy (m * g * h) is converted into kinetic energy (0.5 * m * v^2), where h is the height of the ball. From the given information, the height of the ball is equal to the length of the rope, so h = 11.50 m.
Using the conservation of mechanical energy equation, we can solve for the velocity:
m * g * h = 0.5 * m * v^2
Substituting the given values:
10.25 kg * 9.8 m/s^2 * 11.50 m = 0.5 * 10.25 kg * v^2
Solving for v^2, we get:
v^2 = (10.25 kg * 9.8 m/s^2 * 11.50 m * 2) / 10.25 kg
v^2 = 225.4 m^2/s^2
Taking the square root of both sides, we find:
v ≈ 15 m/s
Now that we have the velocity, we can calculate the tension in the rope at the lowest point:
Tension = Centripetal force + Weight
Tension = m * (v^2 / r) + m * g
Substituting the given values:
Tension = 10.25 kg * (15 m/s)^2 / 11.50 m + 10.25 kg * 9.8 m/s^2
Calculating the equation, we get:
Tension ≈ 170.07 N
Therefore, the tension in the rope when the ball is at the lowest point is approximately 170 N (to 3 significant figures).