In an auditorium, a physics teacher uses a pendulum made by hanging a bowling ball from a wire to the ceiling. At the low point in its swing, a ball with a mass of 7.2-kg has a speed of 6 m/s. Ignoring air resistance, how high will the bob swing above the low point before reversing direction?
m g h = 1/2 m v^2 ... h = v^2 / (2 g)
It's actually so simple. I am waaaaaaaaaay over thinking these problems
thanks!
6^2 /2(9.80)= 1.8m
To find the height to which the bob will swing above the low point, we can use the principle of conservation of mechanical energy. When the bob is at its lowest point, it possesses only kinetic energy. As it swings upwards, this kinetic energy is converted into potential energy.
The conservation of mechanical energy states that the sum of the kinetic energy (K) and potential energy (U) remains constant throughout the motion:
K + U = constant
Initially, at the lowest point, all the energy is in the form of kinetic energy. Kinetic energy is given by the equation:
K = 0.5 * mass * velocity^2
Given that the mass of the ball is 7.2 kg and the velocity at the lowest point is 6 m/s, we can calculate the initial kinetic energy:
K = 0.5 * 7.2 kg * (6 m/s)^2
= 0.5 * 7.2 kg * 36 m^2/s^2
= 129.6 J
At the highest point, when the ball momentarily stops before reversing its direction, all the energy is in the form of potential energy. The potential energy is given by the equation:
U = mass * gravitational acceleration * height
The gravitational acceleration on Earth is approximately 9.8 m/s^2. Let's assume the height to which the bob swings above the low point is h. Substituting the values into the equation and solving for h, we get:
U = 7.2 kg * 9.8 m/s^2 * h
129.6 J = 70.56 kg m^2/s^2 * h
h = 129.6 J / (70.56 kg m^2/s^2)
h ≈ 1.84 m
Therefore, the bob will swing about 1.84 meters above the low point before reversing direction.