A bead slides without friction around a loopthe-loop. The bead is released from a height
of 23.3 m from the bottom of the loop-theloop which has a radius 8 m.
The acceleration of gravity is 9.8 m/ What is its speed at point A answer in m/s^2.
Point A is at the top of the loop-the-loop
To find the speed of the bead at point A, we need to consider the conservation of mechanical energy. At the top of the loop-the-loop, all the gravitational potential energy is converted into kinetic energy.
The initial potential energy of the bead at the starting point (23.3 m from the bottom) can be calculated as:
Potential Energy = mass * gravitational acceleration * height
PE_initial = m * g * h
Given that the acceleration due to gravity (g) is 9.8 m/s^2 and the height (h) is 23.3 m, we can calculate the initial potential energy.
Now, the potential energy is converted into kinetic energy at the top of the loop. The kinetic energy of an object can be expressed as:
Kinetic Energy = 0.5 * mass * velocity^2
KE_final = 0.5 * m * v^2
Since energy is conserved, we can equate the initial potential energy to the final kinetic energy:
PE_initial = KE_final
m * g * h = 0.5 * m * v^2
Simplifying the equation by canceling out the mass (m) on both sides:
g * h = 0.5 * v^2
Rearranging the equation to find the velocity (v):
v^2 = 2 * g * h
Finally, we take the square root of both sides to find the speed at point A:
v = √(2 * g * h)
Plug in the values for g (9.8 m/s^2) and h (23.3 m) into this equation and calculate the speed.