A ball is moving to and fro about the lowest point of a smooth hemispherical bowl. If it is able to rise up to a height of 45cm on either side.it's speed at the lowest point must be?
gain in potential energy=loss in kinetic energy
so 1/2mv^2=mgh
v^2=gh
v=underroot of 2gh
=underroot of 2*10*.2
=2m/s
1/2 m v^2 = m g h
v = √(2 g h) = √(2 * 9.8 * .45) m/s
humm, is it sliding or rolling?
If rolling then some of that potential energy goes into kinetic at the bottom
Ke = (1/2) m v^2 + (1/2) I omega^2
I = (2/5)mr^2 for solid ball
so
Ke = (1/2)mv^2 + (1/2)(2/5)mr^2 omega^2
but v = r omega
so
Ke = (7/10)mv^2
so
9.8*.45 = .7 v^2
by the way if a thin walled sphere, then I = (2/3)mr^2
To find the speed of the ball at the lowest point, we can apply the conservation of energy principle, taking into account the potential and kinetic energy of the ball.
Let's assume that the lowest point of the bowl is at a height of zero, and the maximum height the ball reaches on either side is 45 cm (0.45 m).
At the highest point of its trajectory, when the ball is at a height of 0.45 m, its potential energy is at its maximum, and its kinetic energy is at its minimum (zero). At this point, all the energy is converted into potential energy, given by the equation:
Potential Energy = m * g * h
Where:
m = mass of the ball
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height of the ball above the lowest point (0.45 m)
Let's assume the mass of the ball is 'm'.
Potential Energy = m * 9.8 * 0.45
At the lowest point, all the potential energy is converted into kinetic energy, given by:
Kinetic Energy = (1/2) * m * v^2
Where:
v = velocity of the ball at the lowest point
Equating the potential energy to the kinetic energy, we have:
m * 9.8 * 0.45 = (1/2) * m * v^2
Simplifying the equation, we get:
9.8 * 0.45 = (1/2) * v^2
v^2 = (9.8 * 0.45) / 0.5
v^2 = 9.8 * 0.45 * 2 / 0.5
v^2 = 8.82
Taking the square root of both sides, we find:
v = √(8.82)
v ≈ 2.97 m/s
Therefore, the speed of the ball at the lowest point is approximately 2.97 m/s.