A ball, of mass 0.13Kg, is held on the inside of a smooth hemispherical bowl at a height 0.25 m above the base. It is then released. Assuming there is no friction, the acceleration due to gravity, g, is 10 m/s2 and the potential energy is zero at the base of the bowl - What is the speed of the ball at a height of 0.1 m above the base?
Use conservation of energy. For the correct answer, you need to include the rotational kinetic energy of the ball.
If the bowl is so smooth that the ball slides and does not roll, you can neglect rotational kinetic energy.
Classes that ask you to assume g has the incorrect value of 10 m/s^2 are doing a disservice to physics education and the students, in my opinion.
To find the speed of the ball at a height of 0.1 m above the base, we can use the principle of conservation of energy. Since there is no friction, the only forces acting on the ball are the gravitational force and the normal force from the bowl.
At a height of 0.25 m above the base, the ball has gravitational potential energy given by:
Potential energy = mass * height * gravity
Potential energy = 0.13 kg * 0.25 m * 10 m/s^2
Potential energy = 0.325 J
Since the potential energy at the base of the bowl is defined as zero, the total mechanical energy (consisting of potential energy and kinetic energy) is conserved.
At a height of 0.1 m above the base, the ball has a potential energy given by:
Potential energy = mass * height * gravity
Potential energy = 0.13 kg * 0.1 m * 10 m/s^2
Potential energy = 0.13 J
Since the total mechanical energy is conserved, we can equate the potential energy at the two heights:
0.325 J = 0.13 J + kinetic energy at 0.1 m
Now, we can solve for the kinetic energy at 0.1 m:
kinetic energy at 0.1 m = 0.325 J - 0.13 J
kinetic energy at 0.1 m = 0.195 J
Finally, we can calculate the speed of the ball at a height of 0.1 m above the base using the formula for kinetic energy:
kinetic energy = (1/2) * mass * velocity^2
0.195 J = (1/2) * 0.13 kg * velocity^2
Simplifying the equation, we get:
0.195 J = 0.065 kg * velocity^2
Dividing both sides by 0.065 kg, we have:
velocity^2 = 0.195 J / 0.065 kg
velocity^2 = 3 m^2/s^2
Taking the square root on both sides, the speed of the ball at a height of 0.1 m above the base is:
velocity = √3 m/s