A 5.6 kg ball (I=2/3mr^2) starts on a hill 11.3 m above level surface of the ground. The ball then rolls (without slipping) down the hill. What is the linear speed of the ball when it reaches the bottom of the hill?
To find the linear speed of the ball when it reaches the bottom of the hill, we can use the principle of conservation of energy.
The potential energy of the ball at the top of the hill is given by the formula PE = mgh, where m is the mass of the ball, g is the gravitational acceleration, and h is the height of the hill. In this case, the height of the hill is 11.3 m.
The kinetic energy of the ball at the bottom of the hill is given by the formula KE = (1/2)mv^2, where m is the mass of the ball and v is its linear speed.
According to the conservation of energy, the total mechanical energy of the ball remains constant, so the potential energy at the top of the hill is converted into kinetic energy at the bottom of the hill.
Therefore, we can equate the potential energy at the top of the hill to the kinetic energy at the bottom of the hill:
mgh = (1/2)mv^2
Simplifying the equation by canceling out the mass:
gh = (1/2)v^2
Now we can solve for v:
v^2 = 2gh
v = √(2gh)
Substituting the given values:
v = √(2 × 9.8 m/s^2 × 11.3 m)
v = √(216.04)
v ≈ 14.7 m/s
Therefore, the linear speed of the ball when it reaches the bottom of the hill is approximately 14.7 m/s.