A hollow sphere starts from rest and rolls down a hill without sliding. At the bottom of the hill, it has a linear velocity of 5 m/s. What was the height of the hill the sphere rolled down (in meters)?
PE lost=KE gained
mgh=1/2 mv^2 + 2/3 M r^2*w^2
where w=v/r
gh=1/2 v^2+2/3 v^2
solve for h.
Bobpursley is a genius
Yep!
To find the height of the hill the sphere rolled down, we can use the principle of conservation of energy.
First, let's define the initial and final positions of the sphere on the hill. The initial position is at the top of the hill, and the final position is at the bottom of the hill.
The total mechanical energy of the sphere, E, is the sum of its kinetic energy and potential energy:
E = KE + PE
At the top of the hill, the sphere is at rest, so its initial kinetic energy (KE) is zero. The only form of energy it possesses is potential energy (PE), given by:
PE = mgh
where m is the mass of the sphere, g is the acceleration due to gravity, and h is the vertical height of the hill.
At the bottom of the hill, the sphere has a linear velocity of 5 m/s. Its final kinetic energy is given by:
KE = (1/2)mv^2
where v is the linear velocity of the sphere.
Using the principle of conservation of energy, we equate the initial potential energy to the final kinetic energy:
mgh = (1/2)mv^2
Simplifying and canceling out the mass factor on both sides:
gh = (1/2)v^2
Now we can solve for h, the height of the hill:
h = (1/2)v^2 / g
Plugging in the given values, with v = 5 m/s and g = 9.8 m/s^2, we can calculate the height of the hill:
h = (1/2)(5^2) / 9.8
h = (1/2)(25) / 9.8
h = 12.76 m
Therefore, the height of the hill the sphere rolled down is approximately 12.76 meters.