A solid ball of mass 2 kg, rolls down a hill that is 6 meters high. What is the rotational KE at the bottom of the hill?
To find the rotational kinetic energy (KE) of the solid ball at the bottom of the hill, we need to use the formula:
Rotational KE = (1/2) * I * ω^2
where I represents the moment of inertia and ω is the angular velocity.
First, we can calculate the potential energy (PE) of the ball at the top of the hill using the formula:
Potential Energy = m * g * h
where m is the mass of the ball (2 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the height of the hill (6 meters). Let's compute it:
Potential Energy = 2 kg * 9.8 m/s² * 6 m
= 117.6 Joules
Next, we can equate the potential energy to the sum of the translational kinetic energy and the rotational kinetic energy at the bottom of the hill:
Total Energy at bottom = Translational KE + Rotational KE
Since the ball is rolling down without slipping, we can state that the sum of the translational kinetic energy and the rotational kinetic energy is equal to the total mechanical energy (TE) of the ball, which remains constant throughout the motion.
Thus, we can express TE as:
Total Energy = Translational KE + Rotational KE = Constant (TE)
Since the ball is initially at rest, its translational kinetic energy at the top of the hill is zero.
Therefore, we can rewrite the equation as:
TE = Rotational KE
Hence, the rotational kinetic energy at the bottom of the hill is equal to the total mechanical energy of the system.
So, the rotational KE at the bottom of the hill is 117.6 Joules.
This is a little complicated, it is rolling, and translating .
total energy=mgh
let translational velocity be v.
then w=v/r
translational KE=1/2 m v^2=1/2 m w^2r^2
rotational KE= 1/2 I w^2=1/2*2/5 mr^2*w^2
= 1/5*mv^2
but total energy=sum of above
mgh=(1/5+1/2)mv^2=0.7 mv^2
solving for v
v^2=gh/.7
and then rotational KE= .2*m*v^2
check all that.