A bowling ball is rolled up an incline. If the ball has a speed of 12 m/s at the base of the incline, to what height h on the incline will the center mass travel?
mgh=1/2 mv^2
hg=1/2 v^2
h= v^2/2g
I tried that and did not get the right answer.
I think I need to use KE_i + PE_i = KE_f + PE_f ... which would be:
.5mv_i^2 + 1/2Iw_i^2 + mgh_i = 1/2mv_f^2 + .5Iw_f^2 +mgh_f
--> I= 2/5mr^2 for a solid sphere
I factored out the mass on both sides but my problem is that I need to convert it to an angular speed using w=v/r ... BUT I was not given a radius
I also don't have a mass for when I plug I into the equation above
To determine the height h on the incline that the center of mass of the bowling ball will travel, we can use the principle of conservation of energy.
The initial kinetic energy (KE) of the ball at the base of the incline is given as:
KE initial = 1/2 * mass * velocity^2
As the ball rolls up the incline, its mechanical energy is conserved, meaning the sum of its potential energy (PE) and its kinetic energy remains constant.
At the highest point on the incline, the ball comes to rest, so its final velocity (v) is zero. Therefore, its final kinetic energy is zero:
KE final = 0
The change in potential energy (ΔPE) of the ball as it moves up the incline is given by:
ΔPE = mass * gravity * height
Using the conservation of energy, we can set the initial kinetic energy equal to the change in potential energy at the highest point:
KE initial = ΔPE
1/2 * mass * velocity^2 = mass * gravity * height
Canceling out the mass from both sides of the equation, we get:
1/2 * velocity^2 = gravity * height
To solve for height (h), rearrange the equation:
height = (1/2 * velocity^2) / gravity
Now, substitute the given values into the equation:
velocity = 12 m/s (given)
gravity ≈ 9.8 m/s^2 (acceleration due to gravity)
height = (1/2 * 12^2) / 9.8
height ≈ 7.35 meters
Therefore, the center of mass of the bowling ball will travel to a height of approximately 7.35 meters on the incline.