C) A steel ball bearing the radius 1 cm is rolling along a table with speed 20 cm/sec when it starts to roll up an incline. How high above the table level will it rise before stopping? Ignore friction losses

The kinetic energy of a rolling solid sphere is

(1/2)MV^2 + (1/2)I w^2
= (1/2)M*V^2 + (1/2)(2/5)M(R*w)^2
= (7/10)M*V^2

Set that equal to M*g*H and solve for the height that it rises, H.

H = (0.7) V^2/g

The radius doesn't matter. Neither does the material.

It will only rise about 1.1 cm

20 cm/s is quite slow

it actually equals 0.28 cm I have no clue how you got 1.1 cm

To find the height the steel ball bearing will rise before stopping, we can use the conservation of mechanical energy.

The mechanical energy of a rolling object is the sum of its kinetic energy (KE) and potential energy (PE).
At the start, when the ball is rolling along the table, it only has kinetic energy, given by:

KE = (1/2) * mass * velocity^2

Now, let's determine the mass of the steel ball bearing using its radius:

Mass = density * volume
Density of steel = 7.8 g/cm^3
Volume of the ball = (4/3) * π * radius^3

Substituting the values, we get:

Mass = 7.8 g/cm^3 * (4/3) * π * (1 cm)^3

Once we have the mass, we can calculate the initial kinetic energy:

KE = (1/2) * mass * velocity^2

Next, as the ball rolls up the incline, its kinetic energy is gradually converted into potential energy. At the highest point, when the ball stops, all its initial kinetic energy is converted into potential energy. So we have:

PE = mass * g * height

Setting the initial kinetic energy equal to the potential energy at the highest point, we can solve for the height:

(1/2) * mass * velocity^2 = mass * g * height

Since mass is common on both sides, we can cancel it out:

(1/2) * velocity^2 = g * height

Now, rearrange the formula to solve for height:

height = (1/2) * velocity^2 / g

Plugging in the given values:

height = (1/2) * (20 cm/sec)^2 / (980 cm/sec^2)

Simplifying:

height = (1/2) * 400 cm^2/sec^2 / 980 cm/sec^2

height = 0.204 cm

Therefore, the steel ball bearing will rise approximately 0.204 cm above the table level before stopping.