Sorry for stopping the flow, but I wanted to point out that that answer is still wrong.AlfWeg wrote: ↑October 16th, 2019, 11:50 amOh lol I forgot to say it was rolling. Ignore my stupidityshrewdPanther46 wrote: ↑October 16th, 2019, 10:07 amUmaroth's solution is correct, the sphere has no reason to rollAlfWeg wrote: I think you missed that it's a sphere and it's rolling, so we have to include rotational kinetic energy in our energy equation. Taking that into Consideration, We get [KE (translational) + KE (Rotational) = PE] The formula for rotational KE is (1/2 * Rotational Inertia * Omega^2 ). A solid sphere has a rpt. Inertia of (2/5) * Mass * (Radius^2) and Omega = V/R. Thus, we get the equation
(1/2) * M * V^2 + (1/2) * (2/5) *M * R^2 = M* G*H. Rearranging we get, v = sqrt((10/7)gh), giving us 8.5 m/s^2. This Presentation might help to understand this. Your turn!
When dealing with a ramp problem with a rolling sphere or cylinder, its easier to use the Parallel Axis Theorem. The reason is that if you set your pivot point as the contact point between the ball and the ramp, we can say the torque is only caused by the force of gravity on the sphere. Following this, we get a force diagram like this: https://i.imgur.com/vdR6dHH.png
When calculating moment of inertia, we use the Parallel Axis Theorem which states I_Total = md^2 + I_CoM
Where d is the distance between the pivot point and the center of mass and I_CoM is the moment of inertia of the object around its center of mass.
In the case of this problem I_CoM = 2/5 mr^2 and d = r. We plug in those values and we get I_Total = 7/5 mr^2.
We then continue the problem the same way as before.
mgh = 1/2 mv^2 + 1/2 I_Total*(v/r)^2
mgh = 1/2 mv^2 + 1/2 (7/5mr^2)(v/r)^2
gh = 1/2 v^2 + 7/10v^2
gh = 6/5 v^2
When you plug in the values, v = 6.45m/s. With sig figs, it rounds to 6m/s.
Please carry on with the question above.