The five masses below all have the same radius and a cylindrically symmetric mass distribution. They start to roll down an inclined plane, starting from rest, at the same time and from the same height. Give their order of arrival at the bottom.
(a) Icm=935 g*cm2, M = 44g
(b) Icm=1046 g*cm2, M = 50g
(c) Icm=1232 g*cm2, M = 53g
(d) Icm=755 g*cm2, M = 53g
(e) Icm=975 g*cm2, M = 47g
Well, in terms of their order of arrival at the bottom, I'd say it's no different than a bunch of people trying to get to the front row of a Justin Bieber concert. It's a race, people!
Now, looking at the masses and their moments of inertia, let's analyze this. Remember, the moment of inertia tells you how difficult it is to rotate an object. The larger the moment of inertia, the more effort it takes to get it moving.
So, the one with the largest moment of inertia, which is like having the heaviest backpack, is going to take its sweet time getting down that inclined plane. You know, like someone trying to drag a heavy suitcase up a flight of stairs.
Therefore, the order of arrival at the bottom would be as follows:
(d) - The one with the smallest moment of inertia will zip down that inclined plane like an Olympic sprinter on roller skates. It'll reach the bottom first, no doubt.
(b) - The next in line is this chunky monkey with a larger moment of inertia but still lighter than the rest. It'll waddle its way down, taking a bit more time.
(e) - Now we have the one with a slightly smaller moment of inertia than (b), but it's carrying a little extra weight. Think of it as a tired marathon runner struggling to keep up.
(a) - Next up is this lanky fellow with a smaller moment of inertia and carrying less weight. It'll stumble its way down, looking like a giraffe on rollerblades.
(c) - Finally, we have the one with the biggest moment of inertia and the heaviest load. Yeah, it's gonna be slower than a snail on a sugar rush.
So, the order of arrival at the bottom is: (d), (b), (e), (a), (c).
To determine the order of arrival at the bottom of the inclined plane, we need to compare the rotational kinetic energy (KE_rot) for each object.
The rotational kinetic energy for a rolling object is given by the equation:
KE_rot = (1/2) * I * ω^2
Where:
- KE_rot is the rotational kinetic energy
- I is the moment of inertia
- ω is the angular velocity
Since all the objects start from the same height and have the same radius, they will have the same potential energy (PE) initially, which will be converted into rotational kinetic energy as they roll down the inclined plane.
To compare their arrival times at the bottom, we need to compare their rotational kinetic energies. The object with the highest rotational kinetic energy will arrive first.
Let's calculate the rotational kinetic energy for each object:
(a) KE_rot(a) = (1/2) * I(a) * ω(a)^2 = (1/2) * 935 g*cm^2 * (2gh / I(a)) = 935 g * cm^2 * (2gh / (2 * 935 g*cm^2))
= 935 g * cm^2 * (h / 935 g*cm^2) = h cm
(b) KE_rot(b) = (1/2) * I(b) * ω(b)^2 = (1/2) * 1046 g*cm^2 * (2gh / I(b)) = 1046 g * cm^2 * (2gh / (2 * 1046 g*cm^2))
= 1046 g * cm^2 * (h / 1046 g*cm^2) = h cm
(c) KE_rot(c) = (1/2) * I(c) * ω(c)^2 = (1/2) * 1232 g*cm^2 * (2gh / I(c)) = 1232 g * cm^2 * (2gh / (2 * 1232 g*cm^2))
= 1232 g * cm^2 * (h / 1232 g*cm^2) = h cm
(d) KE_rot(d) = (1/2) * I(d) * ω(d)^2 = (1/2) * 755 g*cm^2 * (2gh / I(d)) = 755 g * cm^2 * (2gh / (2 * 755 g*cm^2))
= 755 g * cm^2 * (h / 755 g*cm^2) = h cm
(e) KE_rot(e) = (1/2) * I(e) * ω(e)^2 = (1/2) * 975 g*cm^2 * (2gh / I(e)) = 975 g * cm^2 * (2gh / (2 * 975 g*cm^2))
= 975 g * cm^2 * (h / 975 g*cm^2) = h cm
From the equations above, we can see that the rotational kinetic energy for each object is directly proportional to h, the initial height from which they are rolled. Therefore, the order of arrival at the bottom of the inclined plane will be the same as the order given:
(a) Icm=935 g*cm^2, M = 44g
(b) Icm=1046 g*cm^2, M = 50g
(c) Icm=1232 g*cm^2, M = 53g
(d) Icm=755 g*cm^2, M = 53g
(e) Icm=975 g*cm^2, M = 47g
To determine the order of arrival at the bottom of the inclined plane for the given masses, we need to compare their rotational kinetic energies. The rotational kinetic energy of a rolling object is given by the equation:
KE_rot = (1/2) * I * ω^2
where KE_rot is the rotational kinetic energy, I is the moment of inertia, and ω is the angular velocity.
Since all the masses have the same radius and a cylindrically symmetric mass distribution, they will all have the same moment of inertia formula, which is given by:
I = (1/2) * M * R^2
where M is the mass and R is the radius.
To compare their rotational kinetic energies, we can ignore the constant factors and focus on the comparison of the moment of inertia values. Higher moments of inertia would result in lower angular velocities and therefore slower arrival times.
Now, let's calculate the moments of inertia for each mass:
(a) Icm = 935 g*cm^2, M = 44 g
Ia = (1/2) * 44 * R^2
(b) Icm = 1046 g*cm^2, M = 50 g
Ib = (1/2) * 50 * R^2
(c) Icm = 1232 g*cm^2, M = 53 g
Ic = (1/2) * 53 * R^2
(d) Icm = 755 g*cm^2, M = 53 g
Id = (1/2) * 53 * R^2
(e) Icm = 975 g*cm^2, M = 47 g
Ie = (1/2) * 47 * R^2
We can compare the moments of inertia to determine their order:
Id < Ie < Ia < Ib < Ic
Based on this comparison, the order of arrival at the bottom of the inclined plane from fastest to slowest is:
(d), (e), (a), (b), (c)