Torque and Rotational motion?

Question

A cylinder is rolling along the ground at 2m/s. It comes to a hill and starts going up. Assuming no losses to friction, how high does it get before it stops?

Answer 1

It cant climb .., due to its weight

Answer 2

You can't get a numerical answer without knowing it's moment of inertia (actually you just need to know I/MR^2, where I is the moment of inertia, M is the mass, and R is the radius).

This is because the cylinder has both translational and rotational kinetic energy.  When it's rolling on the ground, it has kinetic energy 1/2*Mv^2+1/2*I*omega^2 and no potential energy.  When it stops at the top of the hill, it has potential energy Mgh, and no kinetic energy.

Using the conservation of energy,

1/2*Mv^2+1/2*I*omega^2=Mgh

Divide both sides by Mg to get the height at which it stops:

v^2/(2g)+I*omega^2/(2Mg)=h

You can use the fact that omega=v/R to simplify further:

h=v^2/(2g)+I*v^2/(2gMR^2)

h=v^2/(2g)*[1+I/(MR^2)]

You can calculate the value of v^2/(2g):

h=(2 m/s)^2/(2*9.8 m/s^2)*[1+I/(MR^2)]

h=(0.204 m)*[1+I/(MR^2)]

The only thing you're missing is I/(MR^2).  If the cylinder is uniform, then I=1/2*MR^2, so I/(MR^2)=1/2, giving h=0.306.  If it's a hollow cylindrical shell with open ends, I/(MR^2)=1, giving h=0.408.  If it has all its mass concentrated at its axis, I/(MR^2)=0, giving  h=0.204 m.  It could also be some other distribution of mass giving a different value of I/(MR^2).

[COMMENTS]: The problem says it's rolling, not sliding, which implies that the cylinder does have rotational kinetic energy.  Also, static friction, which responsible for the cylinder rolling, does not cause any energy loss.  Only kinetic friction causes energy loss.   To ignore losses due to friction you ignore kinetic friction, but you don't necessarily ignore static friction.

Answer 3

well, lets see, imagine for a moment it rose vertically, of course it can't, if the "hill" was vertical it would just bounce off when it came to to it, but still, imagine.

In that case it would rise for 2m/s / 9.8m/s^2 =  0.2 seconds, and during that time it's speed will uniformly drop to zero, for an average of 1.0m/s, so the height would be 0.2m, if the rise is not vertical, the time will be longer, but the height will be the same since the vertical component of the velocity will be correspondingly smaller.

[EDIT] I have ignored any effects of rotational inertia, because if we are ignoring losses due to friction, is seems reasonable ignore all friction, and if there is no friction with the hill, the rotation makes no difference. If we were to consider the rotation, things get considerably more complicated, and the height would be higher.

I thought my method is a little easier to understand (at least it was the first method I thought of) than 1/2 mv^2 = mgh. "Bomba" used (W being equal to mg),  but being curious why his answer was different, I worked it out that way too:

1/2 * m * 2^2 = m * 9.8 * h

1/2 * 4 = 9.8 * h

2 = 9.8 * h

2 / 9.8 = h = 0.2

perhaps "Bomba" forgot the 1/2 ?

[2nd EDIT] "but you don't necessarily ignore static friction", true enough, and I did notice the question said rolling, but still, I choose to ignore all the friction including static (and therefore the rotational kinetic energy )

The person asking the question would  know better what might be expected, if rotational kinetic energy has been discussed in class or the text book, then "cfiziksh"'s answer is better, but at a lower grade level a simpler answer might be expected,  I have seen questions  here on Y!A, where the "correct" answer was not the one wanted.

Answer 4

it will countinue its movement because there is no frection losses in its movement by the newton's law that in the absence of any external force a body will countinue its motion and a still boday will be stilled forever