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- Homework Statement
- cylinder rolls down a hill

- Relevant Equations
- ##mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2##

A snow cylinder with an initial radius ##R## rolls without slipping down a tall hill with constant slope. As

it rolls, snow sticks onto the cylinder which makes its radius slowly increase. The amount of gathered

snow is proportional to the distance the snow cylinder has traveled. Consider the following graphs

which represent arbitrary units on the ##y##-axis which will be specified according to the problem. (Note:

For graph III, the arrow lies directly on top of the dotted line).

If the dotted line represents the speed of the cylinder if the snow was not sticky, which of the following

graphs represents the speed of the cylinder if the snow sticks to the cylinder, as described above?

The answer says it's Graph III and gives the following explanation,

ince the object is rolling without slipping, we have the relationship ##v = \omega r##. Applying energy conservation allows us to write that

##mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} mv^2 + \frac{1}{2} \beta m R^2 (\frac{v}{r})^2 ##

##v \propto \sqrt{h}##

From this analysis, it is evident that the changing radius of the cylinder does not actually effect

the velocity it rolls down with.

I have some doubt about the given answer. We know that the CM speed is only dependent on the geometry of the object, let the rotational inertia be ##I = \beta m R^2##, for cylinder ##\beta = \frac{1}{2}##, for a ring it's ##\beta = 1##.

Now we can think of the original cylinder having ##\beta = \frac{1}{2}##, if snow does not stick to the cylinder, the final CM speed is

##v = \sqrt{ \frac{2 H}{g (1 + \beta)}} = \sqrt{ \frac{4 H}{3g }}##

If the snow sticks to the cylinder without slipping, we can consider the snow form rings on the cylinder whose ##\beta = 1##. This should change the overall rotational inertia of the snow object and slows it down.

Let me know how you think about this. Thanks,

it rolls, snow sticks onto the cylinder which makes its radius slowly increase. The amount of gathered

snow is proportional to the distance the snow cylinder has traveled. Consider the following graphs

which represent arbitrary units on the ##y##-axis which will be specified according to the problem. (Note:

For graph III, the arrow lies directly on top of the dotted line).

If the dotted line represents the speed of the cylinder if the snow was not sticky, which of the following

graphs represents the speed of the cylinder if the snow sticks to the cylinder, as described above?

The answer says it's Graph III and gives the following explanation,

ince the object is rolling without slipping, we have the relationship ##v = \omega r##. Applying energy conservation allows us to write that

##mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} mv^2 + \frac{1}{2} \beta m R^2 (\frac{v}{r})^2 ##

##v \propto \sqrt{h}##

From this analysis, it is evident that the changing radius of the cylinder does not actually effect

the velocity it rolls down with.

I have some doubt about the given answer. We know that the CM speed is only dependent on the geometry of the object, let the rotational inertia be ##I = \beta m R^2##, for cylinder ##\beta = \frac{1}{2}##, for a ring it's ##\beta = 1##.

Now we can think of the original cylinder having ##\beta = \frac{1}{2}##, if snow does not stick to the cylinder, the final CM speed is

##v = \sqrt{ \frac{2 H}{g (1 + \beta)}} = \sqrt{ \frac{4 H}{3g }}##

If the snow sticks to the cylinder without slipping, we can consider the snow form rings on the cylinder whose ##\beta = 1##. This should change the overall rotational inertia of the snow object and slows it down.

Let me know how you think about this. Thanks,