A string of length `L` and negligible mass is completely wound around a solid cylinder of uniform density, of mass `M` and radius `R` , and it has a small weight of mass `m` attached to its end. If the weight is released from rest under the influence of gravity, what is its velocity when the string is entirely unwound?

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Use conservation of energy.

`E_i=E_f`

Since the string is length `L` , we want to know the speed after the mass has fallen a distance `L` . So the initial potential energy will be `mgL` . The final energy will be the kinetic energy of the falling mass and the...

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Use conservation of energy.

`E_i=E_f`

Since the string is length `L` , we want to know the speed after the mass has fallen a distance `L` . So the initial potential energy will be `mgL` . The final energy will be the kinetic energy of the falling mass and the rotational energy of the cylinder.

`P_i=K_(t r a n s)+K_(rot)`

`mgL=1/2mv^2+1/2I omega^2`

We need to know the moment of inertia of the cylinder. You could look it up, but lets calculate it.

First the total mass of the cylinder of uniform density `rho` is and height `z` is:

`M=int rho dV=rho int_0^z pi R^2 dz=rho pi R^2 z`

`rho=M/(pi R^2 z)`

Now calculate `I` .

`I= int r^2 dm=int r^2 rho dV=int r^2 M/(pi R^2 z) dV`

We need to integrate away from the axis of rotation. so we will integrate over `r` .

`I=int_0^R r^2 M/(pi R^2 z) (2pi r)z dr`

`I=(2M)/R^2 int_0^R r^3 dr=(2M)/R^2*(1/4R^4)=1/2MR^2`

Now we can solve for `v` .

`mgL=1/2mv^2+1/2I omega^2`

`mgL=1/2mv^2+1/2(1/2MR^2) (v/R)^2`

`mgL=1/2mv^2+1/4Mv^2`

`4mgL=2mv^2+Mv^2`

`4mgL=(2m+M)v^2`

`(4mgL)/(2m+M)=v^2`

Therefore, the answer is:

`v=sqrt((4mgL)/(2m+M))`

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In this solution, I am going to assume that the cylinder is fixed so its center of mass can not move, even though this is not explicitly stated in the problem. This means the cylinder will not be moving as a whole. It will, however, rotate due to the string unwinding when the weight is released.

From the consideration of the conservation of energy, the change of the gravitational potential energy of the weight will equal the rotational kinetic energy of the cylinder and the kinetic energy of the weight:

`Delta U_(gr) = DeltaK_(rot) + DeltaK `

The gravitational potential energy of the weight, once the string is unwound, will change by

`DeltaU_(gr) = mgL` .

The change of kinetic energy of the weight at that time, since it starts moving from rest, will be

`DeltaK = 1/2mv^2` , where v is the velocity we are looking for.

The change of the rotational kinetic energy of the cylinder, which also starts moving from rest, is

`DeltaK_(rot) = 1/2Iomega^2` , where I is the cylinder's moment of inertia around its axis

`I = 1/2MR^2`

and

`omega`

is the angular velocity, which can be found considering that the point on the rim of the cylinder has the same linear velocity as that of the weight, v:

`omega = v/R` .

Plugging this into the equation of energy equation, we get

`mgL = 1/2*1/2MR^2*(v/R)^2 + 1/2mv^2`

`mgL = 1/4Mv^2+1/2mv^2 =(M+2m)/4 v^2 `

From here,

`v = sqrt((4m)/(M+2m)gL)` .

This is the velocity of the weight when the string is unwound, in terms of m, M, L and g.

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