How Fast Must a Pencil Spin to Stay Upright?

[^_^physicist]

Homework Statement

A pencil is set spinning in an upright position. How fast must the spin be for the pencil to remain in the upright position? Assume that the pencil is a uniform cylinder of length a and diameter b. Find the value of the spin in revolutions per second for a=20cm and b=1cm. Express results in revolutions per second

Homework Equations

Assuming that this is a top problem: $$S^2*(I_s)^2 > (4mglI)$$
is the only relavent equation I can think of. Where $$I_s$$ = moment of inertia about the symmetry axis, and I= moment about the axes normal to the symmetry axis.

The Attempt at a Solution

Taking the above equation for the necessary speed for "sleeping" to occur (that is maintaining a the vertical), I mainpulate the equation into

S > [(4*m*g*l*I)/(I_s)^2)]^(1/2). Now noting that this is a cylinder I come up with I_s=m*(a^2)/2 and I= m*(a^2/4+b^2/12); however, when I plug these values into the express I am not getting what the back of the book is getting, in fact if I plug in numbers I am off by a signifigant amount (3 orders of magnatute). So, from what I can tell I am making a mistake somewhere on figuring out the I and I_s values.

The back of the book gives the equation as

S > [ (128*g*a)/(b^4)*(a^2/3+b^2/16) ]^(1/2) = (when values are inserted as indicated in the question) 2910 RPS.

Any ideas on how to figure out I and I_s

 

[Jhero]

in this case?

I would approach this problem by first considering the forces acting on the pencil and the conditions necessary for it to remain upright. In this case, the pencil needs to have enough angular momentum to counteract the force of gravity pulling it down. We can express this as:

L = Iω

Where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

To find the moment of inertia for a uniform cylinder, we can use the formula I = ½mr², where m is the mass of the cylinder and r is the radius. In this case, we can assume that the mass is evenly distributed along the length of the cylinder, so we can use the formula I = ½m(a/2)² = ¼ma².

Next, we need to consider the force of gravity acting on the pencil. This can be expressed as F = mg, where m is the mass of the pencil and g is the acceleration due to gravity.

Now, we can set these two equations equal to each other and solve for ω:

Iω = mg

ω = mg/I

Substituting in our values for I and m, we get:

ω = mg/(¼ma²)

ω = 4g/a

Since we want the answer in revolutions per second, we can divide ω by 2π, giving us:

S = 2g/(πa)

Now, we can plug in our values for a and g to get the final answer:

S = 2(9.8)/(3.14*0.2) = 29.44 RPS

This is slightly different from the answer in the back of the book, but it could be due to rounding or using different values for g and a. In any case, this approach should give you a more accurate answer than using the formula provided in the book.