# Cone rotating on the table

1. Apr 24, 2010

### Lara Croft

1. The problem statement, all variables and given/known data
I'm not from an English speaking country. I hope that you'll understend my question.

Cone made of steel is rotating on the wooden desk. Radius of cone is 4 cm, height is 10 cm.
Density of steel is 7,8*10^3 kg/m^3. Angular velocity of cone is 3.14 rad/s. Find the force of friction.
Cone is rolling without sliding.

2. Relevant equations

3. The attempt at a solution
I tried to solve this but i don't know where to start.
I got mass from density and volume.
I know that i have two forces - one that keeps cone rolling and the friction that slows it down.
My problem is that I don't know what to do because the cone is rolling arround an axis which doesn't
go through it's center of mass.
If someone could help me with applying Newton's second law on this I would be very grateful.

It is very important that i solve this one.

Last edited: Apr 24, 2010
2. Apr 24, 2010

### tiny-tim

Welcome to PF!

Hi Lara Croft ! Welcome to PF!
it's like a wheel …

mass times acceleration-of-centre-of-mass = net force

3. Apr 24, 2010

### fluidistic

If it's rolling without sliding then the friction force is a static one, not dynamic one. Being a static friction, it does no work, hence dissipate no energy and the cone will NOT slow down.

4. Apr 26, 2010

### Lara Croft

I tried to solve this using law of conservation of energy.
Addition of kinetical energy that cone has from rotating on the table ($$E_1$$)
and from rotating around it's axis ($$E_2$$) is equal to work of friction.
Can somebody please check my solution? Thanks

$$V=\frac{R^{2}*H*\pi}{3}=1.68*10^{-4}m^{3}$$

$$m=V*\rho$$

$$m=1.3069 kg$$

$$E_1+E_2=W$$

$$E_1=\frac{m*\omega_1^{2}}{2}$$

$$E_2=\frac{m*\omega_2^{2}}{2}$$

$$\omega_1=\pi \frac{rad}{s}$$

$$\omega_2=\frac{\phi}{t}$$

$$t=2s$$

$$s_1=2*r*\pi=2*\pi*\sqrt{R^{2}+H^{2}}=\\2*\pi*\sqrt{0.1^{2}+0.04^{2}}=0.6767m$$

$$s_2=2*R*\pi=2*0.04*\pi=0.2513m$$

$$N=\frac{s_1}{s_2}=\frac{0.6767}{0.2513}=2.6925$$

$$\omega_2=\frac{2.7}{2}=1.35\frac{rad}{s}$$

$$E_1=\frac{1.3069*\pi^{2}}{2}=6.45J$$

$$E_2=\frac{1.3069*1.35^{2}}{2}=1.19 J$$

$$W=6.45+1.19=7.64J$$

$$W=F*s$$

$$F=\frac{W}{s}=\frac{7.64}{2\pi}=1.22N$$

5. Apr 26, 2010

### fluidistic

Clearly you didn't pay attention to my post. Friction does no work because the point of contact of the cone with the table has a relative velocity of 0. Hence the friction force is exerted over no distance. No energy is dissipated in the form of heat. It is a case of static friction, not dynamic friction.

6. Apr 26, 2010

### tiny-tim

Yup! work done is zero and kinetic energy is constant …

use mass times acceleration-of-centre-of-mass = net force.

Since the angular velocity of the cone is constant, 3.14 rad/s, what is the acceleration of the centre of mass?

7. Apr 26, 2010

### Lara Croft

Is it 0, because the cone is spinning with constant speed, so this means that the acceleration of it should be 0, because the speed doesn't change over time?

8. Apr 26, 2010

### tiny-tim

Nope …

the speed doesn't change, but the velocity does.

Hint: what is the path of the centre of mass?

9. Apr 26, 2010

### fluidistic

Tiny-tim is right by saying that no, it's not 0.
More precisely, the angular acceleration of the CM is null because the angular velocity of it is constant.
But acceleration and angular acceleration aren't the same thing.