# Industrial roller

1. Jun 12, 2008

### Buddy123

(a) A heavy industrial mill roller gets loose and rolls down a slope without slipping or sliding. It drops in height 900 mm. it is made of steel and is a hollow cylinder of outer diameter 600 mm, inner diameter 400 mm and length 1.2 m. assuming the density of steel is 7860 kg m-3, determine:
(i) the mass of the roller
(ii) the second moment of mass of the roller
(iii) the energy of the roller at the bottom of the slope
(iv) the speed of the roller at the bottom of the slope

(b) Once on the horizontal floor, the roller slows to a stop. The total frictional force required to bring the roller to a stop is calculated as 145 N.

Calculate, assuming again that the roller doesn’t slide or slip and that the frictional force required to decelerate rotary motion is Iα/r²:

(i) the shortest distance that the roller will cover before stopping
(ii) the time taken for the roller to stop

Thank you

2. Jun 12, 2008

### Staff: Mentor

Welcome to the PF. You must show us your own attempts at the solutions before we can offer tutorial help. How would you approach these questions?

3. Jun 13, 2008

### Buddy123

Can you tell me how to add my workings to this questions please? Also I can't see how I added this thread, there is no simple "add thread" option, I just found it somehow last time. Thanks.

4. Jun 13, 2008

### Staff: Mentor

To start a new thread, you would click on the "New Topic" button up in the main Introductory Physics forum page (or other appropriated Homework Help forum page). To post your work on these questions, you can use the Reply button here in this thread, or just type in the Quick Reply box at the bottom of this page.

5. Jun 13, 2008

### konthelion

Start with the definition of the moments of inertia of the hollow cylinder:
$$I=\frac{1}{2}M(R_{1}^2+R_{2}^2})$$, where R1 is the inner radius, and R2 is the outer radius. You should be able to solve (i) and (ii)

For (iii) and (iv), apply conservation of energy and for the non-slip condition, solve for v.

6. Jun 14, 2008

### Buddy123

Workings so far...
(i) the mass of the roller

density = 7860 kg/m³
OD = 0.3 m ID = 0.2 m
Length = 1.2 m

Surface area of end section = OD – ID

= π0.3² - π0.2 ²

= 0.283 – 0.126

= 0.157

Volume = π r² h

= 0.157 x 1.2

= 0.1884 m³

Mass = Density x Volume

Mass = 7860 x 0.1884

Mass = 1480.8 kg

(ii) the second moment of mass of the roller

second moment of area = I = ½ m (R²+r²)

I = ½ 1480.8 x (0.3² + 0.2²)

Second moment of area = 96.3 m4

7. Jun 24, 2008

### challenger2

I am also stuck on the same question. I got as far as Buddy 123 and then calculated the energy at the bottom of the slope using mxgxh as the potential energy has now converted to kinetic after its travels. I think that in order to get the last section (speed at the bottom of the slope) I require.
Kinetic Energy(total)= Translational Kinetic energy + Rotational Kinetic Energy
or
KT= mv^2 x w^2/2 + Iw^2/2 re-arranged for w, this is where I draw a blank as my mathematical ability is struggling with this part.