# It is well known that for a hollow, cylindrical shell rolling without slipping on a horizontal surface, ?

half of the total kinetic energy is translational and half is rotational. What fraction of the total kinetic energy is rotational for the following objects rolling without slipping on a horizontal surface?

a)A uniform solid cylinder

b)A uniform sphere.

c)A thin-walled hollow sphere.

d)A hollow, cylinder with outer radius R and inner radius R/2.

### 3 Answers

- NCSLv 73 weeks ago
Welcome to Yahoo!Answers.

total KE = trans KE + rotation KE

total KE = ½mv² + ½Iω²

"without slipping" means that

ω = v/R, so

KE = ½mv² + ½I(v/R)² = (v²/2)(m + I/R²)

If we define I = kmR², then

I/R² = km

and so

KE = (v²/2)*(m + km) = (mv²/2)*(1 + k)

This means that the fraction of the total KE that is rotational is

→ k / (1 + k)

for a hollow cylinder, k = 1 and so we get the fraction

1 / (1+1) = ½

as stated.

A) uniform solid cylinder, k = ½, so

½ / (1 + ½) = ½ / (3/2) = 1/3

B) uniform (solid) sphere, k = 2/5, so

(2/5) / (1 + 2/5) = (2/5) / (7/5) = 2/7

C) thin hollow sphere, k = 2/3, so

(2/3) / (1 + 2/3) = (2/3) / (5/3) = 2/5

D) This one is trickier. People often err by finding the difference between the squares of the radii, as the other poster has done -- instead, you have to use the sum; so

I = ½m(R² + (R/2)²) = ½mR²(1 + ¼) = (5/8)mR²

so k = 5/8

and

(5/8) / (1 + 5/8) = (5/8) / (13/8) = 5/13

If you find this helpful, please award Best Answer. You get points too!

- PinkgreenLv 73 weeks ago
Mistake: not "half of" but "part of" the total energy

....etc.

a) For a uniform solid cylinder:

let v=the linear speed of the object(m/s)

.....r=the radius of the cross-section (m)

....T=the time needed for 1 revolution (s)

m=the mass of the solid (kg)

I=the moment of inertia about the axis of

the cylinder=(mr^2)/2

....w=the angular speed(rad./s)

The total kinetic energy=

(mv^2+Iw^2)/2=

[m(2(pi)r/T)^2+Iw^2]/2=

[m(2(pi)r/(2pi/w))^2+Iw^2]/2=

[m(wr)^2+Iw^2]/2

The rotational kinetic energy=

(w^2)[mr^2+I]/2

=>

The rotational k.energy/ The total k. energy=

0.5I(w^2)/[0.5(w^2)(mr^2+0.5mr^2)]=

0.5mr^2/[3m(r^2)/2]=1/3.

b), c), d) can be done similarly with

different I, it is left as your exercise.

- Anonymous3 weeks ago
a)A uniform solid cylinder

KE = 1/2 tr.+1/4 rot = 3/4 tot

rot/tot = 1/4 / 3/4 = 1/3 = 33%

b)A uniform sphere.

KE = 1/2 tr.+1/5 rot = 7/10 tot

rot/tot = 2/10 / 7/10 = 2/7 = 28.6%

c)A thin-walled hollow sphere.

KE = 1/2 tr.+1/3 rot = 5/6 tot

rot/tot = 1/3 / 5/6 = 2/5 = 40.0%

d)A hollow, cylinder with outer radius R and inner radius R/2.

KE = 1/2 tr.+1/2*(4-1)/4 = 3/8 rot = 7/8 tot

rot/tot = 3/8 / 7/8 = 3/7 = 42.9%