10-29-08LectPHY2048

# 10-29-08LectPHY2048 - Chapt 11 examples Rolling(combined...

This preview shows pages 1–8. Sign up to view the full content.

Chapt. 11 examples Rolling (combined rotation & translation) The cylinder , of mass m , radius r , rolls from rest (without slipping) down the incline of angle θ What is its speed after travelling distance L along the incline? This means we can use conservation of mechanical energy without worry that there will be some loss of energy for which we failed to account. L θ m r ω v com Since the cylinder rolls there is friction in the problem (or the cylinder would slip rather than roll) but because it is static friction (no slipping) it is not associated with a displacement (recall that for rolling the point in contact with the surface is at rest) meaning there is no work and thus no thermal energy loss associated with this friction.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
L θ m r ω v com mech E0 Δ= UK 0 Δ+ h In travelling the distance L the cylinder falls through a height h , so, fiif UU K K0 −+− = Since started from rest K i = 0 . Taking U f = 0 at the the end of L , if 0 −= or in terms of the parameters given since, i Um g h = hL s i n = θ i g L s i n = θ We have,
The kinetic energy at the end of L is split between translational and rotational terms, i.e. f Trans rot KK K = + 22 fc o m c o m 11 Km v I =+ ω But, com vr com v r ω= While for a cylinder about its central axis, 2 com 1 Im r 2 = So, 2 com o m v 1 v ( m r ) 2 r ⎛⎞ ⎜⎟ ⎝⎠ 2 f com com com 3 v m v m v 24 4 =+= L θ m r ω v com h

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Then, 2 if c o m 3 U K mgLsin mv 0 4 −= θ = 2 com 4 vg L s i n 3 4 L s i n 3 = θ Or, Alternatively, the textbook derives an expression for the acceleration of the com for a uniform round body rolling down a ramp due to gravity (pp. 279-280) as, L θ m r ω v com h 22 2 1 2 gsin 2 ag s i n 1 I /mr 1 mr /mr 3 θ θ =− θ ++ Then using, oo vv 2 a ( r r ) = +− Solving for the velocity,
L θ m r ω v com h 2 com 2 v0 2 ( g s i n ) ( 0 L ) 3 =+ θ 2 com 4 vg L s i n 3 So that again, com 4 L s i n 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
HITT#1-10/29/08 Suppose that in the previous problem that there is no friction so the cylinder slides down the ramp the distance L without rolling. L θ m r v com With respect to the value when rolling the v com will now be, A. Greater B. Smaller C. The same
HITT#1-10/29/08 L θ m r v com A. Greater with if UK 0 −= i Um g h m g L s i n == θ 2 fc o m 1 Km v 2 = When rolling the gravitational potential energy was split between translational and rotational kinetic energy terms. Now there is no rotation so all the Δ U goes into the translation making the velocity there greater.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 26

10-29-08LectPHY2048 - Chapt 11 examples Rolling(combined...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online