Pure Rolling - Part 2

Classical Mechanics Level 3

The density (mass per unit volume) of a solid sphere of radius R R varies with radial distance ( r r , 0 ≤ r ≤ R 0 \le r \le R ) from the center of the sphere according to the following relation:

ρ ( r ) = r \rho(r) = r

The sphere rolls without slipping on a fixed flat plane with a speed v v . The kinetic energy of the sphere can be expressed as:

T = a π R b v 2 c T = \frac{a \pi R^b v^2}{c}

Here, a a , b b and c c are positive integers with a a and c c being co-prime. Enter your answer as a + b + c a+b+c .

Also try: Pure Rolling


The answer is 35.

Karan Chatrath by Karan Chatrath , 27, Netherlands

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Karan Chatrath
Aug 12, 2019

The solid sphere can be divided into several small thin spherical shells.

Within the solid sphere, consider an elementary spherical shell of radius r r and thickness d r dr . The mass of this elementary hollow sphere is:

d M = ρ ( 4 π r 2 d r ) dM = \rho \left(4 \pi r^2 dr\right)

d I = 2 3 ( d M ) r 2 dI = \frac{2}{3}(dM)r^2

Moment of inertia of a thin spherical shell is a standard result that can be derived using first principles. Integrating both the above expressions from 0 0 to R R gives the relations:

M = π R 4 M = \pi R^4 I = 4 9 π R 6 I = \frac{4}{9} \pi R^6

Given that the sphere is 'pure-rolling', its translation speed of the COG is v v and its angular speed is ω = v / R \omega = v/R . Using this, the total kinetic energy can be found as such:

T = T t r a n s l a t i o n + T r o t a t i o n T = T_{translation} + T_{rotation} T = 1 2 M v 2 + 1 2 I ω 2 T = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2

Substituting expressions and simplifying gives the required answer:

a = 13 \boxed{a = 13} b = 4 \boxed{b = 4} c = 18 \boxed{c = 18}

2 Helpful 1 Interesting 1 Brilliant 0 Confused

You might simply say that the moment of inertia is I = ∫ 0 R d r ∫ 0 π d θ ∫ 0 2 π r 2 sin ⁡ θ × r × ( r sin ⁡ θ ) 2 = 1 3 π R 6 ∫ 0 π sin ⁡ 3 θ d θ = 4 9 π R 6 I \; = \; \int_0^R\,dr\int_0^\pi\,d\theta \int_0^{2\pi} r^2\sin\theta \times r \times (r\sin\theta)^2 \; = \; \tfrac13\pi R^6 \int_0^\pi \sin^3\theta\,d\theta \; = \; \tfrac49\pi R^6

Mark Hennings - 1 year, 10 months ago

Log in to reply

Thank you for the comment. Yes, I cross-checked the expression in the same way while framing this problem. I wanted to present an approach that avoids double/triple integrals.

Karan Chatrath - 1 year, 10 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...