Rotational Kinetic Energy

Question

A solid cylinder of mass 5.30 kg and

radius 0.250 m is rolling along flat

ground at 3.44 m/s. How much

TOTAL KE does it have?

Answer 1

Given:

m = 5.30 kg

r = 0.250 m

v = 3.44 m/s

Find:

KE_tot = ? J

Solution:

I'm going to define some place-holder variables for the 2 kinetic energies:

T = Translational Kinetic Energy (some mass, m, moving at a speed, v)

R = Rotational Kinetic Energy (some mass-moment of inertia, I, rotating at speed ω)

T = 1/2 m v²

R = 1/2 I ω²

Mass-Moment of Inertia of a Solid Cylinder of uniform mass, m, and radius, r:

I = 1/2 m r²

Applying the No-Slip Condition:

v = r ω

∴ ω = v / r

T = 1/2 m v²

R = 1/2 ( 1/2 m r² ) ( v/r )²

∴ R = 1/4 m v²

KE_tot = T + R = 1/2 m v² + 1/4 m v²

KEtot = 3/4 m v² = 3/4 ( 5.30 kg ) * ( 3.44 m/s )²

∴ KE_tot ≈ 47.04 J

I hope this helps! Message me in the comments with any questions, comments, or concerns of how I handled the above problem!

If you are concerned with how I simplified the total kinetic energy, here it is finding each energy and then adding their values:

T = 1/2 ( 5.30 kg ) * ( 3.44 m/s )²

T ≈ 31.36 J

I = 1/2 ( 5.30 kg ) * ( 0.250 m )²

I ≈ 0.166 kg * m²

ω = ( 3.44 m/s ) / ( 0,250 m )

ω = 13.76 rad / s

R = 1/2 I ω² = 1/2 ( 0.166 kg * m² ) * ( 13.76 rad/s )²

R ≈ 15.68 J

KE_tot = T + R = 31.36 J + 15.68 J

∴ KE_tot ≈ 47.04 J