z

from the apex along the cone's symmetry axis. A thin disc at height

z

s (z) = \frac{R}{H} z

d M = ρ π s^{2} d z

. Its moment about a parallel diameter through its centre is

s^{2} d M /4

, while its centre is distance

z

from the required axis. Part (a)(ii) gives

d I = (z^{2} + \frac{s ^{2}}{4}) d M .

Hence

I = ρ π \int_{0}^{H} \frac{R ^{2} z ^{2}}{H ^{2}} (z^{2} + \frac{R ^{2} z ^{2}}{4 H ^{2}}) d z = ρ π (\frac{R ^{2} H ^{3}}{5} + \frac{R ^{4} H}{20}) .

Since

M = ρ π R^{2} H /3

I = M (\frac{3}{20} R^{2} + \frac{3}{5} H^{2}), α = \frac{3}{20}, β = \frac{3}{5} .

ii

Solution

K_{0} = \frac{1}{2} I ω^{2} .

The constant angular speed is

ω = 2 K_{0} / I

, so one full rotation takes

T = \frac{2 π}{ω} = 2 π \frac{I}{2 K _{0}} .

The work law means that friction exerts a constant opposing torque of magnitude

W

. Until the cone stops,

I \overset{ω}{˙} = - W .

Its initial angular speed is

ω_{0} = 2 K_{0} / I

, and uniform angular deceleration gives

0 = ω_{0} - \frac{W}{I} t .

Therefore

t = \frac{I ω _{0}}{W} = \frac{2 K _{0} I}{W ^{2}} .

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