Question 9 A uniform wheel of mass 10.0 kg and radius 0.400 m is mounted rigidly

Question

Question 9 A uniform wheel of mass 10.0 kg and radius 0.400 m is mounted rigidly on an axle through its center (see the figure). The radius of the axle is 0.200 m, and the rotational inertia of the wheel-axle combination about its central axis is o.600 kg-m2. The wheel is initially at rest at the top of a surface that is inclined at angle 10.50 with the horizontal; the axle rests on the surface while the wheel extends into a groove in the surface without touching the surface. Once released, the axle rolls down along the surface smoothly and without slipping. When the wheel-axle combination has moved down the surface by 8.57 m, what are (a) its rotational kinetic energy and (b) its translational kinetic energy? Wheel Axle Axle Groove

Explanation / Answer

Decrease in GPE:
GPE = m*g*L*sin(theta)

Decrease in GPE is equal to gain in total KE, due to conservation of energy.

KE comes in both rotational and translational forms:
KErot = 1/2*I*omega^2
KEtrans = 1/2*m*v^2

GPE = KEtrans + KErot

No-slip condition relates omega and v:
v = omega*r

r is the axle's outer radius

Thus:
KEtrans = 1/2*m*r^2*omega^2

Update:
m*g*L*sin(theta) = 1/2*omega^2*(m*r^2 + I)

solve for omega:
omega = sqrt(2*m*g*L*sin(theta)/(m*r^2 + I))

Substitute in to both equations for modes of KE storage:
KEtrans = 1/2*m*r^2*(2*m*g*L*sin(theta)/(m*r^2 + I))
KErot = 1/2*I*(2*m*g*L*sin(theta)/(m*r^2 + I))

Data: g:=9.8 N/kg; L:=8.57 m; theta:=10.5 deg; I:=0.6 kg m^2; m:=10 kg; r:=0.2 m;

Simplify:
KEtrans = g*L*sin(theta)*m^2*r^2/(m*r^2 + I) = 61.22J
KErot = g*L*sin(theta)*m*I/(m*r^2 + I) = 1311J

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