Question

A uniform wheel of mass 12.0 kg is mounted rigidly on a massless axle through its center, as shown in the figure below. The radius of the axle is 0.200 m, and the rotational inertia of the wheel-axle combination about its central axis is 0.600 kg

Answer 1

Decrease in Gravitational Potential Energy, GPE = m*g*L*sin(theta)

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

KE_rot = 1/2*I*omega²
KE_trans = 1/2*m*v²

GPE = KE_trans + KE_rot

No-slip condition relates omega and v:
v = omega*r
where r is the axle's outer radius

Thus, we have
KE_trans = 1/2*m*r²*omega²

Equating,
m*g*L*sin(theta) = 1/2*omega²*(m*r² + I)
omega = sqrt[2*m*g*L*sin(theta)/(m*r² + I)]

Substitute in to both equations for modes of KE storage:
KE_trans = 1/2*m*r²*(2*m*g*L*sin(theta)/(m*r² + I))
KE_rot = 1/2*I*(2*m*g*L*sin(theta)/(m*r² + I))

On Simplifying, we get
KE_trans = g*L*sin(theta)*m²*r²/(m*r² + I)............(1)
KE_rot = g*L*sin(theta)*m*I/(m*r² + I)..................(2)

Given Data:
g = 9.8 m/s²; L = 2 m; theta = 30 deg; I = 0.6 kg m^2; m = 12 kg; r = 0.2 m;

Substituting these values in (1), we get
KE_trans = 9.8*2*sin(30)*12²*(0.2)²/(12*0.2² + 0.6)
KE_trans = 52.27 J

Substituting these values in (2), we get
KE_rot = 9.8*2*sin(30)*12*0.6/(12*(0.2)² + 0.6)
KE_rot = 65.33 J

Answer 2

Conservation of energy is usually a good starting point. The PE lost must equal the total KE gained. I'm assuming the mass of wheel + axle is 12kg although the question is ambiguous.

PE lost, Ep = m.g.h = 12 x 9.8 x 2sin(30) = 117.6J

The rotational KE, Er = 0.5 Iz w^2
(where Iz is the moment of inertia of the wheel-axle and w is the angular velocity)
The translational KE, Et = 0.5 m (w.r)^2
(where m is the mass of wheel + axle and r is the axle radius and w is angular velocity. w.r is the velocity of the axle in the direction of the slope)

The total KE, Es = Er + Et = 0.5 ( Iz + m.r^2 ).w^2
Es = 0.5 ( 0.6 + 10 x 0.2^2 ) w^2
Es = 0.5 w^2

Conservation of energy dictates that Es = Ep
117.6= 0.5 w^2
w^2 = 235.5

a) rotational KE = 0.5 Iz w^2 = 0.5 x 0.6 x 235.2= 70.56J

b) translational KE = 117.6- 70.56= 47.04J