Step 1

We are given the following data:

The mass of wheel is $m = 50\;{\rm{kg}}$.

The radius of gyration is ${k_G} = 125\;{\rm{mm}}$.

The radius of the wheel is $r = 150\;{\rm{mm}}$.

The height of point $A$ above the horizontal is $h = 25\;{\rm{mm}}$.

We are asked to determine the minimum value of the angular velocity ${\omega _1}$ of the wheel, so that it strikes the step at $A$ without rebounding and then rolls over it without slipping.

 
Step 2

We will draw the angular momentum diagram of the wheel.

Here, ${N_A}$ represent the normal reaction force at point $A$ and $W$ is the weight of the wheel.

The formula to calculate ${h_1}$ is given by,

\[{h_1} = \left( {r - h} \right)\]

Substitute all the known values in the above formula.

\[\begin{array}{c} {h_1} = \left( {150 - 25} \right)\;{\rm{mm}} \times \left( {\frac{{{{10}^{ - 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right)\\ = 0.125\;{\rm{m}} \end{array}\]
 
Step 3

The sum of the angular impulse about point $A$ is zero. Thus, angular momentum of the wheel is conserved about this point.

Since the wheel rolls without slipping, the formula to calculate the velocity ${\left( {{v_G}} \right)_1}$ is given by,

\[{\left( {{v_G}} \right)_1} = r{{\bf{\omega }}_1}\]

Substitute all the known values in the above formula.

\[\begin{array}{c} {\left( {{v_G}} \right)_1} = \left( {150\;{\rm{mm}}\; \times \frac{{{{10}^{ - 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right){{\bf{\omega }}_1}\\ = \left( {0.15\;{\rm{m}}} \right){{\bf{\omega }}_1} \end{array}\]
 
Step 4

The formula to calculate the velocity ${\left( {{v_G}} \right)_2}$ is given by,

\[{\left( {{v_G}} \right)_2} = r{{\bf{\omega }}_2}\]

Substitute all the known values in the above formula.

\[\begin{array}{c} {\left( {{v_G}} \right)_2} = \left( {150\;{\rm{mm}}\; \times \frac{{{{10}^{ - 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right){{\bf{\omega }}_2}\\ = \left( {0.15\;{\rm{m}}} \right){{\bf{\omega }}_2} \end{array}\]
 
Step 5

The formula to calculate the mass moment of inertia of the wheel about its mass center is given by,

\[{I_G} = m{\left( {{k_G}} \right)^2}\]

Substitute all the known values in the above formula.

\[\begin{array}{c} {I_G} = \left( {50\;{\rm{kg}}} \right){\left( {125\;{\rm{mm}}\; \times \frac{{{{10}^{ - 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right)^2}\\ = 0.78125\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}} \end{array}\]
 
Step 6

We will apply the principle of angular impulse momentum on the wheel.

\[\begin{array}{c} {\left( {{H_A}} \right)_1} = {\left( {{H_A}} \right)_2}\\ \left[ {\left( m \right){{\left( {{v_G}} \right)}_1}\left( {{h_1}} \right) + {I_G}{{\bf{\omega }}_1}} \right] = \left[ {\left( m \right){{\left( {{v_G}} \right)}_2}\left( r \right) + {I_G}{{\bf{\omega }}_2}} \right] \end{array}\]

Substitute all the known values in the above equation.

\[\begin{array}{c} \left[ \begin{array}{l} \left\{ \begin{array}{l} \left( {50\;{\rm{kg}}} \right)\left( {0.15\;{\rm{m}}} \right)\\ \left( {{{\bf{\omega }}_1}} \right)\left( {0.125\;{\rm{m}}} \right) \end{array} \right\}\\ + \left\{ {\left( {0.78125\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right)\left( {{{\bf{\omega }}_1}} \right)} \right\} \end{array} \right] = \left[ \begin{array}{l} \left\{ \begin{array}{l} \left( {50\;{\rm{kg}}} \right)\left( {0.15\;{\rm{m}}} \right)\\ \left( {{{\bf{\omega }}_2}} \right)\left( {150\;{\rm{mm}}\; \times \frac{{{{10}^{ - 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right) \end{array} \right\}\\ + \left\{ {\left( {0.78125\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right)\left( {{{\bf{\omega }}_2}} \right)} \right\} \end{array} \right]\\ \left( {1.71875\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right)\left( {{{\bf{\omega }}_1}} \right) = \left( {1.90625\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right)\left( {{{\bf{\omega }}_2}} \right)\\ {{\bf{\omega }}_1} = 1.109{{\bf{\omega }}_2} \end{array}\] ...... (1)
 
Step 7

We will consider wheel 2 as a datum so, ${\left( {{y_G}} \right)_2} = 0$.

With reference to the datum in the diagram, the value of gravitational potential energy can be represented as,

\[{\left( {{V_g}} \right)_2} = W{\left( {{y_G}} \right)_2}\]

Substitute all the known values in the above equation.

\[\begin{array}{c} {\left( {{V_g}} \right)_2} = W\left( 0 \right)\\ = 0 \end{array}\]
 
Step 8

The formula to calculate the gravitational potential energy ${\left( {{V_g}} \right)_3}$ is given by,

\[\begin{array}{c} {\left( {{V_g}} \right)_3} = W{\left( {{y_G}} \right)_3}\\ = \left( {mgh} \right) \end{array}\]

Here, $g$ is the gravitational acceleration $g = 9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$.

Substitute all the known values in the above equation.

\[\begin{array}{c} {\left( {{V_g}} \right)_3} = \left( {50\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {25\;{\rm{mm}}\; \times \frac{{{{10}^{ - 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right)\\ = 12.2625\;{\rm{J}} \end{array}\]
 
Step 9

Since the wheel is required to be at rest in the final position, then ${T_3} = 0$.

The formula to calculate the initial kinetic energy of the wheel is given by,

\[{T_2} = \frac{1}{2}m{\left( {{v_G}} \right)_2}^2 + \frac{1}{2}{I_G}{\left( {{{\bf{\omega }}_2}} \right)^2}\]

Substitute all the known values in the above equation.

\[\begin{array}{c} {T_2} = \frac{1}{2}\left( {50\;{\rm{kg}}} \right){\left\{ {\left( {0.15\;{\rm{m}}} \right){{\bf{\omega }}_2}} \right\}^2} + \frac{1}{2}\left( {0.78125\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right){\left( {{{\bf{\omega }}_2}} \right)^2}\\ = \left[ {\left\{ {\left( {0.5625 + 0.390625} \right){\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right\}{{\left( {{{\bf{\omega }}_2}} \right)}^2}} \right]\\ = \left[ {\left( {0.953125\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right){{\left( {{{\bf{\omega }}_2}} \right)}^2}} \right]\; \end{array}\]
 
Step 10

We will apply the energy conservation principle.

\[{T_2} + {\left( {{V_g}} \right)_2} = {T_3} + {\left( {{V_g}} \right)_3}\]

Substitute all the known values in the above equation.

\[\begin{array}{c} \left[ {\left( {0.953125\,{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right){{\left( {{{\bf{\omega }}_2}} \right)}^2} + 0} \right] = \left[ {0 + \left( {12.2625\;{\rm{J}}} \right)} \right]\\ \left( {0.953125\,{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}} \right){\left( {{{\bf{\omega }}_2}} \right)^2} = \left( {12.2625\;{\rm{J}}} \right)\\ {{\bf{\omega }}_2} \approx 3.587\;{\rm{rad/s}} \end{array}\]
 
Step 11

Substitute all the known values in the equation (1) to obtain ${{\bf{\omega }}_1}$.

\[\begin{array}{c} {{\bf{\omega }}_1} = 1.109\left( {3.587\;{\rm{rad/s}}} \right)\\ \approx 3.98\;{\rm{rad/s}} \end{array}\]