Question

A 6.9-kg wheel with geometric radius m has radius of gyration computed about its mass center given by m. A massless bar at angle is pin-connected to its center and subjected to force N. If the wheel rolls without slipping on the flat stationary ground surface, find its angular acceleration . Consider counterclockwise to be positive when reporting your answer.

Answer 1

Complete Question

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Answer:

The value is
`\alpha =2.538 \ rad/s^2`

Step-by-step explanation:

From the question we are told that

The mass of the wheel is m = 6.9 kg

The radius is
`r = 0.69 \ m`

The radius of gyration is
`k_G = 0.4\ m`

The angle is
`\theta = 47^o`

The force which the massless bar is subjected to
`F = 22.5 \ N`

Generally given that the wheels rolls without slipping on the flat stationary ground surface, it implies that point A is the center of rotation.

Generally the moment of inertia about A is mathematically represented as

`I_a = I_G + M* r^2`

Here
`I_G` is the moment of inertia about G with respect to the radius of gyration which is mathematically represented as

`I_G = M * k_G`

=>
`I_a = k_G* M + M* r^2`

=>
`I_a =0.4 * 6.9 + 6.9 * 0.69^2`

=>
`I_a =6.045 \ kg \cdot m^2`

Generally the torque experienced by the wheel is mathematically represented as

`\tau = F * cos (47)`

=>
`\tau = 22.5 * cos (47)`

=>
`\tau = 15.34 \ kg \cdot m^2 \cdot s^(-2)`

Generally this torque is also mathematically represented as

`\tau = I_a * \alpha`

=>
`15.34 = 6.045 * \alpha`

=>
`\alpha =2.538 \ rad/s^2`