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  $\alpha =2.538 \ rad/s^2$

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

     $\tau = I_a * \alpha$

=>   $15.34 = 6.045 * \alpha$

=>   $\alpha =2.538 \ rad/s^2$