Question

The 12-kg cylinder supported by the bearing brackets at A and B has a moment of inertia about the vertical z0z_{0}-axis through its mass center G equal to 0.080kgm20.080 \mathrm{kg} \cdot \mathrm{m}^{2}. The disk and brackets have a moment of inertia about the vertical z-axis of rotation equal to 0.60kgm20.60 \mathrm{kg} \cdot \mathrm{m}^{2}. If a torque M=16NmM=16 \mathrm{N} \cdot \mathrm{m} is applied to the disk through its shaft with the disk initially at rest, calculate the horizontal x-components of force supported by the bearings at A and B.

Solution

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To solve this, we first observe the sum of moments, with the given picture, around the zz axis to calculate the angular acceleration α\alpha of the system.

For the sum of moments we have:

+Mz=Izα+ \circlearrowleft \sum M_z = I_z \alpha

The moment of inertia IzI_z can be calculated by summing the moment of inertia of the brackets and the disk IbI_b, and using the Steiner's theorem of parallel axes for the moment of inertia around GG, IGI_G, and axis zz.

We have:

Iz=Ib+IG+md2Iz=0.6+0.08+120.22Iz=1.16 kgm2\begin{align*} & I_z = I_b + I_G + md^2 \\ & I_z = 0.6 + 0.08 + 12 \cdot 0.2^2 \\ & I_z = 1.16 \text{ kg} \cdot \text{m}^2 \end{align*}

The sum of moments is then:

M=Izα16=1.16αα=161.16α=13.793 rads2\begin{align*} & M = I_z \alpha \\ & 16 = 1.16 \alpha \\ & \alpha = \dfrac{16}{1.16} \\ & \alpha = 13.793 \text{ } \dfrac{\text{rad}}{\text{s}^2} \end{align*}

The tangential acceleration is then, by definition, equal to:

at=rαat=0.213.793at=2.795 ms2\begin{align*} & a_t = r \alpha \\ & a_t = 0.2 \cdot 13.793 \\ & a_t = 2.795 \text{ } \dfrac{\text{m}}{\text{s}^2} \end{align*}

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