Question

# The 12-kg cylinder supported by the bearing brackets at A and B has a moment of inertia about the vertical $z_{0}$-axis through its mass center G equal to $0.080 \mathrm{kg} \cdot \mathrm{m}^{2}$. The disk and brackets have a moment of inertia about the vertical z-axis of rotation equal to $0.60 \mathrm{kg} \cdot \mathrm{m}^{2}$. If a torque $M=16 \mathrm{N} \cdot \mathrm{m}$ 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 $z$ axis to calculate the angular acceleration $\alpha$ of the system.

For the sum of moments we have:

$+ \circlearrowleft \sum M_z = I_z \alpha$

The moment of inertia $I_z$ can be calculated by summing the moment of inertia of the brackets and the disk $I_b$, and using the Steiner's theorem of parallel axes for the moment of inertia around $G$, $I_G$, and axis $z$.

We have:

\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:

\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:

\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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