Physics, 14.04.2020 19:42 Broov

A uniform rod is 2.0 m long. The rod is pivoted about a horizontal, frictionless pin through one end. The moment of inertia of the rod about this axis is given by (1/3)ML2. The rod is released from rest at an angle of 30° above the horizontal. What is the angular acceleration of the rod at the instant it is released?

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Answer from: JASMINEM8347

The angular acceleration of the rod at the instant it is released is 6.37 rad/s^2 .

Given the following data:

Length = 2 meterAngle = 30°

To calculate the angular acceleration of the rod at the instant it is released:

The moment of inertia of the rod about a horizontal, frictionless pin through one end (axis) is given by the formula:

$(\frac{1}{3})ML^2$ ...equation 1.

Mathematically, torque is given by this formula:

$T = I\alpha$ ....equation 2.

Where:

T represents torque.I is the moment of inertia. $\alpha$ is the angular acceleration.

Making $\alpha$ the subject of formula, we have:

$\alpha =\frac{T}{I}$ ....equation 3.

For the horizontal component, torque is given by the formula:

$Torque = \frac{L}{2}(MgCos\theta)$ ...equation 4

Substituting eqn 1 and eqn 4 into eqn 3, we have:

$\alpha =\frac{\frac{L}{2}(MgCos\theta)}{(\frac{1}{3})ML^2} \\\\\alpha = \frac{\frac{MgCos\theta L}{2}}{\frac{ML^2}{3}} \\\\\alpha = \frac{MgCos\theta L}{2} \times \frac{3}{ML^2} \\\\\alpha = \frac{3gCos\theta}{2L} \\\\\alpha =\frac{\frac{L}{2}(MgCos\theta)}{(\frac{1}{3})ML^2} \\\\\alpha = \frac{\frac{MgCos\theta L}{2}}{\frac{ML^2}{3}} \\\\\alpha = \frac{MgCos\theta L}{2} \times \frac{3}{ML^2} \\\\\alpha = \frac{3gCos\theta}{2L} \\\\\alpha = \frac{3\;\times \;9.8 \times\; Cos30}{2\times 2}$

$\alpha = \frac{29.4 \times\; 0.8660}{4}\\\\\alpha = \frac{25.46}{4}$

Angular acceleration, $\alpha$ = 6.37 rad/s^2

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