Question

a uniform rod of mass m = 2.0, kg and length l = 1.5, m is pivoted at one end. a force of f = 10, n is applied perpendicularly to the rod at its midpoint. what is the resulting angular acceleration of the rod?

Explanation:

Step1: Recall the formula for torque and rotational inertia

The torque \(\tau\) is given by \(\tau = rF\sin\theta\), where \(r\) is the distance from the pivot to the point of force application, \(F\) is the force, and \(\theta\) is the angle between \(r\) and \(F\). For a uniform rod pivoted at one end, the moment of inertia \(I\) about the pivot is \(I=\frac{1}{3}mL^{2}\). Here, the force is applied at the midpoint, so \(r = \frac{L}{2}\), and \(\theta = 90^{\circ}\) (since the force is perpendicular to the rod), so \(\sin\theta=1\). The torque \(\tau=\frac{L}{2}F\). Also, the rotational form of Newton's second law is \(\tau = I\alpha\), where \(\alpha\) is the angular acceleration.

Step2: Substitute the moment of inertia into the torque formula

We know \(I = \frac{1}{3}mL^{2}\) and \(\tau=\frac{L}{2}F\). Substituting into \(\tau = I\alpha\), we get \(\frac{L}{2}F=\frac{1}{3}mL^{2}\alpha\).

Step3: Solve for angular acceleration \(\alpha\)

First, we can cancel out \(L\) from both sides (assuming \(L
eq0\)). We get \(\frac{1}{2}F=\frac{1}{3}mL\alpha\). Then, solve for \(\alpha\): \(\alpha=\frac{3F}{2mL}\).

Now, substitute the given values: \(m = 2.0\space kg\), \(L = 1.5\space m\), \(F = 10\space N\).

\(\alpha=\frac{3\times10}{2\times2.0\times1.5}\)

Calculate the numerator: \(3\times10 = 30\)

Calculate the denominator: \(2\times2.0\times1.5=6.0\)

Then, \(\alpha=\frac{30}{6.0}=5.0\space rad/s^{2}\)

Answer:

The angular acceleration of the rod is \(\boldsymbol{5.0\space rad/s^{2}}\)