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

Torque \(\tau = rF\sin\theta\), here \(\theta = 90^\circ\), so \(\sin\theta = 1\), and \(r=\frac{L}{2}\) (since force is applied at mid - point). The rotational inertia \(I\) of a rod pivoted at one end is \(I=\frac{1}{3}mL^{2}\). Also, from Newton's second law for rotation \(\tau = I\alpha\), so \(\alpha=\frac{\tau}{I}\).

Step2: Calculate the torque

Given \(F = 10\space N\), \(L = 1.5\space m\), so \(r=\frac{L}{2}=\frac{1.5}{2}=0.75\space m\).
Torque \(\tau=rF=(0.75\space m)\times(10\space N) = 7.5\space N\cdot m\)

Step3: Calculate the rotational inertia

Given \(m = 2.0\space kg\), \(L = 1.5\space m\)
\(I=\frac{1}{3}mL^{2}=\frac{1}{3}\times2.0\space kg\times(1.5\space m)^{2}=\frac{1}{3}\times2.0\times2.25= 1.5\space kg\cdot m^{2}\)

Step4: Calculate the angular[SSE onError error]

Answer:

Step1: Recall the formula for torque and rotational inertia

Torque \(\tau = rF\sin\theta\), here \(\theta = 90^\circ\), so \(\sin\theta = 1\), and \(r=\frac{L}{2}\) (since force is applied at mid - point). The rotational inertia \(I\) of a rod pivoted at one end is \(I=\frac{1}{3}mL^{2}\). Also, from Newton's second law for rotation \(\tau = I\alpha\), so \(\alpha=\frac{\tau}{I}\).

Step2: Calculate the torque

Given \(F = 10\space N\), \(L = 1.5\space m\), so \(r=\frac{L}{2}=\frac{1.5}{2}=0.75\space m\).
Torque \(\tau=rF=(0.75\space m)\times(10\space N) = 7.5\space N\cdot m\)

Step3: Calculate the rotational inertia

Given \(m = 2.0\space kg\), \(L = 1.5\space m\)
\(I=\frac{1}{3}mL^{2}=\frac{1}{3}\times2.0\space kg\times(1.5\space m)^{2}=\frac{1}{3}\times2.0\times2.25= 1.5\space kg\cdot m^{2}\)

Step4: Calculate the angular[SSE onError error]