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 applie perpendicularly to the rod at its midpoint. what is the resulting angular acceleration of the rod?

Explanation:

Step1: Recall the formula for torque and moment of inertia for a rod pivoted at one end

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

Step2: Calculate the torque

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

Step3: Calculate the moment of inertia

Given \(m = 2.0\space kg\), \(L = 1.5\space m\), so \(I=\frac{1}{3}mL^{2}=\frac{1}{3}(2.0)(1.5)^{2}=\frac{1}{3}(2.0)(2.25)=1.5\space kg\cdot m^{2}\).

Step4: Calculate the angular acceleration

From \(\tau = I\alpha\), we can solve for \(\alpha\) as \(\alpha=\frac{\tau}{I}\). Substituting \(\tau = 7.5\space N\cdot m\) and \(I = 1.5\space kg\cdot m^{2}\), we get \(\alpha=\frac{7.5}{1.5}=5.0\space rad/s^{2}\).

Answer:

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