Question

a uniform rod of length $l_0$ is free to rotate about an axis through its left end and perpendicular to the horizontal plane of the figure shown. when a force with magnitude $f_0$ is exerted perpendicular to the rod at the right end, the angular acceleration of the rod is $\alpha_1$. the same force is then exerted on the right end of a second uniform rod with the same linear mass density and twice the length as the original rod, as shown on the right side of the figure. the rod of length $2l_0$ has an angular acceleration $\alpha_2$. what is the ratio $\alpha_2 : \alpha_1$?
a $2:1$
b $1:2$
c $1:4$
d $1:8$

Explanation:

Step1: Recall torque-angular acceleration relation

Torque $\tau = I\alpha$, where $I$ is moment of inertia, $\alpha$ is angular acceleration. Also $\tau = rF$ (perpendicular force). So $\alpha = \frac{rF}{I}$.

Step2: Define linear mass density

Let linear mass density be $\lambda$. For first rod: mass $m_1 = \lambda L_0$, moment of inertia about end $I_1 = \frac{1}{3}m_1L_0^2 = \frac{1}{3}\lambda L_0^3$.

Step3: Calculate $\alpha_1$

$\alpha_1 = \frac{F_0 L_0}{I_1} = \frac{F_0 L_0}{\frac{1}{3}\lambda L_0^3} = \frac{3F_0}{\lambda L_0^2}$

Step4: Find $I_2$ for second rod

Second rod: length $2L_0$, mass $m_2 = \lambda \cdot 2L_0$, moment of inertia $I_2 = \frac{1}{3}m_2(2L_0)^2 = \frac{1}{3}(\lambda \cdot 2L_0) \cdot 4L_0^2 = \frac{8}{3}\lambda L_0^3$.

Step5: Calculate $\alpha_2$

$\alpha_2 = \frac{F_0 \cdot 2L_0}{I_2} = \frac{2F_0 L_0}{\frac{8}{3}\lambda L_0^3} = \frac{3F_0}{4\lambda L_0^2}$

Step6: Compute ratio $\alpha_2:\alpha_1$

$\frac{\alpha_2}{\alpha_1} = \frac{\frac{3F_0}{4\lambda L_0^2}}{\frac{3F_0}{\lambda L_0^2}} = \frac{1}{4}$

Answer:

C. 1 : 4