Question

A uniform stick of length L, mass M hinged at one end is released from rest at an angle $\theta_0$ with the vertical. When the angle with the vertical is $\theta$, the hinge exerts a force $F_r$ along the stick and $F_t$ perpendicular to the stick. Calculate $F_r$ and $F_t$.

• A. $F_r = \frac{Mg}{2}(5cos\theta-3cos\theta_0)$
• B. $F_r = \frac{3Mg}{2}(5cos\theta-3cos\theta_0)$
• C. $F_t = \frac{3Mg}{4}sin\theta$
• D. $F_t = \frac{Mg}{4}sin\theta$

Answer 1

Answer: (A), (D)

A, D

Answer 2

Angular Velocity ($\omega$):

The stick is released from rest at angle $\theta_0$ with the vertical. When the angle is $\theta$, the center of mass (CM) at $L/2$ has dropped by a vertical height $h = \frac{L}{2}(\cos\theta - \cos\theta_0)$.

Using conservation of energy:

Loss in Potential Energy = Gain in Rotational Kinetic Energy

$Mg \frac{L}{2}(\cos\theta - \cos\theta_0) = \frac{1}{2} I \omega^2$

The moment of inertia of a uniform stick about one end is $I = \frac{1}{3}ML^2$.

$Mg \frac{L}{2}(\cos\theta - \cos\theta_0) = \frac{1}{2} \left(\frac{1}{3}ML^2\right) \omega^2$

$g(\cos\theta - \cos\theta_0) = \frac{L}{3} \omega^2$

$\omega^2 = \frac{3g}{L}(\cos\theta - \cos\theta_0)$

Angular Acceleration ($\alpha$):

The torque about the hinge is due to gravity acting at the CM.

$\tau = (Mg \sin\theta) \frac{L}{2}$

Also, $\tau = I \alpha$

$Mg \frac{L}{2} \sin\theta = \frac{1}{3}ML^2 \alpha$

$\alpha = \frac{3g}{2L} \sin\theta$

Radial Hinge Force ($F_r$):

The net force in the radial direction (towards the hinge) on the CM causes centripetal acceleration $a_r = \omega^2 \frac{L}{2}$.

The forces along the stick are $F_r$ (towards hinge) and $Mg \cos\theta$ (away from hinge).

$F_r - Mg \cos\theta = M a_r = M \left(\omega^2 \frac{L}{2}\right)$

Substitute $\omega^2$:

$F_r = Mg \cos\theta + M \frac{L}{2} \left[\frac{3g}{L}(\cos\theta - \cos\theta_0)\right]$

$F_r = Mg \cos\theta + \frac{3}{2} Mg (\cos\theta - \cos\theta_0)$

$F_r = Mg \cos\theta + \frac{3}{2} Mg \cos\theta - \frac{3}{2} Mg \cos\theta_0$

$F_r = \frac{5}{2} Mg \cos\theta - \frac{3}{2} Mg \cos\theta_0$

$F_r = \frac{Mg}{2}(5\cos\theta - 3\cos\theta_0)$

Tangential Hinge Force ($F_t$):

The net force in the tangential direction on the CM causes tangential acceleration $a_t = \alpha \frac{L}{2}$.

The component of gravity $Mg \sin\theta$ acts in the tangential direction. The hinge force $F_t$ acts perpendicular to the stick. For the stick to accelerate tangentially downwards, $F_t$ must oppose the motion or be smaller than the gravitational component.

$Mg \sin\theta - F_t = M a_t = M \left(\alpha \frac{L}{2}\right)$

Substitute $\alpha$:

$Mg \sin\theta - F_t = M \frac{L}{2} \left(\frac{3g}{2L} \sin\theta\right)$

$Mg \sin\theta - F_t = \frac{3}{4} Mg \sin\theta$

$F_t = Mg \sin\theta - \frac{3}{4} Mg \sin\theta$

$F_t = \frac{1}{4} Mg \sin\theta$

Both options (A) and (D) are correct.