Determine the force exerted on the vertical wall by the falling dumbbell when the dumbbell axis is at an angle lpha with the horizon. The dumbbell starts moving from a vertical position without initial velocity. Mass of each dumbbell ball m.
Solutions of Savchenko Problems in Physics knowledge must be free
<h3 id="back-link"><a href="/#2.3">$\leftarrow$Back</a></h3>
<h3> Statement </h3>
<p>
$2.3.30.$ Determine the force exerted on the vertical wall by the falling dumbbell when the dumbbell axis is at an angle $\alpha$ with the horizon. The dumbbell starts moving from a vertical position without initial velocity. Mass of each dumbbell ball $m$.
For problem $2.3.30$
<h3>Solution</h3>
<p>
Forces acting on the system
Along the axis $OZ_\parallel$ "The Two-Ball System":
Centripetal acceleration associated with circular rotation:
$$a=\frac{v^2}{R}$$
From the Newton's second law:
$$\frac{mv^2}{R}=2mg\sin\alpha-N_x\cos\alpha-N_y\sin\alpha$$
Law of Conservation of Energy
$$mgR=mgh+\frac{mv^2}{2}$$$$h=R\sin\alpha$$$$\frac{mv^2}{R}=2mg(1-\sin\alpha)$$$$N_y\sin\alpha+N_x\cos\alpha=(4\sin\alpha-2)mg$$
Along the axis $OZ_\perp$ "The Ball System":
$$N_y\cos\alpha=N_x\sin\alpha+mg\cos\alpha$$$$N_y=mg+N_x\tan\alpha$$$N_y$ found from the axis $OZ_\perp$ we substitute into $N_y$ found from the axis $OZ_\parallel$$$mg\sin\alpha\cos\alpha+N_x\sin^2 \alpha+N_x\cos^2 \alpha=2mg(2\sin\alpha-2)\cos\alpha$$$$\frac{N_x}{mg}=4\sin\alpha\cos\alpha-2\cos\alpha-\sin\alpha\cos\alpha$$$$\boxed{N_x=(3\sin\alpha-2)\cos\alpha \cdot mg}$$
When $N_x \leq0$, the shoe will break away from the vertical wall and will no longer act on it:
$$N_x=0$$
Condition of detachment:
$$(3\sin\alpha-2)\cos\alpha \leq0$$$$3\sin\alpha-2\leq0 \Leftrightarrow \boxed{\sin\alpha\leqslant2/3}$$
Savchenko Solutions
