As depicted in Fig. 9-5, a ball B is fastened to one end of...

As depicted in Fig. 9-5, a ball $B$ is fastened to one end of a 24 -cm string, and the other end is held fixed at point $O$. The ball whirls in the horizontal circle shown. Find the speed of the ball in its circular path if the string makes an angle of $30^{\circ}$ to the vertical. The only forces acting on the ball are the ball's weight $m g$ and the tension $F_{r}$ in the cord. The tension must do two things: (1) balance the weight of the ball by means of its vertical component, $F_{T} \cos 30^{\circ} ;(2)$ supply the required centripetal force by means of its horizontal component, $F_{T} \sin 30^{\circ}$. Therefore, we can write $$ F_{T} \cos 30^{\circ}=m g \quad \text { and } \quad F_{T} \sin 30^{\circ}=\frac{m v^{2}}{r} $$ Solving for $F_{Y}$ in the first equation and substituting it in the second gives $$ \frac{m g \sin 30^{\circ}}{\cos 30^{\circ}}=\frac{m v^{2}}{r} \quad \text { or } \quad v=\sqrt{r g(0.577)} $$ However, $r=\overline{B C}=(0.24 \mathrm{~m}) \sin 30^{\circ}=0.12 \mathrm{~m}$ and $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$, from which $v=0.82 \mathrm{~m} / \mathrm{s}$.

As depicted in Fig. 9-5, a ball $B$ is fastened to one end of a 24 -cm string, and the other end is held fixed at point $O$. The ball whirls in the horizontal circle shown. Find the speed of the ball in its circular path if the string makes an angle of $30^{\circ}$ to the vertical. The only forces acting on the ball are the ball's weight $m g$ and the tension $F_{r}$ in the cord. The tension must do two things: (1) balance the weight of the ball by means of its vertical component, $F_{T} \cos 30^{\circ} ;(2)$ supply the required centripetal force by means of its horizontal component, $F_{T} \sin 30^{\circ}$. Therefore, we can write $$
F_{T} \cos 30^{\circ}=m g \quad \text { and } \quad F_{T} \sin 30^{\circ}=\frac{m v^{2}}{r}
$$
Solving for $F_{Y}$ in the first equation and substituting it in the second gives
$$
\frac{m g \sin 30^{\circ}}{\cos 30^{\circ}}=\frac{m v^{2}}{r} \quad \text { or } \quad v=\sqrt{r g(0.577)}
$$
However, $r=\overline{B C}=(0.24 \mathrm{~m}) \sin 30^{\circ}=0.12 \mathrm{~m}$ and $g=9.81 \mathrm{~m} / \mathrm{s}^{2}$, from which $v=0.82 \mathrm{~m} / \mathrm{s}$.

Key Concepts

Centripetal Force

Centripetal force is the inward-directed force required to keep an object moving in a circular path. It is calculated using the relation m*v²/r, where 'm' is the mass, 'v' is the speed, and 'r' is the radius of the circle. Understanding centripetal force is key to analyzing circular motion and determining the conditions required for an object to maintain uniform circular motion.

Trigonometric Relationships in Force Balance

Trigonometric functions like sine and cosine are essential tools in resolving forces into their components. They are used to relate the magnitude of forces to the angles at which they act, making it possible to equate the vertical component of a force to gravitational force and the horizontal component to the centripetal force in problems involving objects moving under the influence of multiple forces.

Conical Pendulum

A conical pendulum is a system where an object is attached to a string and moves in a horizontal circle while the string traces out a cone. Analyzing such systems involves understanding how the forces acting on the object produce both vertical and horizontal components, resulting in a balance of forces in the vertical direction and providing the necessary centripetal force in the horizontal direction.

Force Decomposition

Force decomposition is the process of breaking down a force vector into its orthogonal components. In dynamics problems, particularly those involving inclined forces or circular motion, this technique allows one to separately analyze the effects of a force in different directions, such as balancing gravitational force with the vertical component and generating centripetal force with the horizontal component.

Transcript

Please give Ace some feedback