Talking about the Relationship between Light Rope, Light Rod and Light Spring
August 11 07:18:35, 2025
The light bar can exert both tension and normal force on the object, unlike a light rope that only provides tension. This allows the pendulum ball to perform a complete circular motion in the vertical plane. At the lowest point of the path, the ball must have a minimum velocity $ v_0 $ to maintain the circular motion. At the highest point, the ball can experience either a pulling force or a supporting force from the light bar. Therefore, at the top, the velocity can be zero if the bar is able to provide the necessary support. This means that the ball must have enough speed at the top to stay on the track, and the critical velocity at the top is $ v = \sqrt{gL} $. If the velocity is less than this, the bar will pull downward on the ball. Using the principle of conservation of mechanical energy, from the bottom to the top of the circle, we get:
$$
\frac{1}{2}mv_0^2 = 2mgL \Rightarrow v_0 = \sqrt{4gL}
$$
Thus, for the ball to complete the circular motion, its velocity at the bottom must be greater than or equal to $ \sqrt{4gL} $. This situation is similar to a ball moving inside a vertical circular tube, where the tube can push or pull the ball depending on the position.
In another example, a light spring with original length $ L_0 $ and spring constant $ k $ is attached to an object of mass $ m $, and the other end is fixed at point O on a rotating turntable. The block rotates along with the turntable at an angular velocity $ \omega $. The maximum static friction between the block and the turntable is $ f_m $. We are asked to determine the possible range of positions where the block can remain without slipping.
When the radius of rotation is at its minimum, $ r_1 $, the spring is compressed by $ L_0 - r_1 $. The net centripetal force comes from the static friction and the spring force. So,
$$
f_m - k(L_0 - r_1) = m\omega^2 r_1 \Rightarrow r_1 = \frac{f_m - kL_0}{m\omega^2 - k}
$$
When the radius is at its maximum, $ r_2 $, the spring is stretched by $ r_2 - L_0 $. In this case, the spring force and the static friction both act toward the center, so:
$$
k(r_2 - L_0) - f_m = m\omega^2 r_2 \Rightarrow r_2 = \frac{f_m + kL_0}{k - m\omega^2}
$$
Therefore, the block can rotate within the range:
$$
\frac{f_m - kL_0}{m\omega^2 - k} \leq r \leq \frac{f_m + kL_0}{k - m\omega^2}
$$
This analysis shows how the interplay between spring force and static friction determines the stable positions of the block on the turntable.
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