Question

An object of mass $0.675\mathrm{kg}$ on a frictionless table is attached to a string that passes through a hole in the table at the center of the horizontal circle in which the object moves with constant speed. (a) If the radius of the circle is $0.500\mathrm{m}$ and the speed is $10.0\mathrm{m}/\mathrm{s}$, compute the tension in the string. $(b)$ It is found that drawing an additional $0.200\mathrm{m}$ of the string down through the hole, thereby reducing the radius of the circle to $0.300\mathrm{m}$, has the effect of multiplying the original tension in the string by $4.63.$ Compute the total work done by the string on the revolving object during the reduction of the radius.

Short answer

The tension in the string is $135\mathrm{N}$ and the work done by the string during the radius reduction is $201.64\mathrm{J}$.

Step-by-step solution

Calculate the Tension in the String

For circular motion, the tension in the string provides the necessary centripetal force. The centripetal force can be calculated using the formula $F=ma$, where $m$ is the mass and $a$ is the acceleration. Since this involves circular motion, the acceleration is the centripetal acceleration, $a_c$, which can be computed using $a_c=v^2/r$, where $v$ is the speed and $r$ is the radius. In this exercise, the speed $v$ is provided as $10.0\mathrm{m}/\mathrm{s}$, the radius $r$ is given as $0.500\mathrm{m}$, and the mass $m$ is given as $0.675\mathrm{kg}$. Substituting the above values, we get $a_c=v^2/r=(10.0{)}^2/0.500=200\mathrm{m}/{\mathrm{s}}^2$. Further, the tension $T$ in the string, which provides the centripetal force, is $T=m\cdot a_c=0.675\mathrm{kg}\cdot 200\mathrm{m}/{\mathrm{s}}^2=135\mathrm{N}$.

Calculate the New Tension in the String

In step-2, it is necessary to calculate the new tension when the radius of the circle is reduced to $0.300\mathrm{m}$. This reduction increases the initial tension by a factor of $4.63$. Therefore, the new tension $T_2$ is $T_2=T\cdot 4.63=135\mathrm{N}\cdot 4.63=625.05\mathrm{N}$.

Calculate the Total Work Done

The work done by the tension in the string is calculated using the work-energy theorem stating that work is the change in kinetic energy. The formula for kinetic energy is $KE=0.5\cdot m\cdot v^2$. The initial and final kinetic energies are calculated as $K{E}_1=0.5\cdot m\cdot v^2=0.5\cdot 0.675\mathrm{kg}\cdot (10.0\mathrm{m}/\mathrm{s}{)}^2=33.75\mathrm{J}$ and $K{E}_2=0.5\cdot m\cdot v^{\prime 2}=0.5\cdot 0.675\mathrm{kg}\cdot v^{\prime 2}=0.5\cdot 0.675\mathrm{kg}\cdot (\sqrt{T_2/m\cdot r_2}{)}^2=0.5\cdot 0.675\mathrm{kg}\cdot (625.05\mathrm{N}/0.675\mathrm{kg}\cdot 0.300\mathrm{m})=235.39\mathrm{J}$, where $v^{\prime }$ is the new speed after reduction in the radius and $r_2$ is the new radius $0.300\mathrm{m}$. The change in kinetic energy gives us the work done, i.e., $W=K{E}_2-K{E}_1=235.39\mathrm{J}-33.75\mathrm{J}=201.64\mathrm{J}$.

Key concepts

Centripetal Force

Centripetal force is a fundamental concept in circular motion physics. It's the inward force required to keep an object moving in a circular path. Without this force, an object would continue moving in a straight line due to inertia. In our exercise, the tension in the string provides the necessary centripetal force for the mass to revolve on a frictionless table.

Mathematically, it is calculated using the formula: $F=ma$ where $m$ is the mass, and $a$ is the centripetal acceleration, represented by $a_c=\frac{v^2}{r}$. Here, $v$ is the tangential speed of the object, and $r$ is the radius of the circular path.

The exercise exemplifies this concept by using the provided mass, speed, and radius to find the initial tension, which equals the required centripetal force to maintain circular motion. Understanding how these variables influence the force is key to grasping circular motion.

Work-Energy Theorem

The work-energy theorem is a vital principle in physics that connects work done on an object to its change in kinetic energy. It states that the net work done by forces acting on an object is equal to the change in its kinetic energy. In the context of our problem, this theorem allows us to calculate the total work done on the object as it moves to a smaller radius circle.

When the radius changes, the speed of the object also changes to maintain the centripetal force balance, which is reflected by the tension in the string. The work done is computed by finding the difference in kinetic energies before and after the string is shortened. This gives us a way to quantify the energy transferred to the object by the tension force as it transitions to a smaller circular path. Recognizing how work relates to energy changes is essential for unraveling problems involving force and movement.

Kinetic Energy

Kinetic energy represents the energy an object possesses due to its motion. It is directly proportional to the mass of the object and the square of its velocity, according to the formula: $KE=\frac{1}{2}m{v}^2$.

In the exercise, we use the kinetic energy formula to determine the object's energy at two different points: the initial larger radius and the final smaller radius after the tension is increased. The calculation involves using the mass of the object and the respective speeds before and after the radius change. Understanding kinetic energy and its calculation is crucial in solving various physics problems, especially those involving motion, like the one in our exercise. Recognizing that kinetic energy changes with speed—and indirectly with tension, as influenced by circular motion dynamics—is imperative for fully grasping the concepts at hand.