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Chapter 5: Problem 121

Vertical circular motion is an example of non-uniform motion and for completion of vertical circular motion the maximum velocity at the start of vertical motion from lowest point is (A) \(\sqrt{g R}\) (B) \(\geq \sqrt{5 g R}\) \((\mathrm{C})=\sqrt{5 g R}\) (D) \(\leq \sqrt{5 g R}\)

Short Answer

Expert verified
The maximum velocity at the start of vertical circular motion for the completion of the vertical circular motion is \(v_{start} \geq \sqrt{5 g R}\). The correct answer is (B) \(\geq \sqrt{5 g R}\).

Step by step solution

01

Calculate the minimum tension at the highest point.

At the highest point of the vertical circular motion, the weight of the object is equal to its centripetal force, resulting in a minimum tension of zero. Therefore, we can write the following equation: \(T_{min} - m g = - m \frac{v_{top}^2}{R}\) Since we want \(T_min = 0\), we can rewrite the equation as follows: \(m g = m \frac{v_{top}^2}{R}\) We can now solve for \(v_{top}\), the velocity at the top of the circular path: \(v_{top} = \sqrt{g R}\)
02

Apply conservation of mechanical energy.

As the object moves along the vertical circular path, the mechanical energy is conserved. At the lowest point, the object has only kinetic energy, and at the highest point, it has both kinetic and gravitational potential energy. We can write the conservation of mechanical energy equation as follows: \(K_{bottom} = K_{top} + U_{top}\) At the bottom, kinetic energy is given by: \(K_{bottom} = \frac{1}{2} m v_{start}^2\) At the top, kinetic energy is given by: \(K_{top} = \frac{1}{2} m v_{top}^2 = \frac{1}{2} m(\sqrt{g R})^2 = \frac{1}{2} m g R\) and the gravitational potential energy is given by: \(U_{top} = m g h = m g (2 R)\) Plugging these expressions back into the conservation of mechanical energy equation: \( \frac{1}{2} m v_{start}^2 = \frac{1}{2} m g R + m g (2 R)\)
03

Solve for the maximum velocity at the start.

Now we solve for \(v_{start}\), the maximum velocity required at the start of the circular motion, by rearranging the equation: \(v_{start}^2 = g R + 4 g R\) \(v_{start} = \sqrt{5 g R}\) Therefore, the maximum velocity at the start of vertical circular motion for the completion of the vertical circular motion is: \(v_{start} \geq \sqrt{5 g R}\) The correct answer is (B) \(\geq \sqrt{5 g R}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Non-uniform Motion
When we talk about non-uniform motion, we鈥檙e discussing movement in which the speed of the moving object is not consistent. In vertical circular motion, the speed of an object changes due to the influence of gravity. As the object travels upwards in its path, gravity slows it down until it reaches the top, where its speed is the least. Then, as it descends, gravity accelerates it, increasing its speed until it reaches the bottom. This speed variability with respect to time is the hallmark of non-uniform motion.

In the context of the vertical circular motion problem provided, the varying speeds at different points in the orbit mean that the object is in non-uniform motion. Understanding this concept is critical for students, as it helps in comprehending why the minimum speed required at the lowest point (to complete the circle) depends on the effects of gravity acting throughout the motion.
Conservation of Mechanical Energy
The conservation of mechanical energy principle states that if no external work is done on a system, the total mechanical energy (the sum of potential and kinetic energies) remains constant. This principle is a powerful tool for solving problems involving the motion of objects.

In our vertical circular motion scenario, the mechanical energy at the bottom where the object starts must be enough to reach the top and overcome the gravitational pull. At the bottom, all of the object鈥檚 mechanical energy is kinetic because it is at its lowest gravitational potential point. At the top, the object鈥檚 mechanical energy is a mix of potential energy (due to its height) and kinetic energy (due to its motion). By equating the mechanical energy at these two points, we calculated the necessary starting velocity for the object to just make it to the top of the circle, underlining the importance of the conservation of mechanical energy in determining motion outcomes.
Centripetal Force
The concept of centripetal force is critical when discussing circular motion. It is the inward force necessary for an object to follow a curved path, and it is always directed towards the center of the circle. For vertical circular motion, gravity is part of the centripetal force needed to keep the object moving along its path.

At the highest point in vertical circular motion, the centripetal force is solely provided by gravity's pull (hence, the tension in the string is zero), as demonstrated in the step-by-step solution. The equation for centripetal force, \( m \frac{v^2}{R} \) (where \(m\) is mass, \(v\) is velocity, and \(R\) is the radius of the circle), shows that velocity is crucial for maintaining circular motion. If the speed isn鈥檛 high enough at the top, the object won鈥檛 have the required centripetal force to stay in the circular path, which may result in it falling under the influence of gravity rather than completing the circle.

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Most popular questions from this chapter

Particle A makes a perfectly elastic head-on collision with another stationary particle \(B\). They fly apart in opposite directions with equal velocities. Ratio of their masses \(\frac{M_{A}}{M_{B}}\) will be (A) \(\frac{1}{3}\) (B) \(\frac{1}{2}\) (C) \(\frac{1}{4}\) (D) \(\frac{1}{\sqrt{3}}\)

A gun fires a shell and recoils horizontally. If the shell travels along the barrel with speed \(v\), the ratio of speeds with which the gun recoils, if the barrel is (i) horizontal (ii) inclined at an angle of \(30^{\circ}\) with horizontal, is (A) 1 (B) \(\frac{2}{\sqrt{3}}\) (C) \(\frac{\sqrt{3}}{2}\) (D) \(\frac{1}{2}\)

Two blocks of masses \(3 \mathrm{~kg}\) and \(6 \mathrm{~kg}\) are connected by an ideal spring and are placed on a frictionless horizontal surface. The \(3 \mathrm{~kg}\) block is imparted a speed of \(2 \mathrm{~m} / \mathrm{s}\) towards left. (consider left as positive direction) Column-I (A) When the velocity of \(3 \mathrm{~kg}\) block is \(\frac{2}{3} \mathrm{~m} / \mathrm{s}\). (B) When the speed of 3 \(\mathrm{kg}\) block is \(\frac{2}{3} \mathrm{~m} / \mathrm{s}\). (C) When the speed of \(3 \mathrm{~kg}\) block is minimum. (D) When the velocity of \(6 \mathrm{~kg}\) block is maximum. Column-II 1\. Velocity of centre of mass is \(\frac{2}{3} \mathrm{~m} / \mathrm{s}\). 2\. Deformation of the spring is zero. 3\. Deformation of the spring is maximum. 4\. Both the blocks are at rest with respect to each other.

Two identical billiard balls are in contact on a table. A third identical ball strikes them symmetrically with velocity \(v\) and remains at rest after impact. The speed of balls after collision will be (A) \(\frac{v}{\sqrt{3}}\) (B) \(\frac{v}{3}\) (C) \(\frac{v}{2}\) (D) \(v\)

Two skaters of masses \(40 \mathrm{~kg}\) and \(60 \mathrm{~kg}\) respectively stand facing each other at \(S_{1}\) and \(S_{2}\) where \(S_{1} S_{2}\) is \(5 \mathrm{~m}\). They pull on a massless rope stretched them, then they meet at (A) \(2.5 \mathrm{~m}\) from \(S_{1}\) and \(S_{2}\) (B) \(3 \mathrm{~m}\) from \(S_{1}\) and \(2 \mathrm{~m}\) from \(S_{2}\) (C) \(2 \mathrm{~m}\) from \(S_{1}\) and \(3 \mathrm{~m}\) from \(S_{2}\) (D) \(3 \mathrm{~m}\) from \(S_{1}\) and \(8 \mathrm{~m}\) from \(S_{2}\)
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