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A body of mass \(m\) is tied to one end of a string of length \(l\) and revolves vertically in a circular path. At the lowest point of circle, what must be the \(K . E\). of the body so as to complete the circle (a) \(5 \mathrm{mgl}\) (b) \(4 \mathrm{mgl}\) (c) \(2.5 \mathrm{mgl}\) (d) \(2 \mathrm{mgl}\)

Short Answer

Expert verified
The correct kinetic energy of the body at the lowest point of the circle, in order to complete the circle, should be \(3mgl\). However, this option is not listed among the provided answers.

Step by step solution

01

Apply Conservation of Mechanical Energy

According to the conservation of mechanical energy, the sum of kinetic and potential energy in any part of the isolated system remains constant. Considering this case, the kinetic energy at the bottom (lowest point) of the vertical loop must be equal to the potential energy at the highest point of the loop plus the energy required to maintain the circular motion. So, \( K.E._{bottom} = P.E._{top} + required \quad energy \) .
02

Set up the equation

The potential energy at the highest point of the loop is equal to \(mgh = 2mgl\), because \(h=2l\), which means that the height at the highest point is double the length of the string. The required energy for a vertical circular motion is \(mgl\). Therefore, we set up the equation according to the conservation of energy equation we obtained in the first step: \(K.E._{bottom} = 2mgl + mgl \).
03

Calculating Kinetic Energy

By summing up the energies on the right side of the equation, we get \(K.E._{bottom} = 3mgl \). According to the problem, this energy should be multiplied by a certain constant to find out the correct answer among the four options. From the solution, we can see that the constant is 1. Therefore, the kinetic energy required at the bottom to complete the vertical circular motion is \(K.E._{bottom} = 3 \times 1mgl = 3mgl\). None of the given options match this result.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It depends on two factors: the mass of the object and its velocity. Mathematically, kinetic energy is expressed by the equation: \[ K.E. = \frac{1}{2}mv^2 \] where \( m \) is the mass and \( v \) is the velocity of the object.In the context of vertical circular motion, the kinetic energy is crucial at the lowest point of the circle. At this point, the velocity of the object must be high enough to ensure that it can overcome gravity and travel to the top. The solution shows how kinetic energy at this point must be sufficient to eventually equate to the combination of potential energy and the energy needed for circular motion at the top.When performing calculations that involve kinetic energy as seen in this exercise, remember to address both the mass and velocity. As the mass influences both kinetic and potential energy, understanding this relationship helps us comprehend energy transformations in motion.
Potential Energy
Potential energy is the stored energy in an object due to its position relative to other objects. For an object in a gravitational field, like Earth, it is often associated with its height above the ground. The formula for potential energy is: \[ P.E. = mgh \] where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( h \) is the height above the reference point.In this exercise, the body reaches the highest point of its motion in a vertical circle. Here, its height is maximum, thus providing it maximum potential energy. As calculated, in a vertical circle of radius \( l \), the height \( h \) at the top point is \( 2l \). Therefore, the potential energy at this point is \( 2mgl \).This part of the exercise highlights how potential energy at the highest point in a circular motion connects with kinetic energy to form the basis of energy conservation.
Vertical Circular Motion
Vertical circular motion involves an object moving in a circle in a plane perpendicular to the force of gravity. This motion combines elements of both kinetic and potential energy, and it’s crucial to understand how both forms of energy interchange as the object moves.For an object to maintain vertical circular motion, conservation of energy principles dictate that the energy balance is crucial. The object must have sufficient kinetic energy at the lowest point to convert into potential energy (while also overcoming gravitational force) to reach the top of the circle.The calculated required kinetic energy at the bottom point is actually \( 3mgl \) when adhering to energy conservation laws and accounting for gravity’s role in the motion. This ensures that while it ascends, converting kinetic energy to potential energy, it maintains forward momentum. This is why understanding the principles of vertical circular motion can illuminate how energy conservation helps achieve physical motion goals.

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Most popular questions from this chapter

A wheel of radius \(0.20 \mathrm{~m}\) is accelerated from rest with an angular acceleration of \(1 \mathrm{rad} / \mathrm{s}^{2}\). After a rotation of \(90^{\circ}\) the radial acceleration of a particle on its rim will be (a) \(\pi \mathrm{m} / \mathrm{s}^{2}\) (b) \(0.5 \pi \mathrm{m} / \mathrm{s}^{2}\) (c) \(2.0 \pi \mathrm{m} / \mathrm{s}^{2}\) (d) \(0.2 \pi \mathrm{m} / \mathrm{s}^{2}\)

The speed of a particle moving in a circle of radius \(0.1 m\) is \(v=1.0 t\) where \(t\) is time in second. The resultant acceleration of the particle at \(t=5 s\) will be (a) \(10 \mathrm{~m} / \mathrm{s}^{2}\) (b) \(100 \mathrm{~m} / \mathrm{s}^{2}\) (c) \(250 \mathrm{~m} / \mathrm{s}^{2}\) (d) \(500 \mathrm{~m} / \mathrm{s}^{2}\)

A ball of mass \(0.1 \mathrm{~kg}\) is whirled in a horizontal circle of radius \(1 \mathrm{~m}\) by means of a string at an initial speed of 10 r.p.m. Keeping the radius constant, the tension in the string is reduced to one quarter of its initial value. The new speed is (a) 5 r.p.m. (b) 10 r.p.m. (c) 20 r.p.m. (d) \(14 r . p . m\).

A particle moving along the circular path with a speed \(v\) and its speed increases by ' \(g\) ' in one second. If the radius of the circular path be \(r\), then the net acceleration of the particle is (a) \(\frac{v^{2}}{r}+g\) (b) \(\frac{v^{2}}{r^{2}}+g^{2}\) (c) \(\left[\frac{v^{4}}{r^{2}}+g^{2}\right]^{\frac{1}{2}}\) (d) \(\left[\frac{v^{2}}{r}+g\right]^{\frac{1}{2}}\)

A body of mass of \(100 \mathrm{~g}\) is attached to a \(1 \mathrm{~m}\) long string and it is revolving in a vertical circle. When the string makes an angle of \(60^{\circ}\) with the vertical then its speed is \(2 \mathrm{~m} / \mathrm{s}\). The tension in the string at \(\theta=60^{\circ}\) will be (a) \(89 \mathrm{~N}\) (b) \(0.89 \mathrm{~N}\) (c) \(8.9 \mathrm{~N}\) (d) \(0.089 \mathrm{~N}\)

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