/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 A stone of mass \(m\) is whirled... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 1

A stone of mass \(m\) is whirled in a vertical circle with constant speed \(v\) by attaching it with a string. The difference in tension in the string at lowest point and highest point is (a) \(m g\) (b) \(2 \mathrm{mg}\) (c) \(\frac{2 m^{2}}{r}\) (d) Zero

Short Answer

Expert verified
The correct option is (b) 2mg.

Step by step solution

01

Understanding the Problem

A stone is moving in a vertical circle, and we need to find the difference between the tension in the string at the lowest and highest points. Here, we model the forces involved at these points.
02

Defining Forces

At the lowest point, the forces acting on the stone are tension upwards ( T_{low} ) and weight downwards ( m g ). The centripetal force required for circular motion is provided by the tension minus the weight: T_{low} - mg = rac{mv^2}{r} .
03

Calculating Tension at Lowest Point

Using the formula from the previous step, solve for T_{low} :\[ T_{low} = rac{mv^2}{r} + mg \]
04

Calculating Tension at Highest Point

At the highest point, both tension ( T_{high} ) and weight ( m g ) act in the same direction. The centripetal force is provided by the sum: T_{high} + mg = rac{mv^2}{r} . Solve for T_{high} :\[ T_{high} = rac{mv^2}{r} - mg \]
05

Finding Tension Difference

The difference in tension between the lowest and highest point is:\[ T_{low} - T_{high} = \left( \frac{mv^2}{r} + mg \right) - \left( \frac{mv^2}{r} - mg \right) = 2mg \]
06

Choose the Correct Option

After evaluating the tensions, the difference in tension is found to be 2mg . Therefore, the correct option is (b) 2mg .

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Key Concepts

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

Tension in a String
When an object, like a stone, is tied to a string and spun around in a circle, the tension refers to the force that the string exerts on the stone to keep it moving in its path. In a vertical circle, tension varies at different points in the circle due to gravitational effects.

At the lowest point, the tension is at its highest because it must overcome the force of gravity pulling downwards on the stone and provide sufficient centripetal force to keep the stone on its circular path. Mathematically, this is expressed as:
  • \( T_{low} = rac{mv^2}{r} + mg \)
At the highest point, gravity adds to the force required for circular motion, so tension is less compared to the bottom. This is expressed as:
  • \( T_{high} = rac{mv^2}{r} - mg \)
Centripetal Force
Centripetal force is critical when an object moves in a circle. It is always directed towards the center of the circle, ensuring that the object remains in a circular path instead of moving off in a straight line.

The force is calculated using the formula:
  • \( F_c = rac{mv^2}{r} \)
Where:
  • \( m \) is the mass of the object.
  • \( v \) is the speed of the object.
  • \( r \) is the radius of the circle.
In the context of the stone whirling in a circle, centripetal force changes with the tension and weight of the stone to maintain constant speed and ensure the stone follows its circular path.
Vertical Circle Dynamics
The dynamics of vertical circle motion involves the interplay of gravitational and centripetal forces. In a vertical circle, the gravitational force continuously acts downwards, affecting how forces like tension must adjust to maintain the stone's circular path.

At different points in the circle:
  • At the bottom, additional tension is needed as tension must overcome the gravitational pull, resulting in \( T_{low} = rac{mv^2}{r} + mg \).
  • At the top, gravity aids the centripetal requirement, which reduces the necessary tension, calculated as \( T_{high} = rac{mv^2}{r} - mg \).
The tension difference equation \( T_{low} - T_{high} = 2mg \) highlights how these dynamics play out, showing that extra force is needed at the bottom to maintain circular motion.
Physics Problem Solving
Solving physics problems requires understanding all forces acting on an object and how they interact. For a problem involving circular motion, these forces include tension, gravitational weight, and centripetal force.

Steps in solving such a problem typically include:
  • Determine the forces acting at each point鈥攕eparate calculations for the top and bottom of the circle are often required.
  • Apply Newton鈥檚 second law, which states that the net force is equal to mass times acceleration (which translates into the centripetal force in this case).
  • Solve for unknown forces or variables, typically tension in the case of circular motion.
  • Compare and analyze these forces to find differences or relationships between them, guiding to the correct answer.
These principles make sense of physics concepts and help effectively solve related problems.

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Most popular questions from this chapter

Which of the following force is/are not conservative? (a) \(\vec{F}-a \hat{i}+b \hat{j}\) (b) \(\vec{F}-a x \hat{i}+b y \hat{j}(a \neq b)\) (c) \(\vec{F}-x y \hat{i}\) (d) \(\vec{F}-x^{2} y \hat{i}-2 x y^{2} \hat{j}\) (e) \(\vec{F}-\left(a x+b x^{3}+c x^{4}\right) \hat{i}\) (f) \(\vec{F}-A x^{2} \hat{i}+B x y \hat{j}\)

A block of mass \(m\) moves on a horizontal circle against the wall of a cylindrical room of radius \(R\). The lloor of the room on which the block moves in smooth but the friction coefficient between the wall and the block is \(\mu\). The block is given an initial speed \(v_{0} .\) Find the power developed by the resultant force acting on the block as a function of distance travelled \(s\).

A small disc of mass \(m\) slides down a smooth hill of height \(h\) without any initial velocity and lands upon a plank of mass \(M\) lying along the horizontal plane at the base of the hill. Due to friction between the disc and the plank the disc slows down till the two move together as a single piece with a certain common velocity. Find the work done by the force of friction in the process. Solution Since mass \(m\) falls through hcight \(h\) just before landing on the plank of mass \(M\), so velocity of at this moment is \(v-\sqrt{2 g h}\). Now, the frictional force betwcen \(M\) and \(m\) is the internal force the system so momentum of system remains conserved. I lence, \(m v-(M \mid m) v^{\prime}\) or \(v^{\prime}-\frac{m v}{M+m}\) or \(v^{\prime}-\frac{m \sqrt{2 g h}}{M+m}\) So, initial kinetic cnergy of the system \(K_{i}-\frac{1}{2} m v^{2} \quad\) or \(\quad K_{i}-m g h\) \(\ldots(2)\) and linal kinctic cnergy of the system \(K_{f}-\frac{1}{2}(M+m) v^{\prime 2}-\frac{1}{2}(M+m) \frac{m^{2} \times 2 g h}{(M+m)^{2}} \quad \Rightarrow K_{f}-\frac{m^{2} g h}{M+m} \quad \ldots(3)\) So, change of cnergy \(\Delta K-K_{f} K_{t} \Rightarrow \Delta K-\frac{m M g h}{M \mid m}\) Hence, work-done by friction \(W-\Delta K \quad \therefore W--\frac{m M g h}{M \mathrm{I} m}\)

Statement-1: Work done by a force acting on a particle may be negative, even if kinetic energy of the body remains constant. Statement-2: Work done by conservative force equals negative of change in potential cnergy of Lhe body.

A particle moves with a velocity \(5 \hat{i}-3 \hat{j}+6 \hat{k} \mathrm{~m} / \mathrm{s}\) under the influenec of a constant force \(\vec{F}-10 \hat{i}+10 \hat{j}+20 \hat{k} \mathrm{~N}\). The instantaneous power applied to the particle is (a) \(200 \mathrm{~J} / \mathrm{s}\) (b) \(40 \mathrm{~J} / \mathrm{s}\) (c) \(140 \mathrm{~J} / \mathrm{s}\) (d) \(170 \mathrm{~J} / \mathrm{s}\)
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