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Chapter 7: Problem 77

Cp A small block with mass 0.0500 kg slides in a vertical circle of radius \(R=0.800 \mathrm{m}\) on the inside of a circular track. There is no friction between the track and the block. At the bottom of the block's path, the normal force the track exerts on the block has magnitude 3.40 \(\mathrm{N} .\) What is the magnitude of the normal force that the track exerts on the block when it is at the top of its path?

Short Answer

Expert verified
The magnitude of the normal force at the top of the path is less than 3.40 N due to reduced speed and gravity acting in the same direction.

Step by step solution

01

Understand the Problem

We have a block of mass 0.0500 kg sliding inside a frictionless vertical circle. The radius of the circle is 0.800 m. At the bottom of the track, the normal force is given as 3.40 N. We need to find the normal force at the top of the track.
02

Write the Equations of Motion

At the bottom of the circle, the forces acting on the block are gravity acting downward (mg) and the normal force (N_b) acting upward. Applying Newton's second law for motion in a circle, we have \(N_b - mg = m \frac{v_b^2}{R}\). At the top, the forces are gravity (mg) and normal force (N_t) both acting downward, so \(N_t + mg = m \frac{v_t^2}{R}\).
03

Calculate Velocity at the Bottom

Using the equation from the bottom: \(N_b - mg = m \frac{v_b^2}{R}\). Given \(N_b = 3.40 \, \text{N}\), \(m = 0.0500 \, \text{kg}\), and \(R = 0.800 \, \text{m}\), solve for \(v_b\):\[3.40 - 0.0500 \times 9.81 = 0.0500 \times \frac{v_b^2}{0.800}\].Solving gives \(v_b\).
04

Apply Conservation of Energy

Since there is no friction, mechanical energy is conserved. The energy at the bottom is \(E_{bottom} = \frac{1}{2}mv_b^2\). At the top, the energy is \(E_{top} = \frac{1}{2}mv_t^2 + mg \times 2R\). Set \(E_{bottom} = E_{top}\) to find \(v_t\).
05

Calculate the Normal Force at the Top

Using the derived formula \(N_t + mg = m \frac{v_t^2}{R}\), substitute the value of \(v_t\) obtained from energy conservation. Rearrange to calculate \(N_t\).
06

Compute Final Answer

Substitute all known values and compute \(N_t\). Use \(g = 9.81 \, \text{m/s}^2\). The solution yields the normal force at the top of the path.

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Key Concepts

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

Normal Force in Circular Motion
In circular motion, the normal force is a vital concept to understand. It is the force perpendicular to the surface of the contact, exerted by the surface to support an object. In our exercise, as the block moves in a vertical circle, the normal force changes based on its position in the circle.
  • At the bottom: The force must counteract gravity and provide enough force to keep the object moving in a circle. This is why it is greater.
  • At the top: The normal force works alongside gravity, and therefore, it is generally lesser.
Understanding these variations in the normal force is crucial because they are directly influenced by the velocity and position of the block in its path along the circle. As the block moves upward, the speed is pivotal to achieving the correct force needed for safe rotation.
Newton's Second Law in Circular Motion
Newton's Second Law is a principle that governs how forces affect motion and is written as: \( F = ma \). It plays a central role in circular motion as it connects the mass of an object, its acceleration, and the net force acting on it.
  • At the bottom of the circle: The net force is given by \( N_b - mg = m \frac{v_b^2}{R} \), where \( N_b \) is the normal force, \( mg \) is the gravitational force, and \( v_b \) is the velocity at the bottom.
  • At the top: The forces include gravity and the normal force acting together as \( N_t + mg = m \frac{v_t^2}{R} \).
Both situations emphasize that the net force is not just about pushing or pulling but about maintaining motion along the circular path. The key takeaway is that these forces ensure that the object remains on its curved path by changing direction constantly.
Conservation of Energy in Circular Motion
The conservation of energy principle is fundamental in physics and comes into play when dealing with systems where energy transitions happen, such as a block moving in a circular path.
Energy conservation states that in a closed system with no external forces like friction, the total energy remains constant. This principle has two key expressions in our scenario:
  • At the bottom: The block has kinetic energy given by \( E_{bottom} = \frac{1}{2}mv_b^2 \).
  • At the top: The energy is a combination of kinetic and potential energies, expressed as \( E_{top} = \frac{1}{2}mv_t^2 + mg \times 2R \).
By setting \( E_{bottom} = E_{top} \), we understand that the kinetic energy lost as the block ascends is converted into potential energy. This exchange ensures the block maintains the energy it needs to sustain movement, demonstrating how energy transitions well correlate with physical motion.

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Most popular questions from this chapter

A 62.0 -kg skier is moving at 6.50 \(\mathrm{m} / \mathrm{s}\) on a frictionless, horizontal, snow covered plateau when she encounters a rough patch 3.50 m long. The coefficient of kinetic friction between this patch and her skis is 0.300 . After crossing the rough patch and returning to friction- free snow, she skis down an icy, frictionless hill 2.50 m high. (a) How fast is the skier moving when she gets to the bottom of the hill? (b) How much internal energy was generated in crossing the rough patch?

CALC The potential energy of two atoms in a diatomic molecule is approximated by \(U(r)=a / r^{12}-b / r^{6},\) where \(r\) is the spacing between atoms and \(a\) and \(b\) are positive constants. (a) Find the force \(F(r)\) on one atom as a function of \(r .\) Draw two graphs. one of \(U(r)\) versus \(r\) and one of \(F(r)\) versus \(r\) . \(b\) ) Find the equilibrium distance between the two atoms. Is this equilibrium stable? (c) Suppose the distance between the two atoms is equal to the equilibrium distance found in part (b). What minimum energy must be added to the molecule to dissociate it - that is, to separate the two atoms to an infinite distance apart? This is called the dissociation energy of the molecule. (d) For the molecule CO, the equilibrium distance between the carbon and oxygen atoms is \(1.13 \times 10^{-10} \mathrm{m}\) and the dissociation energy is \(1.54 \times 10^{-18} \mathrm{J}\) per molecule. Find the values of the constants \(a\) and \(b .\)

On a horizontal surface, a crate with mass 50.0 \(\mathrm{kg}\) is placed against a spring that stores 360 \(\mathrm{J}\) of energy. The spring is released, and the crate slides 5.60 \(\mathrm{m}\) before coming to rest. What is the speed of the crate when it is 2.00 \(\mathrm{m}\) from its initial position?

EALE A force parallel to the \(x\) -axis acts on a particle moving along the \(x\) -axis. This force produces potential energy \(U(x)\) given by \(U(x)=\alpha x^{4},\) where \(\alpha=1.20 \mathrm{J} / \mathrm{m}^{4} .\) What is the force (magnitude and direction) when the particle is at \(x=-0.800 \mathrm{m}\) ?

CP A 120 -kg mail bag hangs by a vertical rope 3.5 \(\mathrm{m}\) long. A postal worker then displaces the bag to a position 2.0 \(\mathrm{m}\) sideways from its original position, always keeping the rope taut. (a) What horizontal force is necessary to hold the bag in the new position? (b) As the bag is moved to this position, how much work is done (i) by the rope and (ii) by the worker?
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