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Chapter 2: Problem 6

Mass in cone A particle of mass \(m\) slides without friction on the inside of a cone. The axis of the cone is vertical, and gravity is directed downward. The apex half-angle of the cone is \(\theta\), as shown. The path of the particle happens to be a circle in a horizontal plane. The speed of the particle is \(v_{0} .\) Draw a force diagram and find the radius of the circular path in terms of \(v_{0}, g\), and \(\theta\).

Short Answer

Expert verified
The radius is \(r = \frac{v_{0}^2}{g \sin\theta}\).

Step by step solution

01

Understand the forces involved

First, identify the forces acting on the particle. The particle slides inside the cone, so there are two forces to consider: gravity acting downward, which is the weight of the particle, given by \(mg\), and the normal force exerted by the cone surface on the particle.
02

Determine effective forces in horizontal and vertical directions

Since the path of the particle is a horizontal circle, we focus on forces in the horizontal plane. The components of the gravitational force parallel and perpendicular to the cone鈥檚 surface are \(mg\sin\theta\) and \(mg\cos\theta\) respectively. The normal force balances the perpendicular component \(mg\cos\theta\).
03

Relate horizontal force to circular motion

The horizontal component of the gravitational force, \(mg\sin\theta\), provides the centripetal force necessary for circular motion. According to Newton's second law, the centripetal force \(F_c\) is equal to \(\frac{mv_{0}^2}{r}\), where \(r\) is the radius of the circular path.
04

Equate expressions for centripetal force and solve for radius

Equating the horizontal force component to the centripetal force gives the equation: \(mg\sin\theta = \frac{mv_{0}^2}{r}\). Solving for \(r\), we get \(r = \frac{v_{0}^2}{g\sin\theta}\).
05

Final answer for the radius

Thus, the radius of the circular path is expressed in terms of \(v_{0}\), \(g\), and \(\theta\) as \(r = \frac{v_{0}^2}{g \sin\theta}\).

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Key Concepts

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

Centripetal Force
When an object travels in a circular path, it experiences an inward force known as the centripetal force. This force is crucial for maintaining the object's curved trajectory. Without centripetal force, the object would merely continue in a straight line due to inertia. The formula for centripetal force is given by \( F_c = \frac{mv^2}{r} \), where \( m \) is the mass of the object, \( v \) is its velocity, and \( r \) is the radius of the circular path. To better understand, imagine twirling a ball tied to a string. Your hand pulls the string towards the center, supplying the necessary centripetal force to keep the ball moving in a circle. In our cone scenario, the horizontal component of gravitational force acts as the centripetal force, pulling the particle towards the center of its circular path.
Newton's Laws of Motion
Sir Isaac Newton's laws of motion lay the groundwork for classical mechanics. Particularly, his second law is of significance in our problem. It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass, summed up by the equation \( F = ma \). This principle is key in understanding how forces affect motion. In the cone exercise, Newton's second law assists us in finding the centripetal force necessary for circular motion. The balance of forces, where the normal force opposes the vertical component of gravity, helps maintain the object's horizontal circular path. We've used \( F = ma \) to solve for the radius of the path, demonstrating how these laws are applied in real-life physics scenarios.
Circular Motion
Circular motion involves an object traveling along a circular path. It's a type of motion prevalent in numerous natural and man-made systems, like celestial bodies orbiting a star or wheels spinning on a vehicle. One distinctive feature of circular motion is the continuous change in the object's direction. For an object in uniform circular motion (constant speed), the velocity remains constant in magnitude but alters in direction. This change necessitates an inward force, which is the centripetal force. In analyzing the cone problem, the circular path of the particle is made possible by the horizontal component of gravity acting towards the cone's central axis. Understanding circular motion fosters insights into how objects behave in rotational systems and is fundamental in studying mechanics.

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Most popular questions from this chapter

Concrete mixer' In a concrete mixer, cement, gravel, and water are mixed by tumbling action in a slowly rotating drum. If the drum spins too fast the ingredients stick to the drum wall instead of mixing. Assume that the drum of a mixer has radius \(R=0.5 \mathrm{~m}\) and that it is mounted with its axle horizontal. What is the fastest the drum can rotate without the ingredients sticking to the wall all the time? Assume \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\)

Two blocks on table Two blocks \(m_{1}\) and \(m_{2}\) are in contact on a horizontal table. A horizontal force is applied to one of the blocks, as shown in the drawing. If \(m_{1}=2 \mathrm{~kg}, m_{2}=1 \mathrm{~kg}\), and \(F=3 \mathrm{~N}\), find the force of contact between the two blocks.

Painter on scaffold \(^{*}\) A painter of mass \(M\) stands on a scaffold of mass \(m\) and pulls himself up by two ropes which hang over pulleys, as shown. He pulls each rope with force \(F\) and accelerates upward with a uniform acceleration \(a\). Find \(a\) -neglecting the fact that no one could do this for long.

Disk with catch A disk rotates with constant angular velocity \(\omega\), as shown. Two masses, \(m_{A}\) and \(m_{B}\), slide without friction in a groove passing through the center of the disk. They are connected by a light string of length \(l\), and are initially held in position by a catch, with mass \(m_{A}\) at distance \(r_{A}\) from the center. Neglect gravity. At \(t=0\) the catch is removed and the masses are free to slide. Find \(\ddot{r}_{a}\) immediately after the catch is removed, in terms of \(m_{A}\), \(m_{B}, l, r_{A}\), and \(\omega .\)

Mass on wedge" A \(45^{\circ}\) wedge is pushed along a table with constant acceleration \(A\). A block of mass \(m\) slides without friction on the wedge. Find the block's acceleration. Gravity is directed down.
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