/*! 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 52 A girl, having a mass of \(15 \m... [FREE SOLUTION] | 91Ó°ÊÓ

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A girl, having a mass of \(15 \mathrm{~kg}\), sits motionless relative to the surface of a horizontal platform at a distance of \(r=5 \mathrm{~m}\) from the platform's center. If the angular motion of the platform is slowly increased so that the girl's tangential component of acceleration can be neglected, determine the maximum speed which the girl will have before she begins to slip off the platform. The coefficient of static friction between the girl and the platform is \(\mu=0.2\). Prob. \(13-52\)

Short Answer

Expert verified
The maximum speed which the girl will have before she begins to slip off the platform is approximately 9.9 m/s.

Step by step solution

01

Identify the Forces Acting on the Girl

The forces acting on the girl are the gravitational force due to her weight acting downward and the normal force applied by the platform acting upward. In a situation of equilibrium, the normal force equals the gravitational force: \(N = mg\), where \(m\) is her mass (15 kg) and \(g\) is the acceleration due to gravity (approximately 9.81 m/s²).
02

Determine the Maximum Static Friction

The maximum static friction force (F_static) can be calculated by multiplying the coefficient of static friction (µ) by the normal force (N). Therefore, \(F_{static(max)} = µN = µmg = 0.2 * 15 * 9.81 = 29.43 N\).
03

Calculate the Maximum Speed

This static friction force provides the centripetal force to keep the girl in circular motion. Using the formula for centripetal force, \(F = ma = mv^2 / r\), where \(v\) is the speed and \(r\) is the distance from the center of the platform, we can solve for \(v\). Thus, \(v= \sqrt{Fr / m}= \sqrt{29.43 * 5 / 15} = \sqrt{98.1} = 9.9 m/s.\) So, the maximum speed which the girl can have before she begins to slip off the platform is approximately \(9.9 m/s\).

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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 moves in a circular path, it experiences a force that acts towards the center of the circle. This force is known as the centripetal force. It's crucial for keeping the object in its circular motion. Without enough centripetal force, the object would fly off on a tangent. The centripetal force for an object of mass \(m\) moving at speed \(v\) at a radius \(r\) can be calculated with the formula:
  • \( F = \frac{mv^2}{r} \)
For the girl on the rotating platform, the maximum static friction force acts as the centripetal force. This friction is what keeps her from sliding off. As the platform spins faster, the required centripetal force increases, but it can't exceed the maximum static friction force. Beyond this point, the girl would begin to slip. This balance between centripetal force and static friction is key to determining the maximum speed she can achieve without slipping.
Gravitational Force
Gravitational force is the attraction between two objects with mass. For everyday scenarios on Earth, it's the force that pulls objects toward the ground. It's calculated using the formula \( F = mg \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity (approximately \( 9.81 \mathrm{~m/s^2} \)).In our exercise, the girl's weight is the gravitational force pulling her down. Her gravitational force is balanced by the normal force (the force exerted by the platform upwards) when she is sitting on the platform. The balance between these forces keeps her at rest vertically. Only the horizontally acting forces need to be adjusted to prevent slipping, which is directly managed by static friction.
Normal Force
Normal force acts perpendicular to an object's surface contact. It's a reaction to an object's weight pressing against a surface, such as a girl sitting on a platform. In physics, normal force is crucial for understanding interactions between surfaces and preventing objects from simply passing through each other.In our scenario, the normal force equals the gravitational force, given by \( N = mg \). This ensures that the girl remains stationary vertically while the platform rotates horizontally. The normal force plays a key role in calculating the maximum static friction available, which is the frictional force that can act without causing movement. Given by \( F_{static(max)} = \mu N \), the normal force directly influences the static friction's potential to act as the necessary centripetal force. Without sufficient normal force, the static friction would be inadequate, resulting in slipping at lower speeds.

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Most popular questions from this chapter

The 0.5-lb ball is guided along the vertical circular path \(r=2 r_{c} \cos \theta\) using the \(\operatorname{arm} O A\). If the arm has an angular velocity \(\dot{\theta}=0.4 \mathrm{rad} / \mathrm{s}\) and an angular acceleration \(\ddot{\theta}=0.8 \mathrm{rad} / \mathrm{s}^{2}\) at the instant \(\theta=30^{\circ}\) determine the force of the arm on the ball. Neglect friction and the size of the ball. Set \(r_{c}=0.4 \mathrm{ft}\).

The tractor is used to lift the \(150-\mathrm{kg}\) load \(B\) with the 24 -m-long rope, boom, and pulley system. If the tractor travels to the right with an acceleration of \(3 \mathrm{~m} / \mathrm{s}^{2}\) and has a velocity of \(4 \mathrm{~m} / \mathrm{s}\) at the instant \(s_{A}=5 \mathrm{~m}\), determine the tension in the rope at this instant. When \(s_{A}=0, s_{B}=0\). Probs. \(13-31 / 32\)

A motorcyclist in a circus rides his motorcycle within the confines of the hollow sphere. If the coefficient of static friction between the wheels of the motorcycle and the sphere is \(\mu_{s}=0.4\), determine the minimum speed at which he must travel if he is to ride along the wall when \(\theta=90^{\circ}\). The mass of the motorcycle and rider is \(250 \mathrm{~kg}\), and the radius of curvature to the center of gravity is \(\rho=20 \mathrm{ft}\). Neglect the size of the motorcycle for the calculation. Prob. 13-66

Block \(A\) and \(B\) each have a mass \(m\). Determine the largest horizontal force \(\mathbf{P}\) which can be applied to \(B\) so that it will not slide on \(A\). Also, what is the corresponding acceleration? The coefficient of static friction between \(A\) and \(B\) is \(\mu_{s} .\) Neglect any friction between \(A\) and the horizontal surface. Prob. \(13-33\)

The ball of mass \(m\) is guided along the vertical circular path \(r=2 r_{c} \cos \theta\) using the arm \(O A\). If the arm has a constant angular velocity \(\dot{\theta}_{0}\), determine the angle \(\theta \leq 45^{\circ}\) at which the ball starts to leave the surface of the semicylinder. Neglect friction and the size of the ball. Probs. \(13-100 / 101\)

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