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Chapter 5: Problem 114

A small block with mass \(m\) rests on a frictionless horizontal tabletop a distance \(r\) from a hole in the center of the table (Fig. 5.79\() .\) A string tied to the small block passes down through the hole, and a larger block with mass \(M\) is suspended from the free end of the string. The small block is set into uniform circular motion with radius \(r\) and speed \(v\) . What must \(v\) be if the large block is to remain motionless when released? figure can't copy

Short Answer

Expert verified
The speed \(v\) must be \(\sqrt{Gr}\).

Step by step solution

01

Understand the Forces Involved

The system consists of two blocks and a string connecting them. For the larger block with mass \(M\) to remain motionless, the tension \(T\) in the string must balance its weight, which is given by \(Mg\) where \(g\) is the acceleration due to gravity.
02

Relate Tension to Circular Motion

The tension in the string \(T\) also provides the centripetal force necessary for the small block with mass \(m\) to undergo circular motion. This centripetal force is given by \(\frac{mv^2}{r}\), where \(v\) is the speed of the block and \(r\) is the radius of the circle.
03

Set up the Equation for Motionless Larger Block

For the larger block to remain motionless, set the tension \(T\) equal to its weight: \[ T = Mg \] Similarly, set the tension equal to the centripetal force for the smaller block: \[ T = \frac{mv^2}{r} \]
04

Equate Tension Formulas

Since both expressions equal the tension \(T\), set them equal to each other: \[ Mg = \frac{mv^2}{r} \]
05

Solve for the Speed v

Solve the equation for the speed \(v\) of the smaller block: \[ Mg = \frac{mv^2}{r} \] Rearrange to find \(v\): \[ v^2 = \frac{Mgr}{m} \] \[ v^2 = Gr \] \[ v = \sqrt{Gr} \] Where \(G = \frac{M}{m}g\).

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Key Concepts

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

Centripetal Force
Centripetal force is the necessary force that keeps an object moving in circular motion. When an object is traveling in a circle, it constantly changes direction. To move in a uniform circular motion, a net inward force towards the center of the circle is required.
This force is called centripetal force. The formula for centripetal force can be expressed as \( F_c = \frac{mv^2}{r} \).
  • \(m\) is the mass of the object.
  • \(v\) is the speed of the object.
  • \(r\) is the radius of the circle.
Centripetal force is not an additional force, but it can be the result of other forces. In the exercise, it is the tension in the string that provides this inward force, allowing the block to maintain its circular path.
Tension in String
The concept of tension in a string involves the force transmitted through the string when it is pulled tight by forces acting from opposite ends. Tension is equal to the force required to keep another object at equilibrium when connected via the string.
In our exercise, the tension \(T\) in the string connects two blocks. The larger block is hanging due to gravity, while the smaller block is in circular motion on the table.
When the larger block is motionless, the tension equals its weight, \( T = Mg \), where \( M \) is the mass of the larger block and \( g \) is the acceleration due to gravity.
Simultaneously, this tension provides the centripetal force needed by the smaller block for circular motion: \( T = \frac{mv^2}{r} \). Balancing these equations helps in calculating the necessary speed \( v \) for the circular motion.
Uniform Circular Motion
Uniform circular motion refers to an object moving in a circle at a constant speed. Despite the constant speed, the object has an acceleration due to the continuous change in direction.
The key aspect of uniform circular motion is that the speed of the object remains constant while only its direction changes continuously.
In this exercise, the small block moving on the tabletop exhibits uniform circular motion. Its speed \( v \) must be such that the tension in the string balances the gravitational pull on the larger block. This ensures that the larger block remains stationary, while the smaller block moves consistently around the circle.
Understanding the balance of forces involved is critical. The force required to keep the block in uniform circular motion is provided by the tension in the string, which is set equal to the gravitational force on the larger block to achieve equilibrium.

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Most popular questions from this chapter

You observe a 1350 -kg sports car rolling along flat pavement in a straight line. The only horizontal forces acting on it are a constant rolling friction and air resistance (proportional to thesquare of its speed). You take the following data during a time interval of \(25 \mathrm{s} :\) When its speed is 32 \(\mathrm{m} / \mathrm{s}\) , the car slows down at a rate of \(-0.42 \mathrm{m} / \mathrm{s}^{2},\) and when its speed is decreased to \(24 \mathrm{m} / \mathrm{s},\) it slows down at \(-0.30 \mathrm{m} / \mathrm{s}^{2} .\) (a) Find the coefficient of rolling friction and the air drag constant \(D .(b)\) At what constant speed will this car move down an incline that makes a \(2.2^{\circ}\) angle with the horizontal? (c) How is the constant speed for an incline of angle \(\beta\) related to the terminal speed of this sports car if the car drops off a high cliff? Assume that in both cases the air resistance force is proportional to the square of the speed, and the air drag constant is the same.

Banked Curve I. A curve with a \(120-\mathrm{m}\) radius on a level road is banked at the correct angle for a speed of 20 \(\mathrm{m} / \mathrm{s}\) . If an automobile rounds this curve at 30 \(\mathrm{m} / \mathrm{s}\) , what is the minimum coefficient of static friction needed between tires and road to prevent skidding?

A 75.0 -kg wrecking ball hangs from a uniform heavy-duty chain having a mass of 26.0 \(\mathrm{kg}\) . (a) Find the maximum and minimum tension in the chain. \((b)\) What is the tension at a point three-fourths of the way up from the bottom of the chain?

A \(1125-k g\) car and a \(2250-k g\) pickup truck aproach a curve on the expressway that has a radius of 225 \(\mathrm{m}\) . (a) At what angle should the highway engineer bank this curve so that vehicles traveling at 65.0 \(\mathrm{mi} / \mathrm{h}\) can safely round it regardless of the condition of their tires? Should the heavy truck go slower than the lighter car? (b) As the car and truck round the curve at 65.0 \(\mathrm{mi} / \mathrm{h}\) , find the normal force on each one due to the highway surface.

A steel washer is suspended inside an empty shipping crate from a light string attached to the top of the crate. The crate slides down a long ramp that is inclined at an angle of \(37^{\circ}\) above the horizontal. The crate has mass 180 \(\mathrm{kg}\) . You are sitting inside the crate (with a flashlight); your mass is 55 \(\mathrm{kg}\) . As the crate is sliding down the ramp, you find the washer is at rest with respect to the crate when the string makes an angle of \(68^{\circ}\) with the top of the crate. What is the coefficient of kinetic friction between the ramp and the crate?
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