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Chapter 10: Problem 92

A small block with mass 0.250 \(\mathrm{kg}\) is attached to a string passing through a hole in a frictionless, horizontal surface (see Fig. 10.48 ). The block is originally revolving in a circle with a radius of 0.800 \(\mathrm{m}\) m about the hole with a tangential speed of 4.00 \(\mathrm{m} / \mathrm{s}\) . The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 30.0 \(\mathrm{N}\) . What is the radius of the circle when the string breaks?

Short Answer

Expert verified
The radius when the string breaks is approximately 0.301 m.

Step by step solution

01

Understand the system

We have a block revolving in a circle on a frictionless surface. The only force acting in the radial direction is the tension in the string, which will affect the centripetal force needed to keep the mass moving in a circle.
02

Express centripetal force

The centripetal force required to keep the block moving in a circle is given by the equation:\[ F_c = \frac{m v^2}{r} \]where \( m = 0.250\,\mathrm{kg} \), \( v \) is the tangential speed, and \( r \) is the radius of the circle.
03

Use angular momentum conservation

The angular momentum \( L = m v r \) is conserved because there is no external torque. Initially, \( L_i = 0.250 \times 4.00 \times 0.800 \). When the string breaks: \( L_f = 0.250 \times v_f \times r_f \), equating \( L_i = L_f \).
04

Express final speed in terms of radius

From conservation of angular momentum, \( v_f = \frac{L_i}{m r_f} = \frac{0.8}{r_f} \times 4.00 \).
05

Insert into centripetal force equation

Insert \( v_f \) into the centripetal force equation: \[ \frac{0.250 \times \left(\frac{3.2}{r_f}\right)^2}{r_f} = 30 \]Simplify: \[ \frac{0.250 \times 10.24}{r_f^3} = 30 \]
06

Solve for the radius \( r_f \)

Rearrange and solve: \[ 0.250 \times 10.24 = 30 r_f^3 \] \[ 2.56 = 30 r_f^3 \] \[ r_f^3 = \frac{2.56}{30} \] \[ r_f = \left(\frac{2.56}{30}\right)^{1/3} \] Calculate \( r_f \) using this expression.
07

Calculate final radius

Calculate:\[ r_f = \left(\frac{2.56}{30}\right)^{1/3} \approx 0.301\,\mathrm{m} \]

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Key Concepts

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

Angular Momentum Conservation
When we talk about angular momentum, we're referring to the rotational equivalent of linear momentum. Angular momentum (\(L\)) is conserved in a system where there is no net external torque acting on it. This means that the initial angular momentum of the system will be equal to its final angular momentum. It's a crucial concept, especially in problems involving rotational motion on frictionless surfaces.

In the given scenario, a small block is revolving in a circular path. As the string is pulled, reducing the radius of this path, the no-torque condition kicks in to conserve angular momentum. This principle is expressed mathematically by:\[L_i = L_f\]where \(L_i = m v_i r_i\) and \(L_f = m v_f r_f\). Here, \(m\) is mass, \(v\) is the tangential speed, and \(r\) is the radius.

The exercise illustrates how pulling the string shortens the radius and, due to conservation, increases the tangential speed. It's like a figure skater spinning faster as they pull their arms inward. This is because the product \(mvr\) remains unchanged unless acted upon by an external force.
Tangential Speed
Tangential speed is how fast the block is moving along the circular path. It's the speed at a point in the direction tangent to its path. This speed changes when the radius of the path changes, especially when angular momentum is conserved.

In this exercise, the block's initial tangential speed is 4.00 \(\mathrm{m/s}\). As the radius decreases, it's compensated by an increase in speed to maintain the angular momentum. Think about it as a way of transferring energy from the radius to the speed. This is calculated using \( v_f = \frac{L_i}{m r_f}\), showing that tangential speed is inversely proportional to the radius when mass and angular momentum are constant.
Frictionless Surface
A frictionless surface is an idealized concept where there's no frictional force opposing motion. Hence, movement across such a surface is only influenced by other forces like tension in the string or gravity if we move vertically.

In this problem, the frictionless surface ensures that the only force acting on the block is tension, which provides the centripetal force necessary for circular motion. It's key to note that the lack of friction allows us to conserve angular momentum without losses due to frictional forces. The absence of friction simplifies the mathematics of such problems, focusing mainly on the tensions and forces affecting the tangential path of the block. This setting helps us understand purely the effects of changes in radius and how they influence speed due to tension alone.

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Most popular questions from this chapter

A 392 -N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 25.0 rad/s. The radius of the wheel is \(0.600 \mathrm{m},\) and its moment of inertia about its rotation axis is 0.800\(M R^{2}\) . Friction does work on the wheel as it rolls up the hill to a stop, a height \(h\) above the bottom of the hill; this work has absolute value 3500 \(\mathrm{J}\) . Calculate \(h\) .

Speedometer. Your car's speedometer converts the angular speed of the wheels to the linear speed of the car, assuming standard-size tires and no slipping on the pavement. (a) If your cars's standard tires are 24 inches in diameter, at what rate (in rpm) are your wheels rotating when you are driving at a freeway speed of 60 \(\mathrm{mi} / \mathrm{h} ?\) (b) Suppose you put oversize, 30 -inch-diameter tires on your car. How fast are you really going when your speedometer reads 60 \(\mathrm{mi} / \mathrm{h} ?(\mathrm{c})\) If you now put on undersize, 20 -inch-diameter tires, what will the speedometer read when you are actually traveling at 50 \(\mathrm{mi} / \mathrm{h} ?\)

A hollow, thin-walled sphere of mass 12.0 \(\mathrm{kg}\) and diameter 48.0 \(\mathrm{cm}\) is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by \(\theta(t)=A t^{2}+B t^{4},\) where \(A\) has numerical value 1.50 and \(B\) has numerical value \(1.10 .\) (a) What are the units of the constants \(A\) and \(B ?(b)\) At the time \(3.00 s,\) find \((i)\) the angular momentum of the sphere and (ii) the net torque on the sphere.

A solid, uniform cylinder with mass 8.25 \(\mathrm{kg}\) and diameter 15.0 \(\mathrm{cm}\) is spinning at 220 \(\mathrm{rpm}\) on a thin, frictionless axle that passes along the cylinder axis. You design a simple friction brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of kinetic friction between the brake and \(\mathrm{rim}\) is \(0.333 .\) What must the applied normal force be to bring the cylinder to rest after it has turned through 5.25 revolutions?

A high-wheel antique bicycle has a large front wheel with the foot-powered crank mounted on its axle and a small rear wheel turning independently of the front wheel; there is no chain connecting the wheels. The radius of the front wheel is \(65.5 \mathrm{cm},\) and the radius of the rear wheel is 22.0 \(\mathrm{cm} .\) Your modern bike has awheel diameter of 66.0 \(\mathrm{cm}(26 \text { inches) and front and rear sprockets }\) with radii of 11.0 \(\mathrm{cm}\) and \(5.5 \mathrm{cm},\) respectively. The rear sprocket is rigidly attached to the axle of the rear wheel. You ride your modern bike and turn the front sprocket at 1.00 rev \(/ \mathrm{s}\) . The wheels of both bikes roll along the ground without slipping. (a) What is your linear speed when you ride your modern bike? (b) At what rate must you turn the crank of the antique bike in order to travel at the same speed as in part (a)? (c) What then is the angular speed (in rev/s) of the small rear wheel of the antique bike?
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