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A certain string can withstand a maximum tension of \(40 \mathrm{~N}\) without breaking. A child ties a \(0.37 \mathrm{~kg}\) stone to one end and, holding the other end, whirls the stone in a vertical circle of radius \(0.91\) \(\mathrm{m}\), slowly increasing the speed until the string breaks. (a) Where is the stone on its path when the string breaks? (b) What is the speed of the stone as the string breaks?

Short Answer

Expert verified
The string breaks at the bottom of the circle with a speed of approximately 10.2 m/s.

Step by step solution

01

Identify the Forces at Play

When the stone is at the top of the circle's vertical path, the required centripetal force is the tension in the string minus the weight of the stone. At the bottom, it's the tension plus the weight of the stone. Since the string breaks at maximum tension, consider conditions where tension is maximum.
02

Calculate the Net Force at the Bottom

The tension at the bottom of the circle involves both the gravitational force (weight) and the centripetal force needed to sustain circular motion. The equations are given by:\[T - mg = \frac{mv^2}{r} \quad \text{at the bottom}\]Substituting maximum tension and solving for speed at the bottom where total tension is maximum.
03

Solve for Speed at the Bottom

Rearrange the equation from Step 2 to solve for the speed:\[T = mg + \frac{mv^2}{r} \implies v^2 = \frac{(T - mg)r}{m}\]Substitute the given values \(T = 40 \mathrm{~N}\), \(m = 0.37 \mathrm{~kg}\), \(g = 9.81 \mathrm{~m/s^2}\), and \(r = 0.91 \mathrm{~m}\).
04

Numerical Calculation for Speed

Substitute the numbers into the formula:\[v^2 = \frac{(40 - 0.37 \times 9.81) \times 0.91}{0.37}\]Calculate \(v\) by first finding the expression under the square root and then computing.
05

Determine the Location Where the String Breaks

At the bottom of the circle, the tension needs to counteract both the centripetal force and the gravitational force, reaching its maximum. This implies that the string breaks at the bottom of the circle when it experiences maximum tension.

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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 crucial when studying any kind of circular motion. It is the inward force necessary to keep an object moving in a circle. Imagine twirling a stone tied to a string. The stone wants to fly off in a straight line due to inertia, yet the tension in the string pulls it inward. This tension providing the centripetal force keeps the stone moving along its circular path.
The formula for centripetal force is given by:
  • \( F_c = \frac{mv^2}{r} \)
Here, \( F_c \) is the centripetal force, \( m \) is the mass of the object, \( v \) is the velocity, and \( r \) is the radius of the circle.
Understanding this force is key to solving physics problems involving circular motion, as it explains why objects in motion don't just fly off on a tangent.
Tension in a string
The tension in a string is the pulling force exerted by the string on the object attached to one end. In the context of the stone whirling problem, the tension in the string is crucial in maintaining the stone's circular motion.
When the stone moves in a vertical circle, the tension varies depending on its position. At the top of the circle, tension balances both gravity and the centripetal force. At the bottom, however, tension has to counteract both the weight of the stone and provide the necessary centripetal force. This is why tension is maximum at the bottom.
  • Formula at the bottom: \( T = mg + \frac{mv^2}{r} \)
Understanding how tension changes can help predict where the string might break, essentially at the point where the string can't handle the total force exerted.
Circular motion
Circular motion occurs when an object travels in a circular path with a fixed radius. This type of motion can be uniform, where speed is constant, or non-uniform, where speed changes. The stone tied to the string demonstrates non-uniform circular motion as its speed increases.
To maintain circular motion, a centripetal force is necessary, directed towards the center of the circle. In most real-life scenarios, like the whirling stone, this force is provided by the tension in the string.
The dynamics of circular motion make it a complex topic, but understanding the interplay of forces like tension and centripetal force helps make sense of how objects behave in such motions.
Vertical circle dynamics
Vertical circle dynamics deal with objects moving in a vertical circular path. This motion is similar to swinging on a swing or the path of a looping roller coaster.
The critical point in vertical circle dynamics is recognizing how forces like gravity modify the tension in a string as an object moves around the circle. At the top, gravity helps pull the object down, reducing the tension needed. Conversely, at the bottom, tension must counteract gravity's pull and provide additional centripetal force.
By calculating these forces at different points, you can solve problems involving maximum tension that a string can handle before breaking, or determining speeds at particular points in the motion. Understanding these dynamics is essential for mastering problems relating to vertical circular motions.

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Most popular questions from this chapter

An airplane is flying in a horizontal circle at a speed of \(480 \mathrm{~km} / \mathrm{h}\) (Fig. 6-41). If its wings are tilted at angle \(\theta=40^{\circ}\) to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an "aerodynamic lift" that is perpendicular to the wing surface.

Two blocks, of weights \(3.6 \mathrm{~N}\) and \(7.2 \mathrm{~N}\), are connected by a massless string and slide down a \(30^{\circ}\) inclined plane. The coefficient of kinetic friction between the lighter block and the plane is \(0.10\), and the coefficient between the heavier block and the plane is 0.20. Assuming that the lighter block leads, find (a) the magnitude of the acceleration of the blocks and (b) the tension in the taut string.

A child places a picnic basket on the outer rim of a merrygo-round that has a radius of \(4.6 \mathrm{~m}\) and revolves once every \(30 \mathrm{~s}\). (a) What is the speed of a point on that rim? (b) What is the lowest value of the coefficient of static friction between basket and merry-go-round that allows the basket to stay on the ride?

A circular curve of highway is designed for traffic moving at \(60 \mathrm{~km} / \mathrm{h}\). Assume the traffic consists of cars without negative lift. (a) If the radius of the curve is \(150 \mathrm{~m}\), what is the correct angle of banking of the road? (b) If the curve were not banked, what would be the minimum coefficient of friction between tires and road that would keep traffic from skidding out of the turn when traveling at \(60 \mathrm{~km} / \mathrm{h} ?\)

A warehouse worker exerts a constant horizontal force of magnitude \(85 \mathrm{~N}\) on a \(40 \mathrm{~kg}\) box that is initially at rest on the horizontal floor of the warehouse. When the box has moved a distance of \(1.4 \mathrm{~m}\), its speed is \(1.0 \mathrm{~m} / \mathrm{s}\). What is the coefficient of kinetic friction between the box and the floor?

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