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A puck of mass m 1.50 kg slides in a circle of radius r 25.0 cm on a frictionless table while attached to a hanging cylinder of mass M 2.50 kg by means of a cord that extends through a hole in the table (Fig. 6-45). What speed keeps the cylinder at rest?

Short Answer

Expert verified
The speed is approximately 2.02 m/s.

Step by step solution

01

Understanding the Problem

We have two masses: a puck with mass \(m = 1.50\, \text{kg}\) sliding in a circle on a table, and a hanging cylinder with mass \(M = 2.50\, \text{kg}\). The challenge is to find the speed \(v\) of the puck that keeps the system in equilibrium, meaning the cylinder remains at rest.
02

Identifying Forces in Motion

Because the cylinder hangs at rest, the tension \(T\) in the string holding it is equal to the weight of the cylinder, which is \(T = Mg\), where \(g = 9.81 \, \text{m/s}^2\). Therefore, the tension in the cord is \(T = 2.50 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 24.525 \, \text{N}\).
03

Using Centripetal Force

Since the puck moves in a circle, it experiences a centripetal force. This force is provided by the tension in the string, \(T\). According to the centripetal force formula \(F_c = \frac{mv^2}{r}\), we know that this force is equal to the tension in the string. Therefore, we equate the two: \[ \frac{mv^2}{r} = T \].
04

Solving for Speed

Plug in the known values into the centripetal force equation: \(m = 1.50\, \text{kg}\), \(r = 0.25 \, \text{m}\), and \(T = 24.525 \, \text{N}\). Solve for \(v\):\[ \frac{1.50 \, \text{kg} \cdot v^2}{0.25 \, \text{m}} = 24.525 \, \text{N} \]\[ v^2 = \frac{24.525 \, \text{N} \times 0.25 \, \text{m}}{1.50 \, \text{kg}} \]\[ v^2 = 4.0875 \]\[ v = \sqrt{4.0875} \approx 2.02 \, \text{m/s} \].

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Key Concepts

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

circular motion
When an object moves in a circular path, it experiences a type of motion called circular motion. This motion involves continuous change in the direction of the velocity, which means there is a constant acceleration towards the center of the circle. This inward acceleration is what keeps the object moving along the circular path.
In the case of our puck on the table, the puck is sliding in a circular motion with a radius of 25 cm. This path is maintained by a force that constantly pulls the puck inward. Think of it as how a tetherball is pulled towards the pole when it's swung around. In our problem, once we find the correct speed, the puck maintains this circular motion without deviating because the necessary centripetal force is balanced by the tension in the cord.
tension in cords
Tension is a force exerted along the length of a cord or rope. It is the force that is transmitted through a string when it is pulled tight by forces acting from opposite ends. In our exercise, the tension in the cord plays a crucial role.
Here, the hanging cylinder exerts a downward force equal to its weight, given by the formula \( T = Mg \). This tension serves two purposes: it keeps the puck moving in a circle and also supports the cylinder to remain at rest. Therefore, the tension in the cord is both the centripetal force for the puck and the counteracting force of gravity for the cylinder. This gives us a neat way to link the movement of the puck to the static nature of the cylinder through the concept of tension.
equilibrium
Equilibrium in physics refers to a state where all forces acting on a system balance each other, resulting in no net force. In our problem, equilibrium is achieved when the cylinder remains at rest. This occurs because the upward force (tension in the cord) equals the downward force (gravity on the cylinder).
Since the cylinder hangs motionless, the forces along the vertical direction are balanced. This balance is possible because the movement of the puck provides exactly the right amount of centripetal force through the tension needed to counteract gravity pulling the cylinder downwards. Achieving such equilibrium means the system is perfectly poised — the puck's speed matches what is needed to prevent any movement in the cylinder.
Newton's second law
Newton's Second Law states that the force acting on an object is equal to the mass of that object multiplied by its acceleration, given by the formula \( F = ma \). This is pivotal in analyzing the forces in our exercise.
For the puck moving in a circle, this law helps us understand that the centripetal force required for the circular motion is dictated by the mass of the puck and its velocity. In mathematical terms, this is expressed as \( F_c = \frac{mv^2}{r} \).
Applying Newton's Second Law allows us to relate the motion of the puck to the tension in the cord and the stationary cylinder. By doing so, we can derive the exact speed at which the puck must move to ensure the cylinder stays put, illustrating the elegant balance of forces at play. This approach underscores the power of Newton's law in solving complex motion problems simply by understanding the interplay of mass, acceleration, and force.

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Most popular questions from this chapter

In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallway. If the 3.5 kg book is pushed from rest through a distance of 1.20 m by the horizontal 25 N force from the broom and then has a speed of 1.75 m/s, what is the coefficient of kinetic friction between the book and floor?

A water droplet 4.0 mm in diameter is falling with a speed of 10 km/h at an altitude of 20 km. Another droplet 6.0 mm in diameter is falling at 25% of that speed and at 25% of that altitude. The density of air at 20 km is 0.20 kg/m3 and that at 5.0 km is 0.70 kg/m3 . Assume that the drag coefficient C is the same for the two drops. Find the ratio of the drag force on the higher drop to that on the lower drop.

A 3.5 kg block is pushed along a horizontal floor by a force of magnitude 15 N at an angle 40 with the horizontal (Fig. 6-17). The coefficient of kinetic friction between the block and the floor is 0.25. Calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block’s acceleration.

A circular-motion addict of mass 80 kg rides a Ferris wheel around in a vertical circle of radius 12 m at a constant speed of 5.5 m/s. (a) What is the period of the motion? What is the magnitude of the normal force on the addict from the seat when both go through (b) the highest point of the circular path and (c) the lowest point?

A worker pushes horizontally on a 35 kg crate with a force of magnitude 110 N.The coefficient of static friction between the crate and the floor is 0.37. (a) What is the value of fs,max under the circumstances? (b) Does the crate move? (c) What is the frictional force on the crate from the floor? (d) Suppose, next, that a second worker pulls directly upward on the crate to help out. What is the least vertical pull that will allow the first worker’s 110 N push to move the crate? (e) If, instead, the second worker pulls horizontally to help out, what is the least pull that will get the crate moving?

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