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Chapter 9: Problem 72

A ball having a mass of \(1.00 \mathrm{~kg}\) is attached to a string \(1.00 \mathrm{~m}\) long and is whirled in a vertical circle at a constant speed of \(10.0 \mathrm{~m} / \mathrm{s}\) a) Determine the tension in the string when the ball is at the top of the circle. b) Determine the tension in the string when the ball is at the bottom of the circle. c) Consider the ball at some point other than the top or bottom. What can you say about the tension in the string at this point?

Short Answer

Expert verified
Answer: The tension force acting on the ball is 9.81 N at the top of the circle and 110.81 N at the bottom of the circle.

Step by step solution

01

Identify the forces acting on the ball when it is at the top of the circle

At the top of the circle, the two forces acting on the ball are: 1. Gravitational force (weight), \(mg\), acting vertically downward. 2. Tension force, \(T\), acting along the string towards the center. Since both forces have the same direction at the top of the circle, they are added together.
02

Apply Newton's second law for circular motion at the top of the circle

Newton's second law for circular motion states that the net force towards the center of the circle is equal to the mass of the object multiplied by its centripetal acceleration, \(F_{net} = ma_c\). In our case, \(F_{net} = T + mg\), and the centripetal acceleration, \(a_c = \dfrac{v^2}{r}\), where \(v\) is the constant speed of the ball and \(r\) is the radius of the circle. For the given problem, we have \(m = 1.00 kg\), \(v = 10.0 m/s\), and \(r = 1.00 m\). We can plug in these values and solve for the tension: \(T + mg = ma_c\)
03

Calculate tension at the top of the circle

Plugging in the values for mass, speed, and radius, we get: \(T + (1.00 kg)(9.81 m/s^2) = (1.00 kg)\left(\dfrac{(10.0 m/s)^2}{1.00 m}\right)\) After solving the equation for \(T\), we get: \(T = - 9.81 N\) However, tension cannot be negative. The negative sign indicates that our assumption about the direction of tension at the top of the circle is incorrect. Hence, \(T = 9.81 N\) at the top of the circle.
04

Identify the forces acting on the ball when it is at the bottom of the circle

At the bottom of the circle, the forces acting on the ball are: 1. Gravitational force (weight), \(mg\), acting vertically downward. 2. Tension force, \(T\), acting along the string towards the center. In this case, these forces have opposite directions, so the tension force will be greater than the gravitational force.
05

Apply Newton's second law for circular motion at the bottom of the circle

Using Newton's second law for circular motion, and noting that tension force is greater than gravitational force, we have: \(F_{net} = T - mg = ma_c\)
06

Calculate tension at the bottom of the circle

Plugging in the values for mass, speed, and radius, we get: \(T - (1.00 kg)(9.81 m/s^2) = (1.00 kg)\left(\dfrac{(10.0 m/s)^2}{1.00 m}\right)\) Solving this equation for \(T\), we obtain: \(T = 110.81 N\)
07

Discuss the tension in the string at any other point in the circle

At any point other than the top or the bottom of the circle, the tension force will have a component in the radial direction, which will contribute to the centripetal force. The gravitational force will also have radial and tangential components. Since these components would vary depending on the angle, the tension in the string will have different values at different points in the circle. In general, the tension will be greater at the bottom and decrease as the ball moves towards the top of the circle, reaching its minimum value at the top. Then, the tension will increase again as the ball moves from the top to the bottom of the circle.

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