10. A 260-g ball is tied to a string and it is being swung faster and faster in a nearly horizontal circle. When the speed reaches 16.5 m/s the tension reaches a value of 60.0 N, the string breaks. What is the radius of the circle? cm
10. A 260-g ball is tied to a string and it is being swung faster and faster in a nearly horizontal circle. When the speed reaches 16.5 m/s the tension reaches a value of 60.0 N, the string breaks. What is the radius of the circle? cm
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Question 10: Centripetal Force and Circular Motion**
A 260-g ball is tied to a string and it is being swung faster and faster in a nearly horizontal circle. When the speed reaches 16.5 m/s, the tension reaches a value of 60.0 N, and the string breaks. What is the radius of the circle?
______ cm
---
This problem involves understanding the concepts of centripetal force and circular motion. The tension in the string provides the centripetal force necessary to keep the ball moving in a circular path. The relationship between the tension (centripetal force), mass, velocity, and radius of the circle can be described by the following formula:
\[ F_c = \frac{mv^2}{r} \]
where:
- \( F_c \) is the centripetal force (tension in the string, which is 60.0 N)
- \( m \) is the mass of the ball (260 g or 0.260 kg)
- \( v \) is the velocity of the ball (16.5 m/s)
- \( r \) is the radius of the circle (in meters)
To find the radius (\( r \)), we rearrange the formula to solve for \( r \):
\[ r = \frac{mv^2}{F_c} \]
Substitute the known values into the formula:
\[ r = \frac{(0.260 \, \text{kg}) \cdot (16.5 \, \text{m/s})^2}{60.0 \, \text{N}} \]
From this calculation, you would be able to solve for the radius of the circle. Finally, remember to convert the radius from meters to centimeters by multiplying the result by 100.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcd0a9c9c-b0c8-47bb-88dc-3ac08f167320%2Fe3a954b6-a00a-40ee-b4aa-6a13c521afb9%2F1819m6s_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 10: Centripetal Force and Circular Motion**
A 260-g ball is tied to a string and it is being swung faster and faster in a nearly horizontal circle. When the speed reaches 16.5 m/s, the tension reaches a value of 60.0 N, and the string breaks. What is the radius of the circle?
______ cm
---
This problem involves understanding the concepts of centripetal force and circular motion. The tension in the string provides the centripetal force necessary to keep the ball moving in a circular path. The relationship between the tension (centripetal force), mass, velocity, and radius of the circle can be described by the following formula:
\[ F_c = \frac{mv^2}{r} \]
where:
- \( F_c \) is the centripetal force (tension in the string, which is 60.0 N)
- \( m \) is the mass of the ball (260 g or 0.260 kg)
- \( v \) is the velocity of the ball (16.5 m/s)
- \( r \) is the radius of the circle (in meters)
To find the radius (\( r \)), we rearrange the formula to solve for \( r \):
\[ r = \frac{mv^2}{F_c} \]
Substitute the known values into the formula:
\[ r = \frac{(0.260 \, \text{kg}) \cdot (16.5 \, \text{m/s})^2}{60.0 \, \text{N}} \]
From this calculation, you would be able to solve for the radius of the circle. Finally, remember to convert the radius from meters to centimeters by multiplying the result by 100.
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