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Chapter 9: Problem 7

A ball attached to the end of a string is swung around in a circular path of radius \(r\). If the radius is doubled and the linear speed is kept constant, the centripetal acceleration a) remains the same. b) increases by a factor of 2 . c) decreases by a factor of 2 . d) increases by a factor of 4 e) decreases by a factor of 4 .

Short Answer

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Question: When a ball swings around in a circular path with a constant linear speed and the radius is doubled, the centripetal acceleration: a) stays the same b) increases by a factor of 2 c) decreases by a factor of 2 d) increases by a factor of 4 Answer: c) decreases by a factor of 2.

Step by step solution

01

Recall the formula for centripetal acceleration

The formula for centripetal acceleration is given by: \( a_c = \frac{v^2}{r} \) where \(a_c\) is the centripetal acceleration, \(v\) is the linear speed, and \(r\) is the radius of the circular path.
02

Establish the initial condition

Initially, the ball is swinging with certain centripetal acceleration, \(a_{c1}\), linear speed \(v\), and radius \(r\): \( a_{c1} = \frac{v^2}{r} \)
03

Examine the change in radius

Now the radius is doubled. Let the new radius be \(2r\).
04

Determine the new centripetal acceleration

Since the linear speed remains constant, we can find the new centripetal acceleration, \(a_{c2}\), using the doubled radius \(2r\): \( a_{c2} = \frac{v^2}{2r} \)
05

Compare the initial and new centripetal acceleration

Now that we have the expressions for both the initial and new centripetal acceleration, we can compare them: From step 2, we have \( a_{c1} = \frac{v^2}{r} \) And from step 4, we have \( a_{c2} = \frac{v^2}{2r} \) To find the factor by which the centripetal acceleration has changed, we can divide \(a_{c2}\) by \(a_{c1}\): \( \frac{a_{c2}}{a_{c1}} = \frac{\frac{v^2}{2r}}{\frac{v^2}{r}} = \frac{1}{2} \) This shows that the centripetal acceleration has decreased by a factor of 2. So the correct answer is: c) decreases by a factor of 2.

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