Acceleration of a Tennis Racket

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SUMMARY

The discussion focuses on calculating the total acceleration of a tennis racket during a serve, with a mass of 1.4 kg, an angular acceleration of 163 rad/sec², and an angular speed of 16 rad/sec. The distance from the top of the racket to the shoulder is 1.5 m. The total acceleration is derived using the formulas for tangential and centripetal acceleration, resulting in a total acceleration of approximately 16.39 m/sec². A common mistake noted was the failure to square the angular speed when calculating centripetal acceleration.

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farrah003
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Homework Statement



During a serve, a tennis racket of mass 1.4 kg is given an angular acceleration of 163 rad/sec2. At the top of the serve, the racket has an angular speed of 16 rad/sec.
If the distance between the top of the racket and the shoulder is 1.5 m, what is the total acceleration of the top of the racket?
atotal = m/sec2 ?

Homework Equations



Tangential acceleration = α*r
Centripetal acceleration = ω²*r

The Attempt at a Solution



a = √[at² + ac²]
a = √[(244.5)² + (24)²]
a = 16.3859
 
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farrah003 said:

Homework Statement



During a serve, a tennis racket of mass 1.4 kg is given an angular acceleration of 163 rad/sec2. At the top of the serve, the racket has an angular speed of 16 rad/sec.
If the distance between the top of the racket and the shoulder is 1.5 m, what is the total acceleration of the top of the racket?
atotal = m/sec2 ?

Homework Equations



Tangential acceleration = α*r
Centripetal acceleration = ω²*r
good.

The Attempt at a Solution



a = √[at² + ac²]
yes!
a = √[(244.5)² + (24)²]
a = 16.3859
In calculating ac, you forgot to square the angular speed, ω.
 
ohh hahahah thanks !
 

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