   Chapter 12.4, Problem 67E

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# Centripetal Acceleration An object is spinning at a constant speed on the end of a string, according to the position vector r ( t ) = a cos ω t i + a sin ω t j .(a) When the angular speed ω is doubled, how is the centripetal component of acceleration changed?(b) When the angular speed is unchanged but the length of the string is halved, how is the centripetal component of acceleration changed?

(a)

To determine

To calculate: The change in the centripetal acceleration with the change in angular speed for when angular speed is doubled for the position vector r(t)=(acosωt)i+(asinωt)j.

Explanation

Formula used:

1 mi/hr=1.4667 ft/sec

v(t)=dr(t)dt,a(t)=dv(t)dt

T(t)=v(t)v(t),N(t)=dT(t)dtdT(t)dt

And, aT=a(t)T(t),aN=a(t)N(t)

Calculation:

Use the provided position vector, then calculate the velocity and the acceleration vectors as:

r(t)=(acosωt)i+(asinωt)j

Then,

dr(t)dx=(acosωt)i+(asinωt)jv(t)=(aωsinωt)i+(aωcosωt)j

And,

dv(t)dx=(aωsinωt)i+(aωcosωt)ja(t)=(aω2cosωt)i+(aω2sinωt)j

The tangential and normal component of velocity are calculated as:

T(t)=v(t)v(t)=(aωsinωt)i+(a&

(b)

To determine

To calculate: The change in the centripetal acceleration when the length of the string is halved for the position vector r(t)=(acosωt)i+(asinωt)j.

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