At t1 = 3.00 s, the acceleration of a particle moving at constant speed in counterclockwise circular motion is an = (3.00 m/s?)ì + (6.00 m/s) ^ At t2 = 7.00 s (less than one period later), the acceleration is a2 = (6.00 m/s3)ì - (3.00 m/s2)î The period is more than 4.00 s. What is the radius of the circle?

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At t1 = 3.00 s, the acceleration of a particle moving at constant speed in counterclockwise circular motion is

an = (3.00 m/s?)ì + (6.00 m/s) ^

At t2 = 7.00 s (less than one period later), the acceleration is

a2 = (6.00 m/s3)ì - (3.00 m/s2)î

The period is more than 4.00 s. What is the radius of the circle?

Expert Solution
Introduction:

We are given acceleration in cartesian coordinates. Since speed is constant, there is only centripetal acceleration. We first find the time period from the given cases. We then find the radius of circle.

The centripetal acceleration is given as

a=ω2ra=4π2T2r

Here T,r are time period and radius respectively.

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