A particle P travels with constant speed on a circle of radius r = 9.0 m (see th

Question

A particle P travels with constant speed on a circle of radius r = 9.0 m (see the figure) and completes one revolution in 20.0 s. The particle passes through 0 at t = 0. Find the magnitude and direction of each of the following vectors. With respect to 0, find the particle's position vector at t = 5.0 s. What is it magnitude?

What is its angle with respect to the x-axis?

With respect to 0, find the particle's position vector at t = 7.5 s. What is it magnitude?

What is its angle with respect to the x-axis?

With respect to 0, find the particle's position vector at t = 10.0 s. What is it magnitude?

What is its angle with respect to the x-axis?

For the 5.00 s interval from the end of the fifth second to the end of the tenth second, find the magnitude of the particle's displacement.

For the 5.00 s interval from the end of the fifth second to the end of the tenth second, find the particle's average velocity. What is the x-component of the average velocity?

What is the y-component of the average velocity?

Find its velocity at the beginning and end of that 5.00 s interval. What is the x-component of the velocity at t = 5.0 s?

What is the y-component of the velocity at t = 5.0 s?

Find its acceleration at t = 5.0 s? What is the x-component of the acceleration at t = 5.0 s?

What is the y-component of the acceleration at t = 5.0 s?

Find its velocity at the end of that 5.00 s interval. What is the x-component of the velocity at t = 10.0 s?

What is the y-component of the velocity at t = 10.0 s?

Find its acceleration at t = 10.0 s? What is the x-component of the acceleration at t = 10.0 s?

What is the y-component of the acceleration at t = 10.0 s?

Explanation / Answer

T = 20.0 s
w = 2*.314/T
= .314 rad/s = 18 degree/s



x & y coordinatwes can be written as

x = r cos (wt-90) = r sin wt

y = r + r sin( wt -90) = r - rcoswt

position vector = rsin (wt) i + r(1 - coswt) j

a) t = 5

position vector = 9*( sin 18*5 i + (1-cos 18*5) j )

= 9i + 9j

magnitude = (9^2 +9^2)^0.5

= 12,72 m

direction from x-axis = tan inverse(9/9)

= 45 degree

b) t = 7.5 s

position vector = 3*( sin 18*7.5 i +(1- cos 18*7.5) j )

= 9*(.707 i + 1.707j)

=6.363 i + 15.363j

magnitude = sqrt(6.363^2 + 15.363^2)

=16.6 m



direction from x-axis = tan inverse (6.363/15.363 )

= 22,49 degree

c) t = 10s

position vector = 9*( sin 18*10 i +(1- cos 18*10) j )

= 9j

magnitude = 9 m



direction from x-axis = 90 degree

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