A particle P travels with constant speed on a circle of radius r = 4.30 m (see t

Question

A particle P travels with constant speed on a circle of radius r = 4.30 m (see the figure) and completes one revolution in 20.0 s. The particle passes through O at time t = 0. At t = 5.00 s, what is the particle's position vector? Give (a) magnitude and (b) direction (as an angle relative to the positive direction of x.) At t = 7.50 s, what is the particle's position vector? Give (c) magnitude and (d) direction (as an angle relative to the positive direction of x.) At t = 10.00 s, what is the particle's position vector? Give (e) magnitude and (f) direction (as an angle relative to the positive direction of x.)

Explanation / Answer

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

x & y co-ordinatwes 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 = 4.3*( sin 18*5 i + (1-cos 18*5) j )

= 4.3i + 4.3j

magnitude = 4.3* sqrt(2)

= 6.08 m

direction from x-axis = tan inverse(4.3/4.3)

= 45 degree

b) t = 7.5 s

position vector = 4.3*( sin 18*7.5 i +(1- cos 18*7.5) j )

= 4.3*(.707 i + 1.707j)

= 3.04i + 7.34j

magnitude = sqrt(3.04^2 + 7.34^2)

= 7.94 m

direction from x-axis = tan inverse (7.34/3.04 )

= 67.50 degree

c) t = 10s

position vector = 4.3*( sin 18*10 i +(1- cos 18*10) j )

= 8.6j

magnitude = 8.6 m

direction from x-axis = 90 degree

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