A particle P travels with constant speed on a circle of radius r = 2.40 m (see t

Question

A particle P travels with constant speed on a circle of radius r = 2.40 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 = 3*( sin 18*5 i + (1-cos 18*5) j ) = 3i + 3j magnitude = 3* sqrt(2) = 4.24 m direction from x-axis = tan inverse(3/3) = 45 degree b) t = 7.5 s position vector = 3*( sin 18*7.5 i +(1- cos 18*7.5) j ) = 3*(.707 i + 1.707j) = 2.121i + 5.121j magnitude = sqrt(2.121^2 + 5.121^2) = 5.54m direction from x-axis = tan inverse (5.121/2.121 ) = 67.50 degree c) t = 10s position vector = 3*( sin 18*10 i +(1- cos 18*10) j ) = 6j magnitude = 6 m direction from x-axis = 90 degree

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