A particle P travels with constant speed on a circle of radius r = 3.0 m (see th

Question

A particle P travels with constant speed on a circle of radius r = 3.0 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 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

In 20 second 360 degree is completed.
In 5 second it will complete [360/20]* 5 = 90 degree
ang vel = 2 pie/20 = pie/10

speed = ang vel *r = 3 pie/10 = 0.942 m /s
The coordinate of the point is (3m, 3m)

The magnitude or the distance from the origin is

? (3^2 + 3^2 ) = 3*?2
at t=10 s y component of velocity = 0

acceleration at t=10

= v^2/r = 0.942^2 /3 = 0.2958 m/s^2

x-component of the acceleration at t = 10.0 s is zero as the motion is in -x direction at this time so acceleration is perpendicular to it

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

= v^2/r = 0.942^2 /3 = 0.2958 m/s^2 in -y direction
The angle is given by tan A = 1.2 /1.2 = 1
A = 45 degree.

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