Question

A point on a circle with a diameter of 10m starts at an upright position P at (0,r) and moves clockwise with an acceleration .6m/s^2 .

When t=7 :

-Find position

-Find velocity

-Find acceleration

relative velocity

Answer 1

Its certain from the information given for the initial position that it follows a circular path of radius r with origin as its center. Since the point starts motion at (0,r) we have angular velocity w to be zero initially which means that the radial acceleration is zero thus only tangential acceleration is there initially which is given as at = r.; where is the angular acceleration.

So we have a = r

=> = a/r

=> = 6m/s2/10/2m

=> = 1.2 s-2

Using the circular motion equations;

= 1/2 x x t2

=> = 1/2 x 1.2 x 72 = 29.4 radians

You can use the concept of trigonometry in which the clockwise direction is considered to be negative direction so

= -29.4 rad

Since initial position was (0,r) which means that it has an initially angle of so

= -29.4 + rads

Now using the coordinate system we have,

x = rcos = - 4.513 m

y = rsin = -2.153 m

SO the position will be (-4.513,-2.153)

Now the velocity can be given as rw ; w is the angular velocity at that time

v = rw

=> v = 5 x t

=> v = 5 x 1.2 x 7 = 42 m/s

Radial acceleration is given as ar = w2r

=> ar = (t)2 x 5m

=> ar = (1.2 x 7)2 x 5

=> ar = 352.8 m/s2

and that tangential acceleration will be same 6m/s2.