FE Physics
Formulas & Explanation

Dutch version

Motion in Two Dimensions

Projectile Motion

Example 1

From a tower (height : 50 m) an object is projected  horizontally with a velocity of 20 ms-1

Ignoring the air resistance:

a.  Determine at what place the object hits the ground, ignoring the air resistance ?

b.  With what velocity does that happen ?

c.   With what angle the object hits the ground ?

Solution:

The directions to the right and upward have been chosen as positive.

a.       In the vertical (y)  direction the movement has a constant acceleration due the gravity.

ay = g = - 9.81 m s-2.  Initial velocity in y direction = 0 ms-1

Δ Y =  ½ g t2    - 50 = ½ -9.81 t2      t2 =  10.20     t = 3.2 s

In the horizontal (x)direction the object has a constant velocity

ΔX = vx t         x = 20 x 3.2 = 64 m

The object hits the ground at 64 m from the base of the tower.

b.      The velocity in x - direction remains  constant.

vx van 20 m/s

vy = g t = -9.81 x 3.2 = -31.39 m/s  ( minus because the direction of the velocity is downward)

The magnitude of the final velocity is to be determined using the Pythagorean theorem

v2 = vx2 + vy2        v2 = 202 + 31.392       v2 = 1385.33    v = 37.2 m/s

c.       tan α = vy/vx = 31.39/20 = 1.5695      α = 57.5 o

Uniform Circular Motion

A disc rotates in a horizontal plane with 360 revolutions per minute

At a distance of 20 cm from the center is a block with a mass of 100 g.

frequency f =  360 rev/minute =6 rev /second       f = 6 Hz

period   T=1/f = 1/6 = 0.17 s

speed  of the block   v = (2 π r) / T = (2 π 0.20)/0.17 = 7.39 m/s

angular velocity  ω = (2 π) / r = (2 π) /0.20 = 31.4 rad/s

centripetal acceleration  of the block   ac = v2/r = 7.392/0.20 = 2.73 x 102 ms-2

Remark

In an uniform circular motion is the speed constant

The acceleration is directed toward the center of the circular path. This causes a change in direction of the velocity. The speed (magnitude of the velocity) does not change.

(Centripetal Force : see chapter Force and Torque)