Practice 7A: | 1 | 2 | 3 | 4 | Go up
Angular Displacement  - by Matt Henderson, 2003

1. A girl sitting on a merry-go -round moves counterclockwise through an arc length of 2.5 m. If the girl's angular displacement is 1.67 rad, how far is she from the center of the merry - go- round.

Here's what you know, Dq = 1.67 rad and l = 2.5  use the formula  Dq = Ds/r to find the radius

1.67 = 2.5/r         r = 1.5 m

2. A beetle sits at the top of a bicycle wheel and flies away just  before it would be squashed. Assuming that the wheel turns clockwise, the beetle's angular displacement is (p) rad, which corresponds to an arc length of 1.2 m. What is the wheel's raduis?

We know that Dq = (p) rad and l = 1.2 m  so now we use the formula Dq = Ds/r to find the radius

(p) = 1.2/r         r = .38  m

3. A car on a Ferris wheel has an angular displacement of (p/4) rad, which corresponds to an arc length of 29.8. What is the Ferris wheel's radius?

We know that Dq = (p/4) rad and l = 29.8 m  so now we use the formula Dq = Ds/r to find the radius

(p/4) = 29.8/r         r = 37.94  m

4. Fill in the unknown quantities in the following table:

Dq         Ds        r

a.  ?rad      .25m    .1m      Dq = Ds/r which is Dq = .25/.1 = 2.5 rad

b.   .75 rad   ?         8.5m     Dq = Ds/r which is .75 = Ds/8.5 so  Ds = 6.375m

c.   ? degr  -4.2m    .75m   Dq = Ds/r  which is Dq = -4.2/.75 = -5.6 rad and 1 rad = 57.3 degrees so (57.3)(-5.6) = -320.88 degre

d.  135degr  2.6m   ?    135/57.3 since 1 rad = 57.3 degrees that = 2.356 rad now plug into  Dq = Ds/r so 2.356 = 2.6/r

and r = 1.1 m