Practice 7A: | 1 | 2 |
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**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

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**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

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**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