# Angular displacement and linear displacement – problems and solutions

1. A bike wheel 60 cm in diameter rotates 10 radians. What is the linear displacement of a point on the edge of the wheel?

Known :

Radius (r) = 30 cm = 0.3 m

Angle (θ) = 10 radians

Wanted : linear displacement (l)

Solution :

l = r θ

l = (0.3 m)(10 rad)

l = 3 meters

2. A wheel 50 cm in radius rotates 360o. What is the linear displacement of a point on the edge of the wheel ?

Known :

Radius (r) = 50 cm = 0.5 meters

Angle (θ) = 360o = 6.28 radians

Wanted : linear displacement (l)

Solution :

l = r θ

l = (0.5 m)(6.28 rad)

l = 3.14 meters

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3. A wheel 50 cm in radius rotates 2 revolutions. What is the linear displacement of a point on the edge of the wheel ?

Known :

Radius (r) = 50 cm = 0,5 m

Angle (θ) = 2 revolutions = (2)(6.28 radians) = 12.56 radians

Wanted : linear displacement (l) ?

Solution :

l = r θ

l = (0.5 m)(12.56 rad)

l = 6.28 m

4. A point on the edge of a wheel 2 meters in radius, moves 100 meters. Determine the angular displacement.

Known :

Radius (r) = ½ (diameter) = ½ (2 meters) = 1 meter

linear displacement (l) = 100 meters

Solution :

(a) Angular displacement (in radian)

θ = s / r = 100 / 1 = 100 radians

(b) Angular displacement (in degrees)

1 radian = 360o

100 radians = 100(360o) = 36,000 radians

(c) Angular displacement (in revolution)

6.28 radians = 1 revolution

36,000 / 6.28 = 5732,484 revolutions

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5. A particle round a circle 10 meters and rotates 180o. What is the radius ?

Known :

Linear displacement (l) = 10 meters

Angle (θ) = 180o = 3.14 radians

Wanted : radius (r)

Solution :

r = l / θ = 10 / 3.14 = 3.18 meters

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