1. A wheel 30 cm in radius rotates at constant 5 rad/s2. What is the magnitude of the linear acceleration of a point located at (a) 10 cm from the center (b) 20 cm from the center (c) on the edge of the wheel?
Known :
Radius (r) = 30 cm = 0.3 m
Angular acceleration (α) = 5 rad/s2
Wanted : linear acceleration (a) r = 0.1 m (b) r = 0.2 m (c) r = 0.3 m
Solution :
Relation between linear acceleration (a) and angular acceleration :
a = r α
(a) linear acceleration, r = 0.1 m
a = (0.1 m)(5 rad/s2) = 0.5 m/s2
(b) linear acceleration, r = 0.2 m
a = (0.2 m)(5 rad/s2) = 1 m/s2
(c) linear acceleration, r = 0.3 m
a = (0.3 m)(5 rad/s2) = 1.5 m/s2
2. A pulley 50 cm in radius. If the linear acceleration of a point located on the edge of the pulley is 2 m/s2, determine the angular acceleration of the pulley!
Known :
Radius (r) = 50 cm = 0,5 m
linear acceleration (a) = 2 m/s2
Wanted : the angular acceleration
Solution :
α = a / r = 2 / 0.5 = 4 rad/s2
3. The blades in a blender 20 cm in radius, initially at rest. After 2 seconds, the blades rotates 10 rad/s. Determine the magnitude of linear acceleration (a) a point located at 10 cm from the center (b) a point located at the edge of the blades.
Known :
Radius (r) = 20 cm = 0.2 m
The initial angular velocity (ωo) = 0
The final angular velocity (ωt) = 10 radians/second
Time interval (t) = 2 seconds
Wanted : the linear acceleration of a point located at (a) r = 0.1 m (b) r = 0.2 m
Solution :
ωt = ωo + α t
10 = 0 + α (2)
10 = 2 α
α = 10 / 2
α = 5 rad/s
(a) linear acceleration of r = 0.1 m
a = r α = (0.1 m)(5 rad/s2) = 0.5 m/s2
(b) linear acceleration of r = 0.2 m
a = r α = (0.2 m)(5 rad/s2) = 1 m/s2
4. A wheel 20 cm in radius is accelerated for 2 seconds from 20 rad/s to rest. Determine the magnitude of linear acceleration (a) a point located at 10 cm from the center (b) a point located at 10 cm from the center.
Known :
Radius (r) = 20 cm = 0.2 m
The initial angular speed (ωo) = 20 rad/s
The final angular speed (ωt) = 0
Time interval (t) = 2 seconds
Wanted : The linear acceleration (a) r = 0.1 m (b) r = 0.2 m
Solution :
ωt = ωo + α t
0 = 20 + α (2)
-20 = 2 α
α = -20 / 2
α = -10 rad/s
Negative sign mean the angular speed is decrease.
(a) linear acceleration of r = 0.1 m
a = r α = (0.1 m)(-10 rad/s2) = -1 m/s2
(b) linear acceleration of r = 0.2 m
a = r α = (0.2 m)(-10 rad/s2) = -2 m/s2
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