# Angular acceleration and linear acceleration – problems and solutions

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

[irp]

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

(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

[irp]

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

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

[irp]

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