A 60cm radius, 2.0kg wheel rolls along at an initial angular velocity of 2.4rad/

Question

A 60cm radius, 2.0kg wheel rolls along at an initial angular velocity of 2.4rad/s. Friction applies a constant force perpendicular to the radius as shown causing the wheel to coast to a stop in 10.0s. The moment of inertia of the wheel can be calculated as .5mr2. Find the following and show all work:

1. linear speed =

2. torque applied by friction =

3. angular acceleration =

4. moment of inertia =

5. time to come to rest =

6. angular displacement before stopping =

7. distance it rolls before stopping =

Explanation / Answer

1) linear speed v =w*r = 2.4*0.6 = 1.44 m/s

3) wf = wi + a*t

0 = 2.4 + a*10

a = - 0.24 rad/s^2 (a is the angular accelration)

2) torque due to friction = moment of inertia*a

torque = 0.5mr^2*a = 0.5*2*0.6^2*0.24 = 0.0864 Nm

4) moment of inertia =0.5mr^2 = 0.36 kgm^2

5) time to come to rest = 10s

6) angular displacement = w*t - 0.5*a*t^2 = 12 radians

7) distance it rolls = 12*r = 7.2 metres

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