The 9.9-kg wheel with a radius of gyration of 135 mm about its center O is relea

Question

The 9.9-kg wheel with a radius of gyration of 135 mm

about its center O is released from rest on the incline

and slips as it rolls. If the coefficient of kinetic friction is

?k = 0.35, calculate the magnitudes of the acceleration aO

of the center O of the wheel and its angular acceleration ?.

Answer ao must be in M/s^2 and a must be in rad/s^2

Here is a link to the diagram/question:

http://i1334.photobucket.com/albums/w655/salonemark/6-76_zps2b360def.jpg

The 9.9-kg wheel with a radius of gyration of 135 mm about its center O is released from rest on the incline and slips as it rolls. If the coefficient of kinetic friction is ?k = 0.35, calculate the magnitudes of the acceleration aO of the center O of the wheel and its angular acceleration ?. Answer ao must be in M/s^2 and a must be in rad/s^2

Explanation / Answer

As we know that

In geneal

r X F = I*angular acceleration

Here

F = Force which is Producing Torque

Here the Torque is there due to Frictional Force

Therefore

F = uMgCos(theta)

r = Radius of the Wheel

Also

I = Total Moment of Inertia

= Mk^2

Therefore

Angular Acceleration = uMgrCos(theta)/Mk^2

= ugrCos(theta)/k^2

= (0.35*9.8*0.165*Cos(61 degree))/(0.135*0.135)

= 15.055 rad/sec^2

By Considering the Linear Translation

Met Force = ass*Acceleration

Therefore

Linear Acceleration = (MgSin(theta) - uMgCos(theta))/M

= (9.8*Sin(61 degree) - 0.35*9.8*Cos(61 degree))

= 6.908 m/sec^2

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