Rotational Inertia problem


by John O' Meara
Tags: inertia, rotational
John O' Meara
John O' Meara is offline
#1
Oct1-06, 10:27 AM
P: 330
A lawn roller in the form of a hollow cylinder of mass M is pulled horizontally with a constant force F applied by a handle attached to the axle. If it rolls without slipping, find the acceleration and the frictional force.
Let R1 be the radius of the hollow and R2 the outer radius, and (alpha) the angular acceleration.
Sum Fy= n - M*g
Sum Fx= F - f.
Where n = normal reaction and f = friction force. Applying Sum F = M*a, we get n=M*g ....as ay=0.
F - f = M*ax ....the equation for the translational motion of the center of mass. And where ax and ay are the accelerations in the x and y directions resprctively.
And the equation of rotational motion about the axis through the center of mass is:
f*R2 = I*(alpha)=.5*M*(R1^2+R2^2)
f*R2 = .5*M*(R1^2+R2^2)*ax/R2: let a=ax
f = .5*M*(R1^2+R2^2)*a/R2^2 .......(ii)
F-f = M*a...(i) After adding i and ii
a=F/(M+.5*M*(R1^2+R2^2)/R2^2.
The question is, is this correct or not? And where is it wrong?
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Doc Al
Doc Al is online now
#2
Oct1-06, 11:43 AM
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P: 40,906
Looks OK to me.


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