The figure below shows a wooden cylinder with mass m = 0.225 kg and length L = 0

Question

The figure below shows a wooden cylinder with mass m = 0.225 kg and length L = 0.500 m, with N = 10.0 turns of wire wrapped around it longitudinally, so that the plane of the wire coil contains the long central axis of the cylinder. The cylinder is released on a plane inclined at an angle theta to the horizontal, with the plane of the coil parallel to the incline plane. If there is a vertical uniform magnetic field of magnitude 0.750 T, what is the least current i through the coil that keeps the cylinder from rolling down the plane?

Explanation / Answer

Considering normal motion of the log

After travelling from rest for a hieght h
k.e. = 0.5mv^2 + 0.5Iw^2
but for pure rolling, wR = v
and for a cylinder, I = 0.5mR^2

so, k.e. = 0.5mv^2 + 0.5*0.5mR^2*v^2/R^2 = 0.5mv^2 + 0.25mv^2 = 0.75mv^2
loss in p.e. = mgh
so, mgh = 0.75mv^2
v = sqroot(gh/0.75) -- 1
angular velocity = v/R = sqroot(gh/0.75)/R -- 2
initial angular velocity = 0

hence angular acc, = a
2al/2piR = wf ^2 [ where l is the length along the incline travelled by the log]
hence, h = lsin(theta)
2al/2piR = glsin(theta)/0.75R^2
a = 2*pi*gsin(theta)/1.5R

Torque = Ia = 2*pi*0.5mR^2*gsin(theta)/1.5R = 2*pi*mg*sin(theta)*R/3 = 2*3.14*R*0.225*9.8*sin(theta)/3 = 4.6158*R*sin(theta) -- 4

Now, torque on a current carrying loop = NIABsin(theta) = 4.6158*R*sin(theta)
10*I*0.5*2R*0.75 = 4.6158*R
I = 0.61544 A

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