Acircular plate with diameter of 20 cm is placed over a fixed bottom plate with a 1 mm gap between two plates filled with kerosene at 40 degree c, shown in the following figure. find the torque needed to rotate the top plate at 5 rad/s. the velocity distribution in the gap is linear and the shear stress on the edge of the rotating plate can be neglected.

T = 1.17 x N-m

Explanation:

Given :

Gap between the two plates , dy = 1 mm

                                                  dy = 1 x m

Angular velocity of the top plate , ω = 5 rad/s

Diameter of the plate, D = 20 cm

Radius of the plate, R = 10 cm

                                    = 0.1 m

Temperature of the kerosene = 40°C

Viscosity of kerosene at 40°C = 0.0015 Pa-s

Now let us take a small elemental ring of thickness dr at a radius r.

Therefore, area of this elemental ring of dr = 2πrdr

Now linear velocity at radius r = ω x r

                                                   5r m/s

Now applying Newtons law of viscosity we get,

Shear stress, τ = μ.

Now we know torque due to small strip,

 dT = x r

 dT =

Therefore total torque for r=0 to r=R can be calculated. So by integrating,

T = 1.17 x N-m

answer:

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explanation: