Solving Transfer of Torque with \mu = 0.38 Ns/m^2

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The discussion focuses on calculating the transmitted torque in a clutch system using a 3-mm-thick oil film with a viscosity of 0.38 N·s/m². The system involves two identical 30-cm-diameter disks, with the driving shaft rotating at 1450 rpm and the driven shaft at 1398 rpm. The key equation for torque is T = μ(4π²R³nL/l), where the variables include viscosity (μ), radius (R), and lengths (L and l). The main challenge is determining the value of R, which is essential for calculating the torque difference between the two rotational speeds. The discussion emphasizes the need to derive R's unit from the equation to proceed with the torque calculation.
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Homework Statement


The clutch system shown in Fig. P10-22 is used to transmit torque through a 3-mm-thick oil film with \mu = 0.38 N \bullet s/m^{2} between two identical 30-cm-diameter disks. When the driving shaft rotates at a speed of 1450 rpm, the driven shaft is observed to rotate at 1398 rpm. Assuming a linear velocity profile for the oil film, determine the transmitted torque.

s4c45c.jpg


Homework Equations



T = \mu\left(\frac{4\pi^{2}R^{3}\dot{n}L}{l}\right)

The Attempt at a Solution



\mu = 0.38 N \bullet s/m^{2}
L = 30cm = 0.3m
l = 3mm = 0.003m

This is where I am stuck I am not sure what R is. I think I have to find T for the two rpms and find the difference between them for the transferred torque.
 
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The units should tell you what R is. You have to get N-m out of this equation, so based on what you know (the given values) you can figure out what unit R will have and that will tell you what it's value should be.

Thanks
Matt
 
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