Mindyard | In a disc-type axial clutch, the frictional contact takes place within an annular region with outer and inner diameters 250 mm and 50 mm, respectively. An axial force F1 is needed to transmit a torque by a new clutch. However, to transmit the same torque, one needs an axial force F2 when the clutch wears out. If contact pressure remains uniform during operation of a new clutch while the wear is assumed to be uniform for an old clutch and the coefficient of friction does not change, then the ratio F1/F2 is________

In a disc-type axial clutch, the frictional contact takes place within an annular region with outer and inner diameters 250 mm and 50 mm, respectively. An axial force F1 is needed to transmit a torque by a new clutch. However, to transmit the same torque, one needs an axial force F2 when the clutch wears out. If contact pressure remains uniform during operation of a new clutch while the wear is assumed to be uniform for an old clutch and the coefficient of friction does not change, then the ratio F1/F2 is_________ (round off to 2 decimal places).

Correct Answer :

Correct answer is : 0.87

Given, f(z) = x² – y² + i ψ (x, y)

$ϕ = x^{2} - y^{2}$

∵ f(z) is analytic function

$d ψ = \frac{- \partial ϕ}{\partial y} d x + \frac{\partial ϕ}{\partial x} \cdot d y$

$\frac{\partial ϕ}{\partial y} = - 2 y, \frac{\partial ϕ}{\partial x} = 2 x$

$d ψ = 2 y d x + 2 x d y$

$\int d ψ = 2 \int d (x y)$

ψ = 2 xy

Given, z = 1 + i

Comparing it with z = x + iy, we get :

∴ x = 1, y = 1

$∴ {(ψ)}_{(1, 1)} = 2 \times 1 \times 1 = 2$

ψ = 2 when z = 1 + i

Solution :

The correct answer is 0.87.

To find the ratio of the axial forces needed to transmit the same torque for a new clutch versus an old (worn-out) clutch, we analyze the two different states of the clutch using the Uniform Pressure Theory and the Uniform Wear Theory.
Let the outer diameter be D_o = 250 mm and the inner diameter be D_i = 50 mm.
The corresponding outer radius and inner radius are:
$R_{o} = \frac{250}{2} = 125 mm$
$R_{i} = \frac{50}{2} = 25 mm$

1. Uniform Pressure Theory (New Clutch)
For a new clutch, the contact pressure is assumed to be uniform over the entire contact area. The torque transmitted by the clutch under an axial force F₁ is given by:
$T_{1} = \frac{2}{3} μ F_{1} [\frac{R_{o}^{3} - R_{i}^{3}}{R_{o}^{2} - R_{i}^{2}}]$
where μ is the coefficient of friction.

2. Uniform Wear Theory (Old/Worn Clutch)
For an old or worn-out clutch, the wear is assumed to be uniform across the friction surface. Under this assumption, the torque transmitted under an axial force F₂ is given by:
$T_{2} = \frac{1}{2} μ F_{2} (R_{o} + R_{i})$

3. Calculating the Ratio of Axial Forces
Since the torque transmitted in both cases is the same (T₁ = T₂) and the coefficient of friction μ remains constant, we can equate the two expressions:
$\frac{2}{3} μ F_{1} [\frac{R_{o}^{3} - R_{i}^{3}}{R_{o}^{2} - R_{i}^{2}}] = \frac{1}{2} μ F_{2} (R_{o} + R_{i})$
Simplifying by canceling μ from both sides and rearranging for the ratio F₁/F₂:
$\frac{F_{1}}{F_{2}} = \frac{3 (R_{o} + R_{i}) (R_{o}^{2} - R_{i}^{2})}{4 (R_{o}^{3} - R_{i}^{3})}$

Now, we substitute the values R_o = 125 mm and R_i = 25 mm into the equation:
$R_{o} + R_{i} = 125 + 25 = 150$
$R_{o}^{2} - R_{i}^{2} = 125^{2} - 25^{2} = 15625 - 625 = 15000$
$R_{o}^{3} - R_{i}^{3} = 125^{3} - 25^{3} = 1953125 - 15625 = 1937500$

Substitute these calculated terms back into the ratio formula:
$\frac{F_{1}}{F_{2}} = \frac{3 \times 150 \times 15000}{4 \times 1937500}$
$\frac{F_{1}}{F_{2}} = \frac{6750000}{7750000} = \frac{27}{31} \approx 0.87096$

Rounding off the final value to 2 decimal places yields:
$\frac{F_{1}}{F_{2}} = 0.87$

Get 3 Months FREE trial

Create, conduct, and manage professional online assessments with Mindyard. Perfect for teachers and institutes.

Unlock Our Free Library

You may also like

Mock Tests

GATE 2026 Mechanical Engineering 14/02/2026

GATE Mechanical Engineering (ME) - 2025

Ask AI Tutor