The driving and driven shafts connected by a Hooke’s joint are inclined by an angle α to each other. The angle through which the driving shaft turns is given by θ. The condition for the two shafts to have equal speeds is

This question was previously asked in

ESE Mechanical 2013 Official Paper - 2

Option 3 : \(\tan {\rm{\theta }} = \pm \sqrt {\cos {\rm{\alpha }}} \)

**Concept:**

- Hooke's joint is used
**to connect two non-parallel and intersecting shafts.** - It is also called
**Universal coupling**. - Hooke's joint is used to
**transmit motion and powe**r for shafts having**angular misalignment.**

Referring to the diagram below, a universal joint is used to connect two misaligned shafts at an angle α

__Calculation:__

From ΔOC_{1}M,

\(\tan {\rm{\theta }} = \frac{{OM}}{{M{C_1}}}\;\;\;\;\; \ldots \left( 1 \right)\)

From ΔOC_{2}N,

\(\tan \phi = \frac{{ON}}{{N{C_2}}}\)

∵ NC_{2 }= MC_{1 }

\( ⇒ tan\phi = \frac{{ON}}{{M{C_1}}}\;\;\;\;\; \ldots \left( 2 \right)\)

OM = ON_{1 }× cosα … (3)

∵ ON has Projected along with AB,

⇒ ON_{1 }= ON

By equation (1), (2), and (3),

tanθ = tanϕ × cosα … (4)

Now differentiating equation (1) with respect to time,

⇒ Sec^{2}θ × ω_{1} = cosα × sec^{2}ϕ × ω_{2}

Where,

\({{{\omega }}_1} = \frac{{{{d\theta }}}}{{{{dt}}}} = {{Angular\;velocity\;of\;the\;Driver\;shaft}}\)

\({\omega _2} = \frac{{d\phi }}{{dt}} = {{Angular\;velocity\;of\;the\;Driven\;shaft}}\)

\( \Rightarrow \frac{{{\omega _2}}}{{{\omega _1}}} = \frac{1}{{{{\cos }^2}\theta \; \times \;cos\alpha \; \times \;{{\sec }^2}\phi }}\;\;\;\;\; \ldots \left( 5 \right)\)

By using equation (4),

\(\Rightarrow{\sec ^2}\phi = 1 + \frac{{{{\tan }^2}\theta }}{{{{\cos }^2}\alpha }}\)

\( \Rightarrow {\sec ^2}\phi = \frac{{1 - {{\cos }^2}\theta \times {{\sin }^2}\alpha }}{{{{\cos }^2}\theta \times {{\cos }^2}\alpha }}\;\;\;\;\; \ldots \left( 6 \right)\)

By using equation (5) and (6) to eliminate ϕ,

\(\frac{{{\omega _2}}}{{{\omega _1}}} = \frac{{cos\alpha }}{{1 - {{\cos }^2}\theta \times {{\sin }^2}\alpha }}\;\;\;\;\; \ldots \left( 7 \right)\)

The condition for the two shafts to have equal speeds,

⇒ ω_{1 }= ω_{2}

∴ By using equation (7),

\({\rm{co}}{{\rm{s}}^2}\theta = \frac{{1 - cos\alpha }}{{{{\sin }^2}\alpha }}\) ...(8)

\( \Rightarrow {\sin ^2}\theta = 1 - \frac{{1 - cos\alpha }}{{{{\sin }^2}\alpha }}\)

\( \Rightarrow {\sin ^2}\theta = \frac{{cos\alpha }}{{1 + cos\alpha }}\) ...(9)

By using equation (8) and (9),

\( \Rightarrow \;{\tan ^2}\theta = \frac{{cos\alpha \times {{\sin }^2}\alpha }}{{{{\sin }^2}\alpha }}\)

\( \Rightarrow tan\theta = \pm \sqrt {\left( {cos\alpha } \right)} \)

__Note__**:**

In the Hook joint even when the **input drive shaft axle rotates at a constant speed**, the **output drive shaft axle rotates at a variable speed**, thus causing vibration and wear. This can be understood from the diagram itself, that the angular rotation of the driven shaft (ϕ) is not the same as the rotation of the driver shaft (θ).