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The power is transmitted from a pulley 1 m diameter running at 200 rpm to a pulley 2.25 m diameter by means of a belt.

Find the speed lost by the driven pulley as a result of creep, if the stress on the right and slack side of the belt is 1.4 MPa and 0.5 MPa respectively. The Young’s modulus for the material of the belt is 100MPa.

Subject: Kinematics of Machinery

Topic: Belts, Chains and Brakes

Difficulty: High

1 Answer
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Given :- $d_{1}$ = 1 m; $N_{1}$ = 200 rpm; $d_{2}$ = 2.25 m; $\sigma_{1}$ =1.4 $MP_{a}$=1.4 ×106 N/m$^2$ , $\sigma_{2}$ =0.5 $MP_{a}$=0.5 ×106 N/m$^2$ ; E= 100 $MP_{a}$=100×106 N/m$^2$,

Let $N_{2}$ = speed of the driven pulley.

Neglecting creep, we know that

$\frac{N_{2}}{N_{1}}$ = $\frac{d_{1}}{d_{2}}$ or $N_{2}$ = $N_{1}\frac{d_{1}}{d_{2}}$ …

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