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redhotdevil13l3 · Accepted Answer

We are given following data for a bicycle wheel:  d1=0.64m  m=1.8kg  F1=120N  It is also given that chain sprocket:  d2=0.09m  α=4.5rads2  We can assume that the wheel is a hoop. Hoop moment of inertia is given by:  I=m⋅r12=m⋅(d12)2=1.8⋅(0.642)2=0.184kg m2  Torque equation:  I⋅α=F2⋅r2−F1⋅r1  Solving it for force applied by a chain:  F2=I⋅α+F1⋅r1r2=I⋅α+F1⋅d12d22=0.184⋅4.5+120⋅0.6420.092  =871.7N

Karen Robbins · Accepted Answer

Explanation: The net torque of the sprocket is
$τ_{\neq t} = τ_{a p p l i e d} - τ_{r e s i s t i v e} = I α$ and further it is modified as $r F - R f = I α$ .
Now the moment of inertia of the bicycle wheel is $I = M R^{2}$ and by using these relations, the force that must applied by the chain is found.
The formula for the force must be applied by the chain is,
$F = \frac{M R^{2} α + R f}{r}$
- M is mass of the wheel.
- R is radius of the wheel.
- $α$ is angular acceleration of the wheel.
- f is resistive force.
- r is radius sprocket.
Substitute 1.80kg for M, 64.0cm/2 for R, 120 N for f, 9.00cm/2 for r, and $4.50 r a \frac{d}{s^{2}} f or α$ to find F.
$F = \frac{[(1.80 k g) {(64.0 c \frac{m}{2})}^{2} {(\frac{0.01 m}{1 c m})}^{2}] (4.50 r a \frac{d}{s^{2}}) + (64.0 c \frac{m}{2}) (\frac{0.01 m}{1 c m}) (120 N)}{(9.00 c \frac{m}{2}) (\frac{0.01 m}{1 c m})}$
=872N
Thus, the force must be applied by a chain passing over a sprocket to give the wheel an acceleration of $4.50 \frac{r a d}{s^{2}}$ is 872N.

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