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Gwar [14]
3 years ago
6

The equation for gear ratio

Physics
1 answer:
Ymorist [56]3 years ago
4 0
In a gear train with two gears, the gear ratio is defined as follows
R= \frac{\omega _A}{\omega _B} 

where \omega _A is the angular velocity of the input gear while \omega _B is the angular velocity of the output gear. 

This can be rewritten as a function of the number of teeth of the gears. In fact, the angular velocity of a gear is inversely proportional to the radius r of the gear:
\omega = \frac{v}{r}
But the radius is proportional to the number of teeth N of the gear. Therefore we can rewrite the gear ratio also as
R= \frac{\omega _A}{\omega _B} = \frac{r_B}{r_A} = \frac{N_B}{N_A}

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Second Trial satisfy principle of conservation of momentum

Explanation:

Given mass of ball A and ball B =\ 1.0\ Kg.

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initial velocity of ball A\ is\ u_1

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Now, mu_1+mu_2=mv_1+mv_2

Plugging each trial in this equation we get,

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mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3

momentum before collision \neq moment after collision

Second Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1

moment before collision = moment after collision

Third Trial

mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1

momentum before collision \neq moment after collision

Fourth Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0

momentum before collision \neq moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

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