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3 years ago

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The gear motor can develop to 1/2 hp when it turns at 300 rev/min. If the shaft has a diameter of 1/2 in. determine the maximum

shear stress in the shaft

1 answer:

White raven [17]3 years ago

5 0

Answer:

145.902 psi

Explanation: Given power = 0.5 hp= 225 ft-lb/sec, N= 300 rpm shaft

dia= 0.5 inch

We know that power transmitted by circular shaft is

$P=\frac{2piNT}{60}$

now putting the values to find torque

⇒ $T=\frac{60P}{2piN}$ $=\frac{60\times225}{2\times pi\times300}$

⇒T=7.16 lb-ft

now to calculate shear stress $\tau=\frac{16T}{pi\timesd^{3} }$

⇒ $\tau =\frac{16\times7.16}{pi\times0.5^3}$

⇒ $\tau$ =145.902 psi

hence maximum shear stress in the shaft is 145.902 psi

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2 years ago

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Answer:

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Explanation:

To calculate the values in the question, a deep understanding of perfect inelastic collision is important.

When two bodies undergo inelastic collision, two important parameters must be well understood i.e

Momentum: the momentum is always conserved in perfectly inelastic collision. i.e the total momentum after collision is the sum of the individual momentum before collision

Kinetic energy: Kinetic energy is not conserved due to dissipative force.

a.To calculate the velocity, we first find the total momentum before collision

Momentum of player 1 $p_{1} =mv=95kg*5m/s\\p_{1} =475kgm/s\\$

Momentum of player 2 $p_{2} =mv=90kg*3m/s\\p_{1} =270kgm/s\\$

Hence the total momentum $p_{12}=p_{1}+p_{2}\\$

Note, since the direction of movement before collision is due south and due north respectively we have to represent the velocity using the rectangular coordinate

Hence $p_{12}=(m_{1}+m_{2})v=p_{1}i+p_{2}j\\$

$(95+90)v=475i+270j\\$

$v=2.57i+1.45j\\$

solving for the resultant velocity, we have

$v=\sqrt{2.75^{2} +1.45^{2}}\\ v=3.11mls$

To calculate the direction of movement, we have

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b. to calculate the decrease in total kinetic energy, before collision, the total kinetic was

$K.E_{initial} =\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}.\\K.E_{initial} =((1/2)*95*5^{2})+((1/2)*90*3^{2})\\K.E_{initial} =1187.5+405\\K.E_{initial} =1592.5J\\$

And the final kinetic energy after collision is

$K.E_{final} =\frac{1}{2}(m_{1}+m_{2} )v^{2}\\ K.E_{final} =\frac{1}{2}(95+90)* 3.11^{2}\\ K.E_{final} =894.7J$

The decrease in Kinetic energy is

$K.E =K.E_{final}- K.E_{initial}=894.7-1592.5$

$K.E =-697.8J$

The negative sign indicate a decrease in Kinetic energy

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3 years ago

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