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Tpy6a [65]
3 years ago
7

The 0.4-lb ball and its supporting cord are revolving about the vertical axis on the fixed smooth conical surface with an angula

r velocity of 4 rad /sec. The ball is held in the position b = 14 in. by the tension T in the cord. If the distance b is reduced to the constant value of 9 in. by increasing the tension T in the cord, compute the new angular velocity and the work W1-2 done on the system by T
Physics
1 answer:
solong [7]3 years ago
3 0

Answer:

Explanation:

In the whole process , angular momentum will be conserved because no external torque is acting on the system . The radius of circular path is reduced so angular velocity will be increased

I₁ x ω₁ = I₂ x ω₂

m r₁² x ω₁ = m r₁²ω₂

r₁² x ω₁ =  r₂²ω₂

ω₂ = r₁² x ω₁  /  r₂²

= ( 14 x 14 x 4) / ( 9 x 9 )

= 9.68 rad /s

Work done

= increase in rotational kinetic energy

= 1/2 x I₂ ω₂² - 1/2 x I₁ ω₁²        ( I = m r² )

=  1/2 x .4 x .454 x( 9 x 2.54 x 10⁻² x 9.68 )² - 1/2 x .4 x .454 x( 14 x 2.54 x 10⁻² x 4 )²

=   .4446 - 0.1837

= .261 J .

=

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Answer:

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When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

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y: height from the person jumps = 20.0m

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mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

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you solve the last expression for x:

x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m

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v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}

Next, you calculate x from the equation (1):

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m

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