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Scrat [10]
3 years ago
8

Exactly one turn of a flexible rope with mass m is wrapped around a uniform cylinder with mass M and radius R.

Physics
1 answer:
Dennis_Churaev [7]3 years ago
7 0

Answer:

\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}

Explanation:

The rotational kinetic energy when the cylinder is with the rope is:

E_k=\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^2

where we used the fact that both rope and cylinder hast the same w. This E_k must conserve, that is, E_k must equal E_k when the rope leaves the cylinder. Hence, the final w is given by:

E_{k1}=E_{k2}\\\\\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^{2}=\frac{1}{2}I_c\omega^2\\\\\omega=\sqrt{\omega_0^2(\frac{I_c+I_r}{I_c})} (1)

For Ic and Ir we can assume that the rope is a ring of the same radius of the cylinder. Then, we have:

I_c=\frac{1}{2}MR^2\\\\I_r=mR^2

Finally, by replacing in (1):

\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}

hope this helps!!

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Complete Question

A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 1.60 T in 0.99 s. What is the resulting induced current if the loop has a resistance of 1.20  \ \Omega

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Explanation:

From the question we are told that

   The  area is  A = 8.00 \ cm^2  = 8.0 *10^{-4} \  m^2

   The initial magnetic field at t_o = 0 \ seconds  is B_i = 0.500 \ T

   The magnetic field at t_1 = 0.99 \ seconds is  B_f  = 1.60 \ T

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Generally the induced emf is mathematically represented as

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=>   \epsilon  =  8.0 *10^{-4} * \frac{1.60  - 0.500 }{ 0.99- 0  }

=>   \epsilon  = 0.000889 \ V

Generally the current induced is mathematically represented as

     I = \frac{\epsilon}{R }

=>  I = \frac{0.000889}{ 1.20 }  

=>  I =  0.0007 41 \ A  

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