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Olenka [21]
3 years ago
12

A track is mounted on a large wheel that is free to turn with negligible friction about a vertical axis (Fig. 11-48).A toy train

of mass m is placed on the track and, with the system initially at rest, the train’s electrical power is turned on.The train reaches speed 0.15 m/s with respect to the track.What is the wheel’s angular speed if its mass is 1.1m and its radius is 0.43 m? (Treat it as a hoop, and neglect the mass of the spokes and hub.)

Physics
2 answers:
masya89 [10]3 years ago
5 0

Answer:

0.166 rad/s

Explanation:

See attachment for calculations

katovenus [111]3 years ago
5 0

Answer:

The angular velocity of the wheel is  w = 0.1661 \ rad/ sec

Explanation:

From the question we are told that

     The mass of the toy train is  m

     The speed of the train is v_t = 0.15 m/s

     The radius of the  wheel is  r =  0.43 \ m

     The mass of the wheel is m_w = 1.1 * m

According to the law of conservation of momentum

   L_i = L_f

Where L _i  is the initial angular momentum which is mathematically represented as

                 L_i = rmv

and

L_f is the final angular momentum  which is mathematically represented as

                       L_f = I * w

Where I is the moment of inertia of the wheel which is mathematically represented  as

                 I =  m_w * r^2

   So  

        rmv = m_w r^2 w

      r * m * 0.15  =  1.1 * m  * r^2 * w

      v  =  1.1 * r * w

But we know  the train is moving relative to the wheel so

     v =  v_t - wr

Where wr is the linear velocity component of the wheel so

 Substituting values

        0.15 -  (w * 0.43)   =  1.1 * 0.43 * w

=>     0.15 -  (w * 0.43)   =  0.473 * w

        0.15  =  0.903w

       w = 0.1661 \ rad/ sec

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Two long, parallel wires separated by 3.50 cm carry currents in opposite directions. The current in one wire is 1.55 A, and the
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Answer:

Therefore,

The magnitude of the force per unit length that one wire exerts on the other is

\dfrac{F}{l}=2.79\times 10^{-5}\ N/m

Explanation:

Given:

Two long, parallel wires separated by a distance,

d = 3.50 cm = 0.035 meter

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To Find:

Magnitude of the force per unit length that one wire exerts on the other,

\dfrac{F}{l}=?

Solution:

Magnitude of the force per unit length on each of @ parallel wires seperated by the distance d and carrying currents I₁ and I₂ is given by,

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The magnitude of the force per unit length that one wire exerts on the other is

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