Question

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.)

Answer 1

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

Step-by-step 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`