A 250 g toy car is placed on a narrow 60-cm-diameter track with wheel grooves that keep the car going in a circle. The 1.5 kg track is free

Question

A 250 g toy car is placed on a narrow 60-cm-diameter track with wheel grooves that keep the car going in a circle. The 1.5 kg track is free to turn on a frictionless, vertical axis. The spokes have negligible mass. After the car’s switch is turned on, it soon reaches a steady speed of 0.74 m/s relative to the track. What then is the track’s angular velocity, in rpm?

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MichaelMet 1 week 2021-07-19T23:17:35+00:00 2 Answers 1 views 0

Answers ( )

    0
    2021-07-19T23:19:17+00:00

    Answer:

    the track’s angular velocity = 3.363 rpm

    Explanation:

    Given that ;

    A 250 g toy car is placed on a narrow 60-cm-diameter track

    mass m of the car = 250 g = 0.25 kg

    diameter of the car = 60 cm

    radius of the car = 60/2 = 30 cm = 0.3 m

    The 1.5 kg track is free to turn on a frictionless, vertical axis.

    mass M of the track = 1.5 kg

    steady speed v = 0.74 m/s

    the track’s angular velocity, in rpm =???

    Now; the derived equation for the conservation of angular momentum can be expressed as:

    \omega = \frac{mv}{r(m+M)}

    \omega = \frac{0.25*0.74}{0.3(0.25+1.5)}

    \omega = 0.352 \ rad/ sec

    to rpm ; we have:

    \omega = 0.352 \ rad/ sec (\frac{60 s}{1 m})(\frac{1 rev}{2(3.14)rad})

    \omega = 3.363 \ rpm

    Thus, the track’s angular velocity = 3.363 rpm

    0
    2021-07-19T23:19:22+00:00

    Answer:

    \omega_{t}=-3.9 rpm

    Explanation:

    Let’s use the conservation of angular momentum here. If there is a conservation the addition of both must be zero.

    L_{c}+L_{t}=0

    The momentum of inertia of both are:

    I_{c}=m_{c}R^{2}

    I_{t}=m_{t}R^{2}

    The angular momentum is the product between angular velocity and momentum of inertia.

    I_{c}\omega_{c}+I_{t}\omega_{t}=0

    I_{c}\omega_{c}+I_{t}\omega_{t}=0

    Let’s solve it for ω(t).

    \omega_{t}=-\frac{I_{c}\omega_{c}}{I_{t}}

    \omega_{t}=-\frac{m_{c}R^{2}\omega_{c}}{m_{t}R^{2}}

    But \omega=V/R

    • R is the radius R=D/2 = 30 cm = 0.3 m
    • V is the steady speed V = 0.74 m/s

    \omega_{t}=-\frac{m_{c}R^{2}*V/R}{m_{t}R^{2}}

    \omega_{t}=-\frac{m_{c}V}{m_{t}R}

    \omega_{t}=-\frac{0.25*0.74}{1.5*0.3}  

    \omega_{t}=-0.41 rad/s

     

    Knowing that 2π  rad is a rev. We have.

    \omega_{t}=-3.9 rpm

    The minus sign means the track is moving opposite of the car.      

    I hope it helps you!

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