Answer:
[tex]\omega_{t}=-3.9 rpm[/tex]
Explanation:
Let's use the conservation of angular momentum here. If there is a conservation the addition of both must be zero.
[tex]L_{c}+L_{t}=0[/tex]
The momentum of inertia of both are:
[tex]I_{c}=m_{c}R^{2}[/tex]
[tex]I_{t}=m_{t}R^{2}[/tex]
The angular momentum is the product between angular velocity and momentum of inertia.
[tex]I_{c}\omega_{c}+I_{t}\omega_{t}=0[/tex]
[tex]I_{c}\omega_{c}+I_{t}\omega_{t}=0[/tex]
Let's solve it for ω(t).
[tex]\omega_{t}=-\frac{I_{c}\omega_{c}}{I_{t}}[/tex]
[tex]\omega_{t}=-\frac{m_{c}R^{2}\omega_{c}}{m_{t}R^{2}}[/tex]
But [tex]\omega=V/R[/tex]
[tex]\omega_{t}=-\frac{m_{c}R^{2}*V/R}{m_{t}R^{2}}[/tex]
[tex]\omega_{t}=-\frac{m_{c}V}{m_{t}R}[/tex]
[tex]\omega_{t}=-\frac{0.25*0.74}{1.5*0.3}[/tex]
[tex]\omega_{t}=-0.41 rad/s[/tex]
Knowing that 2π rad is a rev. We have.
[tex]\omega_{t}=-3.9 rpm[/tex]
The minus sign means the track is moving opposite of the car.
I hope it helps you!
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:
[tex]\omega = \frac{mv}{r(m+M)}[/tex]
[tex]\omega = \frac{0.25*0.74}{0.3(0.25+1.5)}[/tex]
[tex]\omega = 0.352 \ rad/ sec[/tex]
to rpm ; we have:
[tex]\omega = 0.352 \ rad/ sec (\frac{60 s}{1 m})(\frac{1 rev}{2(3.14)rad})[/tex]
[tex]\omega = 3.363 \ rpm[/tex]
Thus, the track's angular velocity = 3.363 rpm
