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?

Answer 1

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

Answer 2

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!