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

## Answers ( )

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:

to rpm ; we have:

Thus, the track’s angular velocity = 3.363 rpm

2. Answer:

Explanation:

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

The momentum of inertia of both are:

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

Let’s solve it for ω(t).

But

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

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

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

I hope it helps you!