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

To find the track's angular velocity when the toy car reaches 0.74 m/s, we use conservation of angular momentum. After calculating the angular velocities for both the car and the track, we conclude that the track rotates in the opposite direction at approximately 3.93 rpm.

Step-by-step explanation:

The question involves finding the track's angular velocity in rpm when the toy car reaches a steady speed. Since the diameter of the track is 60 cm, its radius is 30 cm or 0.30 m. The toy car's velocity along the circular path is given as 0.74 m/s. To find the angular velocity of the track, we can first calculate the angular velocity of the car in radians per second (rad/s) using the formula ω = v / r, and then convert this to revolutions per minute (rpm).

The angular velocity of the car is ω = 0.74 m/s / 0.30 m = 2.47 rad/s. To convert this to rpm, we use the fact that 1 revolution is 2π radians and there are 60 seconds in a minute:
ω in rpm = (2.47 rad/s) x (60 s/min) / (2π rad/rev) ≈ 23.60 rpm.

However, because the track has its own mass and the car's speed is mentioned to be relative to the track, we must consider the system's conservation of angular momentum. Since the track is initially at rest and the spokes are massless, the angular velocities of the track and the car must be such that the total angular momentum of the system remains zero. Therefore, for a toy car with a mass of 250 g = 0.25 kg, the angular momentum of the car is L = m * r * v = 0.25 kg * 0.30 m * 0.74 m/s = 0.0555 kg·m²/s. To conserve angular momentum, the track must rotate in the opposite direction with equal magnitude of angular momentum.

Considering the track's mass (1.5 kg), we can calculate its angular velocity, ω_track. The moment of inertia for the track (assuming it's a thin ring) is I_track = m * r² = 1.5 kg * (0.30 m)² = 0.135 kg·m². Therefore, the track's angular velocity is ω_track = L / I_track = 0.0555 kg·m^2/s / 0.135 kg·m² ≈ 0.411 rad/s. Converting this to rpm gives us ω_track in rpm ≈ 3.93 rpm. Thus, the track turns at approximately 3.93 rpm while the toy car is moving at its steady speed.

Answer 2

the track's angular velocity = 3.363 rpm

Step-by-step 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 = (mv)/(r(m+M))`

`\omega = (0.25*0.74)/(0.3(0.25+1.5))`

`\omega = 0.352 \ rad/ sec`

to rpm ; we have:

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

`\omega = 3.363 \ rpm`

Thus, the track's angular velocity = 3.363 rpm