A model train with a mass of #4 kg# is moving along a track at #3 (cm)/s#. If the curvature of the track changes from a radius of #54 cm# to #27 cm#, by how much must the centripetal force applied by the tracks change?

Question

A model train with a mass of #4  kg# is moving along a track at #3 (cm)/s#.  If the curvature of the track changes from a radius of #54 cm# to #27 cm#, by how much must the centripetal force applied by the tracks change?

Leo Abeyta · Accepted Answer

The change in centripetal force is #=0.0066N#

The centripetal force is

#F=(mv^2)/r#

mass, #m=4kg#

speed, #v=0.03ms^-1#

radius, #=(r) m#

The variation in centripetal force is

#DeltaF=F_2-F_1#

#F_1=mv^2/r_1=4*0.03^2/0.54=0.0067N#

#F_2=mv^2/r_2=4*0.03^2/0.27=0.0133N#

#DeltaF=0.0133-0.0067=0.0066N#

Elena Albarran · Answer

undefined The centripetal force applied by the tracks must increase by four times.

Isabella Alvarez · Answer

undefined To calculate the change in centripetal force applied by the tracks when the curvature of the track changes, we can use the formula for centripetal force:

$$F_c = \frac{mv^2}{r}$$

Where:
- $F_c$ is the centripetal force
- $m$ is the mass of the object (in kilograms)
- $v$ is the velocity of the object (in meters per second)
- $r$ is the radius of the curvature (in meters)

First, we convert the velocity from centimeters per second to meters per second:

$$v = 3 \text{ cm/s} = 0.03 \text{ m/s}$$

Then, we calculate the initial centripetal force with the initial radius:

$$r_1 = 54 \text{ cm} = 0.54 \text{ m}$$
$$F_{c1} = \frac{mv^2}{r_1}$$

Next, we calculate the final centripetal force with the final radius:

$$r_2 = 27 \text{ cm} = 0.27 \text{ m}$$
$$F_{c2} = \frac{mv^2}{r_2}$$

Finally, we find the change in centripetal force:

$$\Delta F_c = F_{c2} - F_{c1}$$