Honors Physics

Object Rounding an Unbanked Corner

In rounding a circular curve on a horizontal, unbanked road, your speed is 15 m/s, and you experience a certain centripetal acceleration. At what speed should you drive the same curve to reduce the centripetal acceleration to one-third of the original value?

Typed Out Solution

What is Given

Notes for Original Acceleration

At $15 \frac{m}{s}$

At New Speed

Notes for New Acceleration

A certain centripetal acceleration $a_1 = ?$ $a_2 = ?$ to reduce the centripetal acceleration to one third of the [unknown] original value
your speed is 15 m/s $V_1 = 15\frac{m}{s}$

$V_2 = ? \frac{m}{s}$

at what speed should you drive?
the same curve, so the same unknown radius for both r = ? r = ? the same curve, so the same unknown radius for both
Periods not given, not equal $T_1 = ?$ $T_2 = ?$ Periods not given, not equal
Relationship from the original centripetal acceleration to the new centripetal acceleration $a_2 = \frac{1}{3}a_1$

Calculations

Notes

At $15 \frac{m}{s}$

At New Speed

Starting Point

$a_1 = \frac{V_1^2}{R}$

$a_2 = \frac{V_2^2}{R}$

Substitute
$a_2 = \frac{1}{3}a_1$
in for $a_2$
$\frac{1}{3}a_1 =\frac{V_2^2}{R}$
Multiple each side by 3

$a_1 =3\frac{V_2^2}{R}$

Set these two
equations equal to
each other
$a_1 = \frac{V_1^2}{R}$ $a_1 =3\frac{V_2^2}{R}$

$\frac{V_1^2}{R} = 3\frac{V_2^2}{R}$

Cancel out the Radius in each equation

$V_1^2 = 3V_2^2$

Solve for $V_2$

$V_2 = \sqrt{\frac{V_1^2}{3}}$

Substitute in our known values

$V_2 = \sqrt{\frac{15^2}{3}}$

Calculate the Result

$V_2 = 8.66\frac{m}{s}$