Centripetal Force
Consider a car is traveling 30 m/s on a road with lots of pot holes, and sharply turns to narrowly miss one. If the vehicle’s path is a 150 m curve radius and the force produced by its tires is 1,000 N, what is the car’s weight?
Expand Hint
Centripetal acceleration:
$$$a=\frac{v^2}{r}$$$
where
$$v$$
is the velocity, and
$$r$$
is the radius.
Hint 2
$$$Force=mass\times acceleration$$$
Any motion on a curved path has accelerated motion:
$$$a=\frac{v^2}{r}$$$
where
$$a$$
is centripetal acceleration,
$$v$$
is the velocity, and
$$r$$
is the radius. To stay on the curved path, a force directed towards the curvature’s center is required, and is called centripetal force:
$$$F=ma_{centripetal}=\frac{mv^2}{r}$$$
where
$$m$$
is the mass and
$$a$$
is the acceleration. Therefore,
$$$1,000N=\frac{m(30m/s)^2}{150m}$$$
Solving for mass:
$$$m=\frac{1,000N(150m)}{900m^2/s^2}=\frac{(1,000kg\cdot m)(150m)}{(s^2)(900m^2/s^2)}=167\:kg$$$
167 kg
Time Analysis
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