A scooter weighing $150$ kg together with its rider moving at $36 k m / h r$ is to take a turn of radius $30$ m. What force on the scooter towards the center is needed to make the turn possible? Who or what provides this?

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Solution

$F_{r e t a r d e d} = \frac{m v^{2}}{r}$
$= \frac{150 \times (36 \times \frac{5}{16})^{2}}{30}$
$= \frac{150 \times 100}{30}$
$= 500 N$
Provided centripetal force on scooter.

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together with its rider moving at

36 k m / h r

is to take a turn of radius

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A scooter weighing 150kg together with its rider moving at 36km/hr is to take a turn of radius 30m. What force on the scooter towards the center is needed to make the turn possible? Who or what provides this?

Fretarded=mv2r=150×(36×516)230=150×10030=500NProvided centripetal force on scooter.

A scooter weighing $150$ kg together with its rider moving at $36 k m / h r$ is to take a turn of radius $30$ m. What force on the scooter towards the center is needed to make the turn possible? Who or what provides this?

$F_{r e t a r d e d} = \frac{m v^{2}}{r}$
$= \frac{150 \times (36 \times \frac{5}{16})^{2}}{30}$
$= \frac{150 \times 100}{30}$
$= 500 N$
Provided centripetal force on scooter.