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What is the centripetal force of 280 kg motorcycle is driven around a 12 meter tall vertical circular track at a constant speed of 40 m/s​

Sagot :

[tex]\tt{\huge{\blue{Explanation:}}}[/tex]

The centripetal force exerted on an object is given by

[tex]\boxed{F_{c} = \frac{mv^{2}}{r}}[/tex]

where:

Fc = centripetal force

m = mass

v = velocity

r = radius

[tex]\tt{\huge{\red{Solution:}}}[/tex]

Solving for r

Note: The height of the vertical circular track is equal to the diameter of the track

[tex]r = \dfrac{d}{2}[/tex]

[tex]r = \dfrac{\text{12 m}}{2}[/tex]

r = 6 m

Solving for Fc

[tex]F_{c} = \dfrac{mv^{2}}{r}[/tex]

[tex]F_{c} = \dfrac{(\text{280 kg})(\text{40 m/s})^{2}}{\text{6 m}}[/tex]

[tex]\boxed{F_{c} = \text{74.666.67 N}}[/tex]

Therefore, the centripetal force of the motorcycle is 74.666.67 N.

[tex]\\[/tex]

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