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Suppose you start an antique car by exerting a force of 270 N on its crank for 0.275 s. What angular momentum is given to the engine, in kg · m2/s, if the handle of the crank is 0.300 m from the pivot and the force is exerted to create maximum torque the entire time? kg · m2/s

Answer :

Answer:

[tex]L_f=22.27\ kg-m^2/s[/tex]

Explanation:

Given that,

Force exerted on an antique car, F = 270 N

Time, t = 0.275 s

Distance, d = 0.3 m

To find:

The angular momentum given to the engine.

Solution,

We know that the relation between the angular momentum and the torque is given by :

[tex]\tau=\dfrac{dL}{dt}[/tex]

Since, [tex]\tau=d\times F[/tex]

[tex]dF=\dfrac{dL}{dt}[/tex]

[tex]d.F=\dfrac{L_f-L_i}{t}[/tex]

[tex]d.F=\dfrac{L_f}{t}[/tex]

[tex]L_f=d\times F\times t[/tex]

[tex]L_f=0.3\ m\times 270\ N\times 0.275\ s[/tex]

[tex]L_f=22.27\ kg-m^2/s[/tex]

So, the angular momentum given to the engine is [tex]22.27\ kg-m^2/s[/tex].

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