Question

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 1

Answer:

$L_f=22.27\ kg-m^2/s$

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 :

$\tau=\dfrac{dL}{dt}$

Since, $\tau=d\times F$

$dF=\dfrac{dL}{dt}$

$d.F=\dfrac{L_f-L_i}{t}$

$d.F=\dfrac{L_f}{t}$

$L_f=d\times F\times t$

$L_f=0.3\ m\times 270\ N\times 0.275\ s$

$L_f=22.27\ kg-m^2/s$

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