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Average torque on a projectile

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Average stock on a projectile of mass M initial speed u and angle of projection $\theta $ between initial and final position about the point of projection.

what I did:
$r×F=r×mg
=ucos\theta .t .mg $(because we take perpendicular component)

then what's wrong as answer is in terms of m,u, sin or cos only ?

asked Feb 1, 2017 in Physics Problems by physicsapproval (2,320 points)
What is $t$? What is the "correct" answer?

1 Answer

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Best answer

You are correct so far. You have to substitute a value for $t$, the average time during flight, ie the time taken to reach maximum height, which is $\frac{u\sin\theta}{g}$. Then average torque is $\frac12mu^2\sin2\theta$.

answered Feb 1, 2017 by sammy gerbil (28,178 points)
edited Feb 2, 2017 by sammy gerbil
ohh! I didn't solve that ahead
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