Question

Average torque on a projectile of mass m, initial speed u and angle of projection between initial and final positions p and q, about the point of projection is

Answer 1

Answer:

Average Torque = mu^2sin2θ/2

Explanation:

1st please have a look on the attached pic which will clear you the idea about the situation and different angles.

τ→.Δt=ΔL→…(i)  

Here Δt=time of flight = 2usinθ/g  

Change in angular about point of projection (initially it is zero)  

∣ΔL=∣Lf ∣ = ∣ Li ∣= (m*u*sinθ)  

=(m*u*sinθ)(u*2sin2θ)/g

=m*u^2sinθsin2θ/g

 

Now ∣τ→av∣=∣ΔL/Δt∣

=(mu^3sinθsin2θ/g)×(g/2usinθ)  

=mu^2sin2θ/2