Question

A particle of mass m is projected with an initial velocity u at angle theta to horizontal.The torque of gravity on projectile at maximum height about the point of projection is

Answer 1

See attached diagram to get some idea of the projection.

• First of all we have to find the time(t) for projection.

      By using a linear motion equation for upward vertical motion,

      $S = ut + (1/2)at^{2}$

     $0 = U Sin (theta)*t + (1/2)*(-g)*t^{2}$

     $t = 2U Sin (theta)/g$--------(1)

• Then we can find range(R) of the projection by using a linear motion equation for horizontal motion,

     $S = ut + (1/2)at^{2}$

    $R = U Cos (theta)*t + 0$

    Substituting equation (1),

    $R = U Cos (theta) * 2U Sin (theta)/g$

    $R = 2U^{2}  Cos (theta)Sin (theta)/g$

    $R = U^{2} Sin (2*theta)/g$---(2)     (∵ 2SinΘCosΘ = Sin2Θ)

Now we can calculate torque at maximum height.

Torque= Force × perpendicular distance between force and the point of projection.

$Torque = mg * R/2$

Substituting equation (2),

$Torque = mU^{2} Sin (2*theta)/2$

Answer :Torque = mU^{2} Sin (2*theta)/2