Question

A particle of mass m=1kg is projected with speed u=20sqrt(2)m//s at angle theta=45^(@) with horizontal find the torque of the weight of the particle about the point of projection when the particle is at the highest point.

Answer 1

The torque of weight of particle about the point of projection is 100 N-m.

Particle of mass m=1 kg is projected with speed u=20 m/s at angle theta=45 degree.

1. Here m=  kg

           u= 20 m/s

         theta= 45 degree.

2. When the particle reaches the maximum height then the the height is given by H = u*sin(Theta)^2/2*g

3. Now sin(theta) =sin(45)

                             =1/ $\sqrt{2}$

4. Therefore H = (20*(1/$\sqrt{2}$))^2/2*9.81

                         = 10.1937 m  (approx)

5. Now Torque = F*r

                         = Weight of particle*r   ( Here r = H)

                         = 1*9.81*10.1937

                         = 100 N-m